Cast the variables to a long int or long long int type giving at the same time the range of unsigned int (0..2^32-1) and sign.

avr,signed,avr-gcc,16-bit,atmega

Your arithmetic seems flawed. As written, you would allow a jump from 10000 to -10000 as okay. Instead, you want to use abs() only once. difference = abs(newvalue - lastvalue); You can also avoid abs() altogether. difference = newvalue - lastvalue; if (difference < 0) difference *= -1; if (difference...

This difference between unsigned and signed number is that one of the bit is used to indicate positive or negative number. So in your example you have 8 bits. If I treat is as signed, then I have 7 bits to work with: 2^7 000 0000 = 0 111 1111...

c,signed,modulus,ida,and-operator

The optimization of x % N to x & (N-1) only works if N is a power of two. You also need to know that x is positive, otherwise there is a little difference between the bitmask operation and the remainder operation. The bitmask operation produces the remainder for the...

The C++11 standard (draft) §3.9.1.2 has this to say; There are ﬁve standard signed integer types : “signed char”, “short int”, “int”, “long int”, and “long long int”. The C11 standard (draft) §6.2.5.4 has this to say; There are ﬁve standard signed integer types, designated as signed char, short int,...

No. Integer types are permitted to have padding bits which do not contribute to the value. For example, a 32-bit signed integer type with 8 padding bits would have a maximum value of 223-1, or 16777215, rather than 231-1, or 2147483647. This is stated explicitly in the C standard. I...

The bit's are not changing between b and sb at all. This behavior is coming from Convert.ToString(). There just isn't an overload of Convert.ToString() that takes an sbyte and a base. The closest match would be Convert.ToString Method (Int16, Int32). So sb is being sign extended to 16 bits.

java,byte,signed,primitive-types,type-mismatch

At first, it seems puzzling: b = 'b'; Takes a wider type char which takes 2 bytes in memory, and puts it into a single byte. This shouldn't work. But it does. b = -b; Takes a negative byte and puts it in a byte. This should work, but it...

android,android-studio,apk,release,signed

You need to be in release mode to generate a signed APK. On the lower left corner of your Android Studio should be a tiny square icon. Click on it and a menu shall pop up. Now choose Build Variants and you'll get a table with two columns: Module and...

binary,overflow,addition,signed,twos-complement

Overflow occurs only when correct answer cannot be interpreted using given number of bits.An extra bit is required to interpret the answer right.Consider the following example of byte numbers addition: +70 in binary 01000110 +80 in binary 01010000 On addition :10010110 1 in the MSB(most significant bit) indicates that answer...

Let's see how the computer stores the value b: 5 is 00000101, so -5 will be 11111011, so, when you convert it to unsigned char, it will became some positive number with value 11111011 in binary, which is larger than 00000101. So, that's why a = 00000101 is smaller than...

It seems to me that a big problem is that your recursive call to add2() ignores the return value: pushl %eax movl 4(%ebx), %esi pushl %esi call add2 addl $4, %esp popl %eax ; <-- overwrites what the add2 call returned Also, your C equivalent code doesn't seem to be...

android,facebook-graph-api,hash,signed

According with FB doc: https://developers.facebook.com/docs/android/getting-started?locale=es_LA#release-key-hash It's the same way to generate the debug key hash, but now you have to use the release key store. Good Luck!...

Cordova 5 uses gradle to compile now, so the old ant.properties no longer works. You can work the same trick by creating a release-signing.properties file in platforms/android and adding the following: storeFile=<path to="" .keystore="" file=""> storeType=jks keyAlias=<your key="" alias=""> // optional : keyPassword=<your-key-password> storePassword=<your-store-password> ...

assembly,hex,unsigned,addition,signed

binary addition has no notion if signed vs unsigned. exactly the same the bit patterns. The meaning of those bit patterns is in the eyes of the beholder (multiply and divide this is not true). This is the beauty of twos complement. 0xF0 + 0x02 = 0xF2 = 240 +...

java,integer,byte,unsigned,signed

When you or a byte with an int the byte will be promoted to an int. By default this is done with sign extension. In other words: // sign bit // v byte b = -1; // 11111111 = -1 int i = (int) b; // 11111111111111111111111111111111 = -1 //...

vhdl,multiplication,equation,signed

Since x and y are of datatype signed, you can multiply them. However, there is no multiplication of signed with real. Even if there was, the result would be real (not signed or integer). So first, you need to figure out what you want (the semantics). Then you should add...

c#,java,encoding,base64,signed

It's the same. In the first case bytes are unsigned and Java bytes are signed. What we see in array is different interpretation of the same value, eg 189 and -67 = 0xBD byte b = (byte) 189; now b = -67...

One way of doing it: >> typecast(uint8(bi2de(byte, 'left-msb')), 'int8') ans = -126 ...

overflow,verilog,addition,signed,subtraction

{OFAdd, AddAB} <= A + B; In the example the MSB (OFAdd) is not an overflow bit. If you had access to the carry out of the final bit of an adder this could act as an overflow, but in RTL you do not have access to this only another...

You could subtract 128 from the value: byte x = (byte) (value - 128); That would be order-preserving, and reversible later by simply adding 128 again. Be careful to make sure you do add 128 later on though... It's as simple as: int value = x + 128; So for...

c++,memory,integer,signed,signed-integer

With the help of a lot of the posts above, I've come up with this solution: inline void reduceSignedIntWidth(int32_t& destInt, int width) { //create a value mask, with 1's at the masked part uint32_t l_mask = (0x01u << width) - 1; destInt &= l_mask; } It will return the reduced...

It's true that 9223372036854775808 becomes -9223372036854775808, but it's not so simple as just "integers wrap around". Here are a few more mappings (found by experimentation): 9223372036854775809 → still -9223372036854775808 (not -9223372036854775807) 9223372036854775810 → still -9223372036854775808 (not -9223372036854775806) So you might think that numbers outside the int range all get mapped...

c++,bit-manipulation,language-lawyer,signed

You're quite correct -- the expression ~'q' << 6 is undefined behavior according to the standard. Its even worse than you state, as the ~ operator is defined as computing "The one's complement" of the value, which is meaningless for a signed (2s-complement) integer -- the term "one's complement" only...

Pre-C++11 you can use this implementation of std::conditional: template<bool B, class T, class F> struct conditional { typedef T type; }; template<class T, class F> struct conditional<false, T, F> { typedef F type; }; Then we can write a struct to extract the signedness of a type: template <typename T>...

c,types,casting,unsigned,signed

That answer is precisely correct for 32-bit ints. unsigned int a = -4; sets a to the bit pattern 0xFFFFFFFC, which, interpreted as unsigned, is 4294967292 (232 - 4). Likewise, b is set to 232 - 5. When you add the two, you get 0x1FFFFFFF7 (8589934583), which is wider than...