End while loop with ctrl+d, scanf?
The biggest problem with your program is that scanf will not read an EOF into a variable. However, fixing just this problem is not going to make your program entirely correct, because there are other issues in your code: Your code repeats itself - when possible, you should unify the...
why after looping there's a newline
On the 2nd time through the loop, scanf("%c",&penal1); is scanning in the '\n' from the previous user input. Add a preceding space like code used in other places to consume all preceding white-space. scanf(" %c",&penal1); // added space. Code should check not only for 'O' and 'o', but 'X' and...
How to make fgets() ignore a \n at beginning?
Ok, the original idea does not seem to work, so another try: scanf("%d",&s.rollNumber); printf("Name: "); fgets(s.name, 20, stdin); /* capture the new line */ fgets(s.name, 20, stdin); Original idea: Simply tell scanf to also capture the newline: scanf("%d\n",&s[i].rollNumber); ...
Checking input types with scanf() in a while loop [duplicate]
int check = scanf("%d", &x); does not consume the "input is a character not a number", leaving that input in stdin for the next input function. Since the next input function is check = scanf("%d", &x);, it does not consume the offending data and so the cycle repeats. Code needs...
Specifying the ']' character in a C/C++ scanf() scanset
The trick is to put ']' as the first character in the pattern: scanf(" %80[^]]", &buf); ...
Program immediately terminates with segfault when using scanf
c,pointers,segmentation-fault,scanf
In your code, stack* st; st->top=-1; you're using st uninitialized which in turn invokes undefined behaviour. You need to allocate memory to st before using it. Try something like stack* st = malloc(sizeof*st); //also, check for malloc success That said, Please see why not to cast the return value of...
Segmentation Fault after 5th scanf
assumes sizeof (int) and sizeof (int*) are the same m=(int**)malloc(n*sizeof(int)); Try this m = malloc(n * sizeof *m); ...
scanf crops last byte of numbers
If you read e.g. this scanf reference, you will see that the "%[" format is for reading a string which of course includes writing the string terminator character. Since you only provide a single character to store the read string, the scanf function will always write out of bounds and...
Storing a string of inputs into a struct in C
c,string,struct,user-input,scanf
You can specify complex formats using scanf, like: scanf("%79[^,],%c,%d,%d", apple.item, &apple.letter, &apple.x, &apple.y); %79[^,] means scan anything that is not a comma character, up to 79 characters. Note that this does no error handling if the user enters a poorly formatted string, like "aaa;b;1;2". For that, you'll need to write...
How do you type something on the same line of text in C [closed]
From your tags, I'm guessing you're using printf(). Simply leave out the "\n" character, which means "newline.". In other words. printf("> "); printf("text\n"); This will print: > text ...
How to get several value in once enter ?(Number of uncertainty)
Have you considered doing this using a combination of scanf(), fgets() and strtok()? For example, you can use scanf() to simple query the number of inputs expected. Then use a loop around `strtok() to tokenize user's input: int main() { char line[1024]; char *token = {0}; char delim[]={" \r\n\t"}; long...
Dynamically allocate user inputted string
c,arrays,user-input,scanf,dynamic-memory-allocation
First of all, scanf format strings do not use regular expressions, so I don't think something close to what you want will work. As for the error you get, according to my trusty manual, the %a conversion flag is for floating point numbers, but it only works on C99 (and...
Why passing string to scanf() compiles fine in C?
The code is behaving correctly. Indeed, scanf is declared int scanf(const char *format, ...); Luckily, your format does not contain any % for which there would be no correspondence in ..., which would invoke UB. Further, format is a string literal which allows the compiler to go through it ensuring...
Logical && statement
char c, implies scanf("%c", &c) can only read a single character into char variable c. Hence input 197 will read as 1 in to c, making logical && true. Look at this compilation
Why doesn't isdigit() work?
isdigit() takes the ascii value of a character and returns 0 if it's not a digit and non-0 if it is. You are passing to it an integer value which is not necessarily an ascii value, you don't need to check if it's a digit since you read it with...
