Discussion on: How can you swap two variables without using a third?

[Zuodian Hu]

Very contrived, only works on ARM, not even acceptable by all C compilers, but you never constrained the problem that way!

void swap(int* a, int* b) {
    __asm volatile (
        "MOV R2, R1 \n"
        "MOV R1, R0 \n"
        "MOV R0, R2 \n"
    );
}

[Zuodian Hu]

Oh and as prplz mentioned, any self-respecting compiler nowadays will use only registers any way you decide to implement this little stub of a function. Yes yes, the question is very contrived too, but hey.

[Nick Karnik]

Aren't you using a third register though?

Are you're thinking that you can do that because it's not in RAM?

[Zuodian Hu]

The way the question is formulated, you're limited by having a certain amount of memory. Any register use doesn't count against that at all.

[Zuodian Hu]

I went to bed, and realized that my answer is wrong. Exchanging the values in register is completely invisible to the function caller. Here's the correct answer.

void swap(int* a, int* b) {
    __asm volatile {
        "LDR R2, [R0] \n"
        "LDR R3, [R1] \n"
        "STR R2, [R1] \n"
        "STR R3, [R0] \n"
    }
}

Further, as R0-R3 are all caller-save registers in the ARM ABI, the function body, excluding the function entry stack shenanigans, uses zero memory.