Why is the RAX Register Pushed onto the Stack During Tail Calls with std::function?

Stack Alignment in Tailcall Situations

The question arises as to why the RAX register is first pushed to the stack in the assembly code generated for the C code that interacts with a std::function object.

The Necessity of Stack Alignment

The 64-bit ABI mandates that the stack be aligned to 16 bytes prior to any call instruction. When a call is made, it pushes an 8-byte return address onto the stack, disrupting this alignment. To rectify this, the compiler must take steps to realign the stack to a multiple of 16 before the subsequent call.

Pushing a Disposable Value for Alignment

Instead of executing "sub rsp, 8," pushing a "don't-care" value, such as RAX, proves more efficient on CPUs equipped with a stack engine. This is because a simple push instruction often requires less processor overhead than a sub rsp, 8 instruction.

Comparison with a Tailcall without std::function Wrapper

When there is no std::function wrapper present, as in the "void g(void (*a)())" example, the tailcall is straightforward: a simple jump (jmp) instruction to the target function. No additional steps are necessary for stack alignment since the tailcall will naturally maintain the proper stack alignment.

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