# Char addition does not have expected result in C

Can someone explain me how is this at the end

``````a=?,b=?,c=-124
``````

and not

``````a=?,b=?,c=132
``````

This is code:

``````#include <stdio.h>

int main() {
char a = 'D', b='C', c='A';
a = c + 'D';
b = a + b + 'A';
c = b - a;
a = a - b + 'C';
b = b - 68;
printf("a=%c,b=%c,c=%d", a, b, c);
}
``````

### >Solution :

Your C implementation has a signed eight-bit `char`, likely uses ASCII, and, when converting an out-of-range value to a signed integer type, wraps modulo 2w, where w is the width of (number of bits in) the type. These are all implementation-defined; they may differ in other C implementations, with certain constraints.

`char a = 'D', b='C', c='A';` initializes `a` to 68, `b` to 67, and `c` to 65.

`a = c + 'D';` assigns 65 + 68 = 133 to `a`. Since 133 does not fit in `char`, it wraps to 133 − 256 = −123.

`b = a + b + 'A';` assigns −123 + 67 + 65 = 9 to `b`.

`c = b - a;` assigns 9 − −123 = 132 to `c`. This wraps to 132 − 256 = −124.

`a = a - b + 'C';` assigns −123 − 9 + 67 = −65 to `a`.

`b = b - 68;` assigns 9 − 68 = −59 to `b`.

`printf("a=%c,b=%c,c=%d", a, b, c);` prints `a=?,b=?,c=-124` because `a` and `b` are codes for abnormal characters and the value of `c` is −124.