negating characters in C

Question

Is this undefined behavior? It prints -128 as the result:

#include<stdio.h>
int main()
{
    char i=-128;
    i=-i;
    printf("%d",i);
}

Please explain.

Answer 1

The two's complement of -128 in an 8-bit signed value is -128.

Look at the binary values:

Original value: 10000000

Complement: 01111111

Increment: 10000000

Answer 2

This is not undefined behavior. Assuming type char is signed in your platform, it is implementation defined behavior. (C99, 6.3.1.3p3)

i = -i;

the i in -i is first promoted to int, so -i is 128 and then 128 is converted to char by the integer conversions.

Here is the paragraph of the Standard that says the conversion of 128 to char is implementation defined:

(C99, 6.3.1.3p3) Otherwise, the new type is signed and the value cannot be represented in it; either the result is implementation-defined or an implementation-defined signal is raised.

EDIT: while it is implementation defined, there is a common implementation behavior between most implementations. Here is what gcc (and most other compilers do) documents to do:

For conversion to a type of width N, the value is reduced modulo 2^N to be within range of the type; no signal is raised

http://gcc.gnu.org/onlinedocs/gcc/Integers-implementation.html

Note that not all compilers behave this way. Some (DSP) compilers just saturate. In this case (and still assuming a signed char), after i = -i; the value of i would be 127.