# Computing the absolute value of a signed 32-bit two’s complement integer with bitwise operators

solution:

int iabs(int a) {
int t = a >> 31;
a = (a^t) - t;
return a;
}


Can someone explain at the math level why shifting 31 bits to the right, xoring this result with t and subtracting t from this work?

## 1 Answer

This is more of a programming question than a math one. Note that the right shift >> is sign-extending in the C language. Then, assuming an int is 32-bit wide:

                   // if a >= 0          // if a < 0

int t = a >> 31;   // t = 0              // t = 0xFFFFFFFF = -1
a = (a^t) - t;     // a = (a^0)-0 = a    // a = (a^(-1))-(-1) = ~a + 1 = -a


[ EDIT ]  Strictly speaking, the right-shift behavior is implementation defined in both C and C++, though (emphasis mine):

For negative a, the value of a >> b is implementation-defined (in most implementations, this performs arithmetic right shift, so that the result remains negative).

This is, however, a matter better suited to discuss next door at stackoverflow.com.

• So if we have a = -2 = 110 then 110>> 2 = 001 , 110 xor 001 = 111 and 7 - 1 != 2. Why did I do wrong. I think it has something to do with your sign extending claim. – ElChavoDelOcho Dec 14 '17 at 1:15
• @Dan It does. Assuming a 3-bit int as you have here, 0x110 >> 2 produces 0x111. It’s an arithmetic shift, not a logical shift. – amd Dec 14 '17 at 1:21
• @Dan In a 3-bit two's-complement representation 110 >> 1 = 111 under most implementations (see the edit at the end). In 32-bit -2 = 1111 1111 1111 1110 and -2 >> 31 = 1111 1111 1111 1111. – dxiv Dec 14 '17 at 1:31