将变量保存在内存中,C

extern void some_function(char * s,int l);

void do_it(void) {
     char str[] = "I'm doing it!";
     some_function(str,sizeof(str) );
}

这将变成类似的东西(在psudo asm中为组合处理器)：

.data
local .do_it.str       ; The contents of str are stored in a static variable
.text   ; text is where code lives within the executable or object file
do_it:
    subtract (sizeof(str)) from stack_pointer        ; This reserves the space for str,sizeof(str)
                                                     ; and a pointer to str on the stack

    copy (sizeof(str)) bytes from .do_it.str to [stack_pointer+0]   ; this loads the local variable
                                               ; using to the memory at the top of the stack
                                               ; This copy can be a function call or inline code.

    push sizeof(str)         ; push second argument first
    push stack_pointer+4     ; assuming that we have 4 byte integers,; this is the memory just on the other side of where we pushed
                             ; sizeof(str),which is where str[0] ended up

    call some_function

    add (sizeof(str)+8) to stack_pointer          ; reclaim the memory used by str,sizeof(str),; and the pointer to str from the stack
    return

从中您可以看出,关于如何创建局部变量str的假设并不完全正确,但这仍然不一定有效.

如果你这样做了

void do_it(void) {
     static str[] = "I'm doing it!";

然后编译器不会为堆栈保留堆栈上的空间,然后将其复制到堆栈中.如果some_function要改变str的内容,则对do_it的下一个(或并发)调用(在同一个进程中)将使用str的更改版本.

如果some_function被声明为：

extern void some_function(const char * s,int l);

然后,由于编译器可以看到在do_it中没有更改str的操作,即使str未被声明为静态,它也可以在堆栈上不生成str的本地副本.