Different Output for b=+1 in java [closed]
java,post-increment,pre-increment
Reverse the = and + symbols. Unary + isn't what you want. b+=1; or b++; or ++b; Unary plus is b = (+1); or just b = 1....
Undefined Behavior of Postfix or Prefix Increment in Function Calls in C
c,function,post-increment,pre-increment
That is undefined behaviour and as such it is entirely up to the implementation of the compiler in which order the following operations are done: submit argument a submit argument a++ submit argument ++a increment a for ++a increment a for a++ The only thing that the compiler knows is:...
post and pre increment in for loop. string length without library function in C
c,post-increment,pre-increment
First, you need a longer char array for a 16 digit number: char card[17]; // 16 digits + `\0` Then try this: for(len=0; card[len] != '\0'; len++); Do not rely on the value of i outside the loop....
Increment Operator in arrays C++ [duplicate]
c++,arrays,post-increment,pre-increment
My initial reaction: C++ does not guarantee the order of evaluation on a statement like cout<<a[t]<<a[++t]<<a[++t]; You need to do something along the lines of: cout<<a[t]<<a[t + 1]<<a[t + 2]; t += 3; Upon further review, in response to R Sahu's comments (thanks): The order of evaluation may actually be...
How are java increment statements evaluated in complex expressions
java,post-increment,order-of-evaluation
Its easier to see the what is going on with x = 1 instead of 2. The output for x=1 is 7. The key to the understanding of this is in JLS 15.7.2 which states that the every operand is fully evaluated before any part of the operation is preformed....
Is increment/decrement operators Undefined Behaviour in Bash?
c,linux,bash,posix,post-increment
You seem to be seeking to find a defined standard for bash, as there is for C. Unfortunately, there isn't one. There are really only two guides: Volume 3 (Shell and Utilities) of the Open Group Base Specifications, commonly known as "Posix", which is deliberately underspecified. It does state that...
sequence point and side effects
c,post-increment,sequence-points
i++ Evaluates to previous value of i. As a side effect value of i is incremented by 1. So yes the sequence point is there but the expression i++ evaluates to 0 (though value of i is 1 at the same time) For the expected results use ++i instead of...
Javascript $('li').length call throws uncaught exception
javascript,html,post-increment,uncaught-exception,pre-increment
Change $('li').length in your moveRight() and moveLeft() functions to use your slideCount variable. Right now $('li').length returns the total number of li elements on the page (which is greater than the number of carousel slides). This results in your index variable being set to values that are out-of-bounds for your...
Bash arithmetic expression's side effects not executed
bash,post-increment,arithmetic-expressions,side-effects
It gets incremented in the subshell created by the $(command substitution). When that process exits, the modified value is lost. Here are similar ways of seeing the same effect: i=0 bash -c 'let i++' # Subprocess ( let i++ ) # Explicit subshell let i++ & wait # Backgrounded process...
Why doesn't a post-increment/decrement operator have any effect on a variable inside a loop?
javascript,operator-keyword,post-increment
j++ loads the current value of j, then increments the variable, then returns the original value. j = j++ reassigns the original value of j, which is 0, back to j with every iteration of the loop. If you just put j++; on a line by itself inside the loop...
Prefix and postfix increment in for loop
c#,post-increment,pre-increment
If you have a loop in the form for (a; b; c) { d; }, you can treat this as the following while loop: a; while(b) { d; c; } As you can see, the increment doesn't occur until after the first iteration, regardless of whether it's a pre- or...
Is post-increment operator guaranteed to run instantly?
This code is broken for two reasons: Accessing a variable twice between sequence points, for other purposes than to determine which value to store, is undefined behavior. There are no sequence points between the evaluation of function parameters. Meaning anything could happen, your program might crash & burn (or more...
Confusing output after use of increment operator
c,post-increment,pre-increment
I have written the values of the variables before and after the statement gets executes and all side effects have taken place. int a[5] = {5, 1, 15, 20, 25}; int i, j, m; // before : i is indeterminate, a[1] = 1 i = ++a[1]; // after: i =...
Combined preincrement and postincrement in C
c,post-increment,pre-increment
Read up on Undefined Behavior and Sequence Points. This is a slightly different, yet similar example (thanks Zan): 2) Furthermore, the prior value shall be accessed only to determine the value to be stored. C++ example: std::printf("%d %d", i,++i); // invokes Undefined Behaviour because of Rule no 2 ...
Copy character using pointers with post-increment
When using the postfix ++ unary operator, the increment is sequenced after the computation of the value of the operand. So the expression is equivalent to: *to = *from; to++ ; from++ ; In your example: *to++ = *from++;, the values of *to and *from are obtained and then the...
post and pre increment not working as expected for REPL variable in scala
java,scala,compiler-errors,integer,post-increment
Quick answer: you declared a a var, meaning its value can be changed. On the other hand, the REPL res<x> are read-only vals, so they can't be changed. Now, to the error message. In Scala, Ints are not special at language level, and neither are operators. So a += 1...
behavior of --it and it— in [duplicate]
c++,iterator,post-increment,pre-increment
Most of these examples don't have defined behaviour (call 1 is the sole exception). The evaluation of function arguments is unsequenced, meaning that their order of evaluation is unspecified, so whether the side effect of operator++ has kicked in on the other argument by the time the function is called...
C - Strange behavior with a post-increment on isdigit() call inside while loop
Program is behaving as it should. In first code, isdigit(*s++) will return 0 and s will be incremented and the statement if (*s == '\0') return true; will return true. In second snippet, *s++ will not be evaluated and the statement return false; will return false.
Difference between pre and post decrement in recursive function argument
c++,recursion,post-increment,pre-increment
This happens because function(c--) will be called with the same value of c and when it finishes c will be decremented. But as it is recursively called, hence it will be called with same value with no chance of return and eventually you will hit stack overflow error. Assume this,...
Difference between b=b++ and b++ [duplicate]
In Java, the expression b = b++ is equivalent to int tmp = b; b = b + 1; b = tmp; Hence the result. (In some other languages, the exact same expression has unspecified behaviour. See Undefined Behavior and Sequence Points.)...
i++ vs. ++i in a JavaScript for loop
javascript,for-loop,post-increment,pre-increment
In JS and PHP it does not make any difference, I think even in Java it does not make any difference but in pure c when compiler is not optimizing code it does, and that is why a lot of people use ++i because they are used to it from...
What are the historical reasons C languages have pre-increments and post-increments?
c++,c,post-increment,decrement,pre-increment
Incrementing and decrementing by 1 were widely supported in hardware at the time: a single opcode, and fast. This because "incrementing by 1" and "decrementing by 1" were a very common operation in code (true to this day). The post- and predecrement forms only affected the place where this opcode...
Why is “ref” not assigning updated value to the next field?
c#,.net,pass-by-reference,ref,post-increment
Since you wrote it as; return i++ This will still return 2 as a value but it will increase a value to 3 after the expression. If you wrote it as; return ++i This will return incremented value which is 3, and since a will be 3 after the execute...
x = x++ doesn't increment because the ++ is applied after the assignment?
...the ++ is applied after the assignment has occurred. Okay, hold on. This is actually confusing and perhaps suggests an incorrect behavior. You have the expression†: x = ( x++ ) What happens is (JLS 15.26.1): The expression on the left-hand side of the assignment x is evaluated (to...