Question

I just started to learn a bit assembler from compiler output.

test(1);

This simple function call creates following asm output (compiled with x64)

000000013FFF2E76  mov         ecx,1  
000000013FFF2E7B  call        test (13FFF33C0h) 

But why isn’t it:

000000013FFF2E76  push        1  
000000013FFF2E7B  call        test (13FFF33C0h) 

I thought a function parameter will be pushed to the stack and then poped in the function. Can somebody explain why VS prefer the top one?

Answer 1

It’s because that’s the ABI on x64 Windows.

On Windows x64, the first integer argument is passed in RCX, the second in RDX, the third in R8 and the fourth in R9. The fifth and following are passed through the stack.

Because your function has a single argument, only RCX is used.

The compiler issued a write to ECX because all writes to 32-bit registers result in zeroing the higher part of the 64-bit register, and 32-bit immediates are obviously shorter than 64-bit ones (instruction cache anyone?).