Why value of second variable is irrelevant after adding space or new line?
Because 215- 15 can only match one number: 215. As soon as scanf() reads a -, it stops processing the first match since - can't possibly be a digit of the current number, so num1 is matched with 215. Then, no more numbers can match, because you are left with...
scanf in while loop only works on first iteration
Remove the '\n' left by the previous scanf() like this scanf(" %c %x", &read_size, &address); /* ^ tell scanf to skip white spaces with the %c specifier */ also do not ignore scanf()'s return value, it might cause very weird things in case it fails and you fail to detect...
Reading from txt file in C using scanf
sscanf takes first argument of type const char *.You are passing the first argument as FILE *. You can use: fscanf OR fgets with sscanf ...
Various questions about arrays in C
Why can I not use N to define my array X? Prior to the 1999 standard, array sizes had to be specified using a constant expression whose value could be computed at compile time; this means either a numeric literal such as 10, a preprocessor macro that would expand...
Replacing gets with scanf in simple equation crashes the program
Why spend time converting the answer when scanf() can do it for you? #include <stdio.h> int main(void) { float height_in_inches; printf("Enter your height in inches: "); if (scanf("%f", &height_in_inches) == 1) { float height_in_cm = height_in_inches * 2.54; printf("You are %.2f inches or %.2f centimetres tall.\n", height_in_inches, height_in_cm); } else...
Why does my program accept one integer too many and input one too few?
Your loops should be for (count=0;count<SIZE;count++) Array indexing is 0 based in C. Since you have a whitespace chatacter (\n) in scanf() call, it waits for you input a non-whitespace character to complete each call. Remove the \n: for (count=0;count<SIZE;count++){ scanf("%d",&myArray[count]); } ...
Why isn't scanf( ) waiting for input from keyboard?
c,printf,structure,scanf,fflush
The C11 standard explains how %c works: §7.21.6.2/8 Input white-space characters (as specified by the isspace function) are skipped, unless the specification includes a [, c, or n specifier. As such, the linefeed produced by the enter key is consumed by %c. You can fix this by adding a space...
Program blocks after input
I tried you code and manage to reproduce the weird behaviour. If you remove the " " around "%li" it doesn't block anymore. The problems were the blank spaces because scanf was expecting the input to match the empty space as well. From the scanf documentation: All conversions are introduced...
How to loop until user input N in C
c,loops,input,while-loop,scanf
Change scanf("%c", &choice); to scanf(" %c", &choice); // note the space before %c This is done to discard all white-space characters like \n and spaces from the stdin. When you enter data for a scanf,you enter some data and press the enter key. scanf consumes the data entered and leaves...
Why does command prompt show numbers before I begin?
Substitute printf("%d",&n); with scanf("%d",&n); printf Writes the C string pointed by format to the standard output (stdout) scanf Reads data from stdin In your code you are printing n, that is not initialized, that a random number is printed out after "Enter numbers to be stored in file" string....
C sscanf (fscanf) behaving differently row from row
Scanning with %i and values with a leading zero are assumed to be octal. 08 and 09 are not octal values. Use %d instead as leading zero's are ignored and values are in base 10. For the sscanf you might try this. It will scan the five items present in...
Is there any way in C to terminate scanf() inputting in array without ctrl+d?
You can do something like that: for(i=0;a[i]<10000;i++) { // chack that input correct if((scanf("%d",&a[i])==1) { count++; } else // if input is incorrect { // read first letter int c = getchar(); if( c == 'q' ) { break; // stop the loop } // clean the input buffer while(...
How to read input in C
Let us look at this step by step: "... read a line with scanf("%[^\n]");". scanf("%[^\n]", buf) does not read a line. It almost does - sometimes. "%[^\n]" directs scanf() to read any number of non-'\n' char until one is encountered (that '\n' is then put back into stdin) or EOF...
How to break a string, that was read using fgets(), using scanf(“%s”,..)
