In binary, convert from signed-magnitude to two’s-complement. All in C

I have an interesting puzzle. I need to convert from a signed number, say -5 (0x80000005) to two’s complement -5 (0xFFFFFFFB). The only operators I can use are, !, ~, &, ^, |, +, <<, >> and at most 15 of them in any combination with as many variables as I want. = may be used and does not count towards the total number of operators. Also, no “if” statements, loops or calls to functions of any kind may be used.

the function will look like

int sign2two(int x){
//... some stuff here
return x;
}

The problem I am having is that I can find out if the number is negative. But I want to say, if num is negative then flip bits and add 1. Otherwise return number. And I can’t figure out how to do that. Thanks.

These problems are extremely dumb in my opinion...

#include <stdint.h>
#include <stdio.h>

int32_t test1(uint32_t x) {
    const uint32_t sign  = (x & 0x80000000); // Total operations: 1
    const uint32_t value = (x & 0x7FFFFFFF); // Total operations: 2

    // Let's create a value N, where N is equal to:
    // if (sign)
    //     N = 0xFFFFFFFF;
    // else
    //     N = 0x00000000;
    uint32_t N = sign;  // Total operations: 2
    N = N | (N >> 1);   // Total operations: 4 
    N = N | (N >> 2);   // Total operations: 6
    N = N | (N >> 4);   // Total operations: 8
    N = N | (N >> 8);   // Total operations: 10
    N = N | (N >> 16);  // Total operations: 12

    // Let's create a value MaybeOne, where MaybeOne is equal to:
    // if (sign)
    //     MaybeOne = 1;
    // else
    //     MaybeOne = 0;
    uint32_t MaybeOne = N & 0x1; // Total operations: 13

    // Now, Let's perform the following step:
    // if (sign)
    //     return (value ^ 0xFFFFFFFF) + 1;
    // else
    //     return (value ^ 0x00000000) + 0;
    return (value ^ N) + MaybeOne; // Total operations: 15
}

int32_t test2(uint32_t x) {
    int32_t sign = (x & 0x80000000);
    int32_t value = (x & 0x7FFFFFFF);

    if (sign)
        return -value;
    else
        return value;
}

int main() {
    uint32_t i;
    for (i=0; i<0xFFFFFFFF; ++i) 
        if (test1(i) != test2(i))
            printf("0x%08x\n", i);
}

Additionaly, using a comment from Eric Postpischil, we can reduce the number of operations significantly:

int32_t test3(uint32_t x) {
    const uint32_t sign  = (x >> 31);        // Total operations: 1
    const uint32_t value = (x & 0x7FFFFFFF); // Total operations: 2

    // Let's create a value N, where N is equal to:
    // if (sign)
    //     N = 0xFFFFFFFF;
    // else
    //     N = 0x00000000;
    const uint32_t N = ~sign + 1;  // Total operations: 4

    // Let's create a value MaybeOne, where MaybeOne is equal to:
    // if (sign)
    //     MaybeOne = 1;
    // else
    //     MaybeOne = 0;
    uint32_t MaybeOne = sign; // Total operations: 4

    // Now, Let's perform the following step:
    // if (sign)
    //     return (value ^ 0xFFFFFFFF) + 1;
    // else
    //     return (value ^ 0x00000000) + 0;
    return (value ^ N) + MaybeOne; // Total operations: 6
}