C函数调用原理 - c语言编程

函数调用处理过程

The C calling convention in 16-bit programs is as follows. In the following description, the words caller and callee are used to denote the function doing the calling and the function which gets called.

The caller pushes the function's parameters on the stack, one after another, in reverse order (right to left, so that the first argument specified to the function is pushed last).The caller then executes a CALL instruction to pass control to the callee. This CALL is either near or far depending on the memory model.The callee receives control, and typically (although this is not actually necessary, in functions which do not need to access their parameters) starts by saving the value of SP in BP so as to be able to use BP as a base pointer to find its parameters on the stack. However, the caller was probably doing this too, so part of the calling convention states that BP must be preserved by any C function. Hence the callee, if it is going to set up BP as a frame pointer, must push the previous value first.The callee may then access its parameters relative to BP. The word at [BP] holds the previous value of BP as it was pushed; the next word, at [BP+2], holds the offset part of the return address, pushed implicitly byCALL. In a small-model (near) function, the parameters start after that, at [BP+4]; in a large-model (far) function, the segment part of the return address lives at [BP+4], and the parameters begin at [BP+6]. The leftmost parameter of the function, since it was pushed last, is accessible at this offset from BP; the others follow, at successively greater offsets. Thus, in a function such as printf which takes a variable number of parameters, the pushing of the parameters in reverse order means that the function knows where to find its first parameter, which tells it the number and type of the remaining ones.The callee may also wish to decrease SP further, so as to allocate space on the stack for local variables, which will then be accessible at negative offsets from BP.The callee, if it wishes to return a value to the caller, should leave the value in AL, AX or DX:AX depending on the size of the value. Floating-point results are sometimes (depending on the compiler) returned in ST0.Once the callee has finished processing, it restores SP from BP if it had allocated local stack space, then pops the previous value of BP, and returns via RETN or RETF depending on memory model.When the caller regains control from the callee, the function parameters are still on the stack, so it typically adds an immediate constant to SP to remove them (instead of executing a number of slow POPinstructions). Thus, if a function is accidentally called with the wrong number of parameters due to a prototype mismatch, the stack will still be returned to a sensible state since the caller, which knows how many parameters it pushed, does the removing. 引用自NASM手册
调用过程示例及测试用例
```
/*
	 
以三个参数为例子，AT&T32汇编,栈向低地址增长
fun(p0, p1, p2);
1.参数入栈，right->left
	gcc的做法：
       其中p0 p1 p2姑且认为是立即数
	movl $p2 8(%esp)
	movl $p1 4(%esp)
	movl $p0 (%esp)


 	 |_______|
ebp->|_______|
 	 |_______|
 	 |_______|
 	 |_______|
 	 |__p2___|
 	 |__p1___|
 	 |__p0___|<-esp
 	 |_______|
 	 |_______|
 	 |_______|


2.call fun


 	 |_______|
ebp->|_______|
 	 |_______|
 	 |_______|
 	 |_______|
 	 |__p2___|
 	 |__p1___|
 	 |__p0___|
 	 |__ret__|<-esp
 	 |_______|
 	 |_______|


3.跳转到fun代码
	push %ebp
	movl %esp, %ebp
 
 	 |_______|
 	 |_______|
 	 |_______|
 	 |_______|
 	 |__p2___|
 	 |__p1___|
 	 |__p0___|
 	 |__ret__|
ebp->|__ebp__|<-esp
 	 |_______|
 	 |_______|
 
其中存储的ebp是第2步中的ebp，即之前的ebp，不是图左面的ebp
如果栈向低地址增长
那么
old ebp = [ebp]
ret = [ebp + 4]
p0 = [ebp + 8]
p1 = [ebp + 12]
p2 = [ebp + 16]
 
4.
decrease ESP further, so as to allocate space on the stack for local variables,
which will then be accessible at negative offsets from EBP.
 
	subl $24, %esp
	根据具体需要分配栈空间,这里示例24/4=6个4B局部变量


 	 |_______|
 	 |_______|
 	 |_______|
 	 |_______|
 	 |__p2___|
 	 |__p1___|
 	 |__p0___|
 	 |__ret__|
ebp->|__ebp__|
 	 |_______|
 	 |_______|
 	 |_______|<= esp + 12 = ebp - 12
 	 |_______|
 	 |_______|
 	 |_______|<-esp
 	 |_______|
 	 |_______|
 
 
5.返回
	5.1 调整栈指针
	方式1：addl	$24, %esp
	方式2：movl %ebp, %esp
 	 |_______|
 	 |_______|
 	 |_______|
 	 |_______|
 	 |__p2___|
 	 |__p1___|
 	 |__p0___|
 	 |__ret__|
ebp->|__ebp__|<-esp
 	 |_______|
 	 |_______|
 	 |_______|
 	 |_______|
 	 |_______|
 	 |_______|
 	 |_______|
 	 |_______|


	 5.2弹出旧frame指针=销毁当前栈帧
	popl	%ebp


 	 |_______|
ebp->|_______|
 	 |_______|
 	 |_______|
 	 |_______|
 	 |__p2___|
 	 |__p1___|
 	 |__p0___|

            
```

C函数调用原理(一)

函数调用处理过程

调用过程示例及测试用例