You can use sscanf() to advance through a buffer read by fgets() %n will receive the number of characters used by the scan. Adding used to the offset will advance the scan through the buffer. int offset = 0; int used = 0; char line[1000]; char item[200]; fgets ( line,...
C++ — error C2664: 'int scanf(const char *,…)' : cannot convert argument 1 from 'int' to 'const char *'
Change scanf('%d', &row); to scanf("%d", &row); '%d' is a multicharacter literal which is of type int. "%d" on the other hand is a string literal, which is compatible with the const char * as expected by scanf first argument. If you pass single quoted %d, compiler would try to do...
Parse string separated with semicolon using a regex
scanf doesn't support regular expressions. It supports strings of a given set of characters. When your format contains %[^';'] it matches any sequence of one or more characters except ' and ;. When your format contains a comma (,) it matches a comma. So when you say: sscanf(multiple, "%[^';'], %[^';'],...
Add and print multiple char in C [closed]
Judging by this statment: where the user rolls 3 dice and gets some random outputs (up to integer 6). My next step is to add those 3 values obtained and get its sum I assume you want this: first initialize sumDice to 0: int sumDice=0; then, in every for loop...
Why doesnt getchar() stop reading strings in C?
Because you have an operator precendence problem here chr = getchar() != EOF this is evaluated as chr = (getchar() != EOF) because the != operator has higher precendence than the assignment operator =, so you just need to add parentheses like this (chr = getchar()) != EOF Tip: Check...
C program skips a line when asking for user input [duplicate]
Point 1 Do not use fflush( stdin );, it is undefined behaviour. Related: from C11 standard docmunet, chapter 7.21.5.2, (emphasis mine) int fflush(FILE *stream); If stream points to an output stream or an update stream in which the most recent operation was not input, the fflush function causes any unwritten...
C program loops infinitely after scanf gets unexpected data
When scanf() fails, the argument is not automatically initialized, and uninitialized values could be any value, so it might be less than 2 or greater than 64 no one knows. Try this int initialBase; /* some default value would be good. */ initialBase = 2; do { printf("Initial base: ");...
how to scan a string input
c,arrays,string,segmentation-fault,scanf
In your code, char * n = ""; makes n point to a string literal which is usually placed in read-only memory area, so cannot be modified. Hence, n cannot be used to scan another input. What you want, is either of below a char array, like char n[128] =...
I'm not able to use scanf() in else statment
When you enter the data in the first read of this scanf: scanf("%c", &ms); A newline character remains in the keyboard. To solve this put a space in your second scanf: scanf(" %c",&s); This is to consume any trailing character in the stdin that might have been left by previous...
Spaces with scanf in a client to server message. C
Use fgets() to get your input into a string and sscanf() to evaluate it. Since you just want what the user entered, you don't really need sscanf() in this case anyway See this post...
scanf in a while loop reads on first iteration only
You need to use int c; while((c=getchar()) != '\n' && c != EOF); at the end of the while loop in order to clear/flush the standard input stream(stdin). Why? The answer can be seen below: The scanf with the format string(" (%d,%d)\n") you have requires the user to type An...
scanf issue when reading double
In your code, change scanf("%lf \n", &radius); to scanf("%lf", &radius); Otherwise, with a format string having whitespace, scanf() will behave as below (quoted from C11, chapter §7.21.6.2, paragraph 5) A directive composed of white-space character(s) is executed by reading input up to the first non-white-space character (which remains unread), or...
Nonportable Pointer Conversion
The problem is are with Point 1: if(gender == "M") it should be if(gender == 'M') Reason: "M" represents a string, 'M' represents a char. Point 2: scanf("&s", gender); should be scanf(" %c", &gender); Reason: Need to use proper signature of scanf(). %c is the format specifier for scanning a...
In C, Why does my float print out as 0 even though I input 13.5 for it using scanf?
c,arrays,floating-point,printf,scanf
Just change this: float price[10000]; to this: float price; Because you use it as a single variable and not as an array...
How to read and write files using printf and scanf in C? [closed]
You basically have three options. Option I Redirect stdin/stdout (those are the streams scanf reads from and printf writes to) in the console, when you start your program. On both Windows and Linux, this can be done like so: < in.txt - redirects all reads from stdin to in.txt >...
scanf() to get in the string on the second time
fflush(stdin); is undefined behaviour in standard C. To flush the newline character, you could simply use getchar() instead. printf("\nEnter staff name \t> "); getchar(); scanf("%[^\n]s", staff_name2); I would also use fgets() instead of scanf to read a line and trim the newline if necessary, which offers better control over invalid...
How to introduce string by keyboard in child after fork
c,concurrency,process,signals,scanf
Let me see if I got it right, your problem is that child process is failing to read stdin, right? You may use fscanf(stdin, stringa); I've tested with a simple example and worked (just changing your scanf by fscanf). Let me know if that was your problem....
How to scanf to just one dimension of a multidimensional array in C?
c,arrays,loops,multidimensional-array,scanf
You want something like this. int i=0; do{ scanf("%d", &mat[i][1]); i++; }while (getchar() != '\n'); ...
scanf a big hexadecimal value
You must use "%llx" for both scanf format and printf. See the manual page for additional details.
Simple Unit Conversion Program - scanf issues with do-while loop (C programming)
Two advices when using scanf (especially for numbers): 1) use fflush(stdin); before each call of scanf; 2) check the result of unputs - is number received or not? For example, if you cannot work without correct input: do{ printf("Please, enter number: "); fflush(stdin); } while( 1 == scanf("%d", &number)); ...
Scanf text overflow in C
I have some advices for string (char*) usage in C. First, as sumithdev said, use fflush(stdin) to clean input stream buffer before input any piece of information. Second, try to use uniform style for storing all string data. For example, your middleInitial also can be defined as array of char...
C - Using file as database
When the format string in scanf_s contains %s or %S, the function expects another parameter that indicates the size of the string. See https://msdn.microsoft.com/en-us/library/w40768et.aspx. Instead of scanf_s("%[^\n]",std.studentNumber); you'll need to use: scanf_s("%[^\n]",std.studentNumber, sizeof(std.studentNumber)); Similar changes need to be to the other calls to the function....
How does the data flow from input stream into the input buffer during scanf() in C?
How is it determined? It's determined by the format string itself. The scanf function will read items until they no longer match the format specifier given. Then it stops, leaving the first "non-compliant" character still in the buffer. If you mean "how is it handled under the covers?", that's a...
Simple scanf does not set variable value
For var4, you declare it as unsigned char, but read it using a %d format specifier, which invokes undefined behavior. I would recommend using %hhu as the format specifier for that field, if your compiler supports it. Or you could declare var4 as an int, and then assign it to...
program using the rand () function
Some issues with your code: Point 1: scanf("%d\n",&x); should be scanf("%d",&x); Point 2: for(i=0;i<1;i++) this for loop is practically useless. It only iterates one. either use a longer counter, or get rid of the loop. Point 3: It's better to provide a unique seed to your PRNG. You may want...
Difference between fgets and gets
Drop gets() and scanf(). Create a helper function to handle and qualify user input. // Helper function that strips off _potential_ \n char *read1line(const char * prompt, char *dest, sizeof size) { fputs(prompt, stdout); char buf[100]; *dest = '\0'; if (fgets(buf, sizeof buf, stdin) == NULL) { return NULL; //...
scanf int8_t corrupts stack
You have run into a bug of Microsoft's C/C++ runtime library, cf. http://mfctips.com/tag/format/ or https://gcc.gnu.org/bugzilla/show_bug.cgi?id=63417 (which reports this bug for gcc under mingw, which links against the Microsft libs). "%hhd" just doesn't work; you must program around it (which isn't too hard, but sad). I am not aware of a...
Why fgets takes cursor to next line?
The reason printf is outputting a newline is that you have one in your string. fgets is not "adding" a newline --- it is simply reading it from the input as well. Reading for fgets stops just after the newline (if any). Excerpt from the manpage, emphasis mine: The fgets()...
Please explain this output of the code
c,function,return,printf,scanf
In the following function: int GetPositiveInt(void) { int n; do { printf("Enter a positive number : "); scanf("%d",&n); }while(n<=0); } the function is expected to return an int value, whereas there is no return statement. You probably wanted to return the value of n. This can be achieved by adding...
Why compiler is not issuing error while passing an int (not int *) as the argument of scanf()?
As per C11 standard, Chapter §6.3.2.3 An integer may be converted to any pointer type. Except as previously specified, the result is implementation-defined, might not be correctly aligned, might not point to an entity of the referenced type, and might be a trap representation. So, the compiler will allow this,...
What does a # sign after a % sign in a scanf() function mean?
c,input,printf,scanf,format-specifiers
TL;DR; - A # after a % sign in the format string of scanf() function is wrong code. Explanation: The # here is a flag character, which is allowed in fprintf() and family, not in fscanf() and family. In case of your code, the presence of # after % is...
C: Format %s expects argument of type char* in funny strings program
Apparently you have a char * and you are passing it's address, which is wrong, scanf() wants a char pointer for each "%s" specifier, and the fix for your code is to use char string[10]; scanf("%s", string); the array automatically becomes a char pointer when passed to scanf() like above,...
Trouble with delimiting colon with scanf
It seems the formatting is wrong, you want to read the colon and there are no commas. Try this: scanf("%m[^':']:%m[^':']", &str, &i); Hope this helps....
if statement for string comparison is not executing properly
Point 1: Use strcmp() for string comparison, not == operator. Point 2: Anyways, for an array of char yourchoice[40]="";, using yourchoice[40] is out of bounds which in turn invokes undefined behaviour. Array index in C starts from 0. Point 3: printf() does not need to have a pointer to data...
Why doesn't multiple `\n` characters end up in the standard input stream?
A skipped character is not "left" in the stream. The input stream can only be read once and in one direction,1 so once a character is skipped, it's gone. 1 without extra fanciness, such as buffering...
What does %[^<] (and friends) mean in the formatted string family?
From the C11 standard document, chapter §7.21.6.2, Paragraph 12, conversion specifiers, (emphasis mine) [ Matches a nonempty sequence of characters from a set of expected characters (the scanset). .... The conversion specifier includes all subsequent characters in the format string, up to and including the matching right bracket (]). The...
Scanf() working differently with \n and '\n' supplied in format specifier
So you had a scanf("%f'\n'%f'\n'%s",&lati,&longi,info) in your code. When you break this code up: %f - expect a float ' - expect a literal ' in the input \n - expect a line break ' - expect a literal ' in the input ... and so on. \n is a...
Why does scanf returns control back to the program on pressing Enter key?
Most OSes buffer keyboard input so that they can process backspaces properly -- the OS keeps input in a buffer and only gives it to the program when Enter is hit. Most OSes also provide ways to control this, but the way is different for different OSes. On POSIX systems,...
Find the lonely integer in an array
You'd want to use: #include <stdlib.h> /* ... */ int *arr; scanf("%d", &n); arr = malloc(sizeof(int) * n); This way, arr gets dynamically allocated at runtime, so it can be of any size depending on the input n. What you were originally doing (i.e. declaring arr[n] after recieving n via...
Reading from a CSV file and solving problems with “;”
You can change your code fscanf(fp,"%4s %7s %9s %16[^\n] %11[^\n] %2s %5s",... to the style fscanf(fp,"%4s;%7[^;];%10[^;];...... to have the ;s as a part of the format string. Then it won't be considered as a part of the string input. Note: The above method is just a workaround and will break...
scanf takes extra parameter why?
You have a space in your format after the second %i. scanf will read extra data to match the space. Remove the space and it should work as you expected.
Strange C++ behavior. Value gets overriden
Your scanf to read in op is invalid. op is a single char, so you should read with scanf("%c",&op); Not scanf("%s",&op); ...
