在c中添加char值

Question

#include<stdio.h>
int main()
{

  char c =118,a=117;
  c=c+10;
  a=a+10;
  printf("c:%d, a:%d\n", c,a);

}

The answer is c:-128, a:127.

Can someone explain me why c+10 is -128 and a+10 is 127?

Thanks in advance.

Answer 1

Signed 8-bit values range from -128 to 127. The bits for 127 are 0111 1111, and the bits for 128 would be 1000 0000. The problem is, in a signed number, the high order (leftmost) bit is the sign flag (0 is +, 1 is -). So, because it is signed, the computer interprets this as a negative number with the result -128 (This is called signed overflow, if I remember correctly, and everyone runs into this when programming at one point or another) (check out 2's complement to see why the low 7 bits are 128, not 0). You can get around this "problem" by declaring c and a as unsigned char instead of char .

BTW, you could save a variable this way:

char a=117; printf("c:%d, a:%d\\n", a+11,a+10);

Answer 2

117 + 10 = 127 which is in range of the char type ( -128 to 127 )

If you open the calculator of your windows in programmer mode, you can see that 118 is represented as 1110110 in binary. also, 10 is represented as 1010 . if we add this two, the result is ‭ 10000000‬ . it is not in the range of the char type, and this number is equivalent to -128 . so the -128 is printed.

Answer 3

because char is 8-bit signed in your compiler. So 118+10 is out of range (max is 127 ).

The implementation of your compiler "wraps" it around and you get -128 instead.

Answer 4

In your case, a char is being represented by the compiler on your particular platform as a signed 8-bit value , and is in 2's complement reoresentation . That means the highest bit is the sign bit (if it's a 1, then the number is negative). So the range in binary for non-negative values is (in binary) 00000000-01111111 which is 0-127 in decimal. The negative values range (in binary) from 10000000 to 11111111 which is -128 to -1 in decimal.

If you start with 118, in binary that's:

01110110

If I add decimal 10, that's adding 1010 in binary:

 01110110
+00001010
---------
 10000000

You can see now the highest bit is set, meaning the number overflowed (became greater than the maximum 127 in decimal) and now represents a negative number. The 8-bit binary value 10000000 happens to represent -128 in decimal.

So adding (in decimal) 10 to the char value 118, yields -128.

Your value of a is 117, so 10+117 = 127 still fits in the 7 bits for a positive value of 127. You can do the above binary analysis as an exercise to see how that works.