Closed. This question is off-topic。它当前不接受答案。
想改善这个问题吗? Update the question,所以它是on-topic,用于堆栈溢出。
2年前关闭。
我在理解如何将此汇编代码转换为C方面遇到麻烦。它非常简短,只有几行,而且答案应该是一个衬里。
正如Lashane所说,此汇编代码等效于以下C代码:
或者,我想你也可以说:
但是C编译器会为您隐式地执行此操作,因此不必显式显示这种扩展转换。
想改善这个问题吗? Update the question,所以它是on-topic,用于堆栈溢出。
2年前关闭。
我在理解如何将此汇编代码转换为C方面遇到麻烦。它非常简短,只有几行,而且答案应该是一个衬里。
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
}
最佳答案
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
正如Lashane所说,此汇编代码等效于以下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编译器会为您隐式地执行此操作,因此不必显式显示这种扩展转换。
关于c - 如何将汇编代码转换为C? ,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/43573694/