字符添加在 C 中没有预期结果

Question

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);
}

Answer 1

It appears on your system that char is signed. With ASCII, 'D' , 'C' and 'A' are the same as the integers 68 , 67 , and 65 respectively.

Add 68 to 65 and you get 133 . The binary representation of 133 is 10000101 . Notice that the most significant bit is 1 . As you're using signed chars, twos complement comes into play, and the result is actually -123 .

Remember that a signed char can hold values ranging from -128 to 127 , rather than 0 to 255 .

Answer 2

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 2 ^w , 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.

Answer 3

It would seem your compiler treats char as signed char , so:

a = 'D', b='C', c='A'

a = c + 'D' = 'A' + 'D' = 65 + 68 = 133

but since a is a signed char , it becomes -123 (133-256)

b = a + b + 'A' = -123 + 'C' + 'A' = -123 + 67 + 65 = 9

c = b - a = 9 - -123 = 9 + 123 = 132

a = a - b + 'C' = -123 - 9 + 66 = -66

b = b - 68 = 9 - 68 = `-59``

So at the end, a = -66 , b = -59 , and c = 132