Storing scanf values into a 2d string array
scanf returns the number of items scanned. In this case it would be 2 instead of 1. Here a return of 1 would indicate a problem during the scan. The %m specifier allocates memory to the pointers. Using a single pair of pointers, they should be freed in eadh iteration...
Specifing the maximum string length to scanf dynamically in C (like “%*s” in printf)
Basic answer There isn't an analog to the printf() format specifier * in scanf(). In The Practice of Programming, Kernighan and Pike recommend using snprintf() to create the format string: size_t sz = 64; char format[32]; snprintf(format, sizeof(format), "%%%zus", sz); if (scanf(format, buffer) != 1) { …oops… } Extra information...
How to limit input length with scanf
As pointed out by @SouravGhosh, you can limit your scanf with "%3s", but the problem is still there if you don't flush stdin on each iteration, you can do it in this way: printf("\n enter the names \n"); for(i = 0; i < 3; i++) { int c; scanf("%3s", name[i]);...
control scanf() duration for taking any input
#include <windows.h> #include <stdio.h> int main(void){ int num = 0; DWORD waitCode; printf("input number : "); //wait console input 10,000 Millisecond waitCode = WaitForSingleObject(GetStdHandle(STD_INPUT_HANDLE) , 10*1000); switch(waitCode){ case WAIT_TIMEOUT: fprintf(stderr, "\n10 seconds are over no more input\n"); return -1; case WAIT_OBJECT_0://normal status scanf("%d", &num);//input from stdin buffer if(num)//not zero printf("input...
How does scanf determine whether to block?
scanf is just reading from its input stream. If the input stream is a pipe, and the other end of that pipe is associated with a tty (which is usually the case if you are interactively entering data by pressing keys on a keyboard), scanf will return as soon as...
Why can't I input values?
Try this if (scanf_s("%99s", name, _countof(name)) == 1) printf("Your Name is: %s", name); Two things scanf_s() is a buffer overflow safe function, and it expects a length argument for "%s" specifier. You should only proceed to printf() if you actually succeeded scannig the value, for which the check (scanf(...) ==...
Shortening strings in C
Code has input and output issues; char home[15]="HOME"; ... printf("Do you want to enter teams' names?\n1-Yes\n2-No\n:>>"); scanf("%d",&choice1); ... printf("\nPlease enter HOME team's name(max 15 characters including space)\n:>>"); scanf(" %[^\n]s",&home); ... Input1: The 's' in " %[^\n]s" is certainly not needed. 's' is not part of the "%[^\n]" format specifier. Input2:...
Do white spaces take space in txt files?
c,file-io,whitespace,scanf,fscanf
To your first question: No, white space does not need to be considered in fscanf(). To your second question: Neither of these will work. You seem to be misunderstanding how information is stored in a computer. In the program, an integer is 32 bits (4 bytes), so sizeof(int) will always...
Printing an int value obtained from user
Here is your problem. char gender; scanf(" %s",&gender); gender is a char. That is, it only has memory for a 1 byte character. But you are using it as a string. You probably have the same problem for name1 since you are using & for that as well but can't...
How to use scanf for a char array input?
scanf() with format specifier %s scans the input until a space is encountered and in your case what you are seeing is undefined behavior your array can hold 10 char's but you are writing into array out of bounds and still getting desired answer which is not guaranteed and might...
Stop scanf loop if user enters a specific number (Not working) C
if (array[i] == '5') You're checking whether array[i] is equal to the ASCII value of the character '5'. Remove the '' to make it compare against the integer 5....
printf after scanf always displays 1 (same unexpected value)
You are assigning the returned value from scanf_s() to the variable n, that means that the program will print 1 in case a successful read happened. What you should do is int numberOfItemsMatched; int readValue; numberOfItemsMatched = scanf_s("%d", &readValue); if (numberOfItemsMatched == 1) printf("%d\n", readValue); I hope the variable names...
Stopping an infinite while loop using scanf in c?
You got it all wrong scanf() will return the number of arguments matched and you are not checking for that. Also, the isdigit() function takes an integer, and returns non 0 if the ascii value of the passed argument corresponds to a digit. To make your program stop when a...
C Programming: How to check if user has entered numeric values only in the following expression?
c,validation,date,exception,scanf
There are a couple of issues. If using scanf check its return. In addition to the numeric validity checks, you also have the issue of needing to flush the input buffer in the event something other than an integer is entered by the user. While there are several ways to...
C Wrong Answer when squaring
To add to the above answers ... look at this example: int number = 5; printf(" %d\n", number); // Prints value of 'number'. printf(" %p\n", &number); // Prints address of 'number'. printf(" %d\n", *(&number)); // Prints value at address of 'number' ... ...
Inserting a word(char[]) into a structure word(char[]) in C
Remove the & fscanf(finput, "%s", &temp[i].word); change it to fscanf(finput, "%s", temp[i].word); Or take the address of the first element, which is equivalent to the previous line fscanf(finput, "%s", &temp[i].word[0]); the passed array decays to a pointer to the first element of the array, so these both are equivalent fscanf(finput,...
Getting a char* with spaces in C from sscanf
Most scanf() field descriptors implicitly cause leading whitespace to be skipped and expect the field to be whitespace-terminated. To scan a string that may contain whitespace, however, you can use the %[] field descriptor with an appropriate scan set. Thus, you might scan sequence of lines following the pattern you...
Trying to get scanf to only look at first n number(s) of the input
You could write: if ( num != 3 || getchar() != '\n' ) If you intend to use this code in a loop (or indeed have any other input after this) then you probably want to flush the buffer inside the { } for this if: int ch; while (...
scanf get multiple values at once
c,char,segmentation-fault,user-input,scanf
I'm not saying that it cannot be done using scanf(), but IMHO, that's not the best way to do it. Instead, use fgets() to read the whole like, use strtok() to tokenize the input and then, based on the first token value, iterate over the input string as required. A...
getchar() not working in c
c,while-loop,char,scanf,getchar
That's because scanf() left the trailing newline in input. I suggest replacing this: ch = getchar(); With: scanf(" %c", &ch); Note the leading space in the format string. It is needed to force scanf() to ignore every whitespace character until a non-whitespace is read. This is generally more robust than...
Print each word in a separate line from an input string
In your code, printf("%s", s[i]); is wrong. Change it to printf("%c", s[i]); as, you're trying to print a char value. The conversion specifier for a char is %c. Note: Always remember, using wrong conversion specifier will lead to undefined behaviour. Also, while scan()-ing with %s, you cannot read the whole...
scanf doesn't work (integer)
Your scanf is perfect. I think that your mistake is the loop for from esPrimo. Actually you have an infinite loop, because sqrt(n) has always the same value and it isn't a boolean expression. Change your loop: for(divisor = 2; sqrt(n); divisor++) { if(n <= 0) { return falso; }...
second scanf to recognize keyword to exit iteration of integers? C
first issue: Try to read to string and then translate the string to integer by using sscanf, if it is not equal to "end". (compare by strcmp not by ==). second issue: when you read numbers, you never reach to c++. the continue make it skip it. for example, you...
How to get each element of a array from user input then pass it to a function in another file in C
c,arrays,function,scanf,elements
The problem is that you are iterating from 1 and you should iterate from 0, in the inner_product() function for( i=1; i<count; i++) { /* ^ this should be 0 */ also, don't use global variables specially because you got the rest of it right, you are passing the arrays...
Scanning values in C till hit a new-line char, '\n'
c,arrays,algorithm,scanf,getchar
You are trying to read integers as characters so once read you need to convert it to integers. Read the line to a buffer using fgets() then parse the input buffer to get integers. Store the integers to the array. The code looks like char buf[300]; int a[5],i=0; fgets(buf,sizeof(buf),stdin);...
Why does inputting an unexpected value type via scanf() cause this program to enter an infinite loop
Because the scanf function will only extract the input if it's correct. If the input is not correct, the input will still be in the input buffer when the loop iterates, and the next call to scanf will read the exact same input. You might want to either use the...