C 如何使用静态变量在递归函数中保持状态

我编写了以下玩具函数，以了解堆栈如何在递归函数调用时向下增长，然后在返回时展开：

void countdown(int n) {

    // trying to get initial address so the math is a bit simpler 
    // on how much the stack grows down/up
    static intptr_t initial_address = (intptr_t) &n;

    if (n!=0) {
        printf("%d...at address: %p\n", n, &n);
        countdown(n-1);
        printf("%d...at address: %p\n", n, &n);
    } else 
        printf("Liftoff!\n");
}

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有没有一种方法可以在不向

倒计时

函数传递另一个参数的情况下执行类似操作？如果是，正确的方法是什么？

一种方法是使用的地址或以前的地址的

max

值，知道顶级函数调用将具有最高的堆栈地址。例如：

void countdown(int n) {

    // will only be initialized once, and first value is throw-away, since
    // the memