Why is 1^-1 = -2 in the C language?

Question

Can anyone explain how x^-1 is being executed in below code. I tried but no luck. I understood how x^1 is being executed.

#include <stdio.h>
int main(void) 
{
int a=220, b=221, c=3;
printf("a^1= %d ,, a^-1= %d \n", a^1, a^-1);
printf("b^1= %d ,, b^-1= %d \n", b^1, b^-1);
printf("c^1= %d ,, c^-1= %d \n", c^1, c^-1);
return 0;
}
/* output: a^1= 221 ,, a^-1= -221
           b^1= 220 ,, b^-1= -222
           c^1= 2   ,, c^-1= -4    */

Given the tags, you seem to understand what the `^` operator does. If so, you should learn about the representation of negative numbers in memory. — ForceBru, Jul 18 '17 at 19:26
`x^-1` means `x xor -1` in both C & C++ : http://en.cppreference.com/w/cpp/language/operator_arithmetic — Jesper Juhl, Jul 18 '17 at 19:27
If by `^-1` you meant "multiplicative inverse", that was never going to work for even numbers anyway. There's no built-in operator for it in C sadly (nor most other languages) — harold, Jul 18 '17 at 20:47
Possible duplicate of [Why is my power operator (^) not working?](https://stackoverflow.com/questions/4843304/why-is-my-power-operator-not-working) — phuclv, Jul 19 '18 at 14:56

Weather Vane · Answer 1 · 2017-07-18T19:38:53.997

3

The ^ operator is the XOR or exclusive-or operator in C. To make it simple, consider 8-bit signed values, using the typical two's-complement encoding. The int type will work in the same way.

Decimal   Binary
      1   00000001
     -1   11111111
          -------- XOR
     -2   11111110

Note that the unary operator - has higher precedence that the ^ bitwise operator.

edited Jul 18 '17 at 19:38

answered Jul 18 '17 at 19:33

Weather Vane

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I especially like the terseness of this post. – chux - Reinstate Monica Jul 19 '17 at 20:06

score 0 · Accepted Answer · answered Jul 18 '17 at 19:35

Signed ints use Two's complement in order to represent negative numbers.

1 represented in int32:

0000 0000 0000 0000 0000 0000 0000 0001

-1 represented in int32, using Two's complement:

1111 1111 1111 1111 1111 1111 1111 1111

XORing the two results in:

1111 1111 1111 1111 1111 1111 1111 1110

which, for signed ints, is -2.

for unsigned ints, that would be 2^32 - 2.

Why is 1^-1 = -2 in the C language?

2 Answers2