如何将汇编代码转换为C？

Question

我无法理解如何将此汇编代码转换为C.它非常短，只有几行，答案应该是一行代码。

char_out: 
    subq $8, %esp 
    movsbl %dil, %edi # Parameter 1 in %dil/%edi 
    call putchar  # put char is a C function that writes (prints) 
    addq $8, %rsp  # a single char to stdout (the screen). 
    ret 

void char_out(char P1) 
{ 
    //insert here 
} 

Answer 1

char_out: 
    # Allocate 8 bytes on the stack by subtracting 8 from the stack pointer. 
    # 
    # This is done for conformance to the calling convention expected by the 
    # 'putchar' function, which we're about to call. 
    subq $8, %esp 

    # This is the weirdo AT&T mnemonic for the MOVSX instruction, which does a 
    # move with sign extension. This means that it extends a small value 
    # (here, a BYTE-sized value in DIL) into a larger value (here, a 
    # DWORD-sized value in EDI), in a way that properly accounts for the 
    # value's sign bit. 
    # 
    # This is necessary because the 'putchar' function expects to be passed a 
    # 32-bit 'int' parameter. 
    movsbl %dil, %edi 

    # It's obvious what this does: it calls the 'putchar' function. 
    call putchar 

    # Clean up the stack, undoing what we previously did to the stack pointer 
    # at the top of the function (the 'subq' instruction). 
    addq $8, %rsp 

由于Lashane已经评论说，这汇编代码相当于下面的C代码：

void char_out(char P1) 
{ 
    putchar(P1); 
} 

或者，我想你也可以说：

void char_out(char P1) 
{ 
    int temp = (int)P1;  // movsbl 
    putchar(temp); 
} 

但是C编译器会为你隐式执行，所以没有必要明确地显示扩展转换。