I feel like you're probably right, but I can't figure out the exact reason. The multiplication of (n-1)(n-2) already expands to 64 bits. It has to: Even when

I feel like you're probably right, but I can't figure out the exact reason. The multiplication of (n-1)(n-2) already expands to 64 bits. It has to: Even when the loop accumulator doesn't overflow when counting up to n(n-1)/2 in the abstract machine, (n-1)(n-2) will be larger (and possibly overflow 32 bits) for most n. (The highest valid n is 92682, I think.) So surely once you're doing that, I'd think n(n-1)/2 can be done in 64 bits just as easily. unsigned int sum2(unsigned int n) { unsigned long ln = n; return ln * (ln - 1) / 2; } a.out <+0>: pushq %rbp a.out <+1>: movq %rsp, %rbp a.out <+4>: movl %edi, %ecx a.out <+6>: leaq -0x1(%rcx), %rax a.out <+10>: imulq %rcx, %rax a.out <+14>: shrq %rax a.out <+17>: popq %rbp a.out <+18>: retq