int main() { printf("%c\n",~('C'*-1)); return 0; } I tried...">
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How does printf ("% c \ n", ~ ('C' * - 1)) evaluate to c?

#include<stdio.h>
int main()
{
printf("%c\n",~('C'*-1));
return 0;
}

I tried the above source code and executed without any warnings.

Yield B . I'm excited about how the above code is handled and what the printf value means ("% c \ n", ~ ('C' * - 1))

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3 answers


In C, 'C'

there is int

, it is a small integer with a value 67

(assuming ASCII). You can get each step:

#include<stdio.h>
int main()
{
    printf("%d\n", 'C');            //67
    printf("%d\n", 'C' * -1);       //-67
    printf("%d\n", ~('C' * - 1));   //66
    printf("%c\n",~('C' * -1));     //B
    return 0;
}


In 2's complement, the value ~(-67)

is 66

.

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The only important part is the expression:

~('C' * -1)

Let's break it down:



  • 'C'

    ASCII code 67.
  • ('C' * -1)

    - -67.
  • -67 in binary, 10111101
  • Bitwise negation of this (s ~

    ) and you have 01000010, which is 66.
  • 66 is the ASCII code for "B".

More generally, most computers use " two's complement " arithmetic, where a numeric negation followed by a bitwise negation is equivalent to subtracting 1. And, of course, B

less than C

ASCII.

On a computer that does not use two's complement, the result may be different. Such computers are rare.

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It is well known that "making a number negative" (multiplying it by -1

) in two's complement is equivalent to inverting its bit representation ( ~

) and adding one ( +1

). So the above is equivalent ~(~'C' + 1)

, which in turn is equivalent to a simple decrement if the original number, as here, the ASCII code 'C'

, was odd.

That is, if LSB is installed 'C'

:

  • LSB is ~'C'

    not installed, which means that
  • after adding one ( ~'C' + 1

    ) LSB is set again.
  • Now, inverting the whole expression ( ~(~'C' + 1)

    ) will give us the original number - only when the LSB is no longer set, because it was set before the final inversion (which inverted the LSB like any other bit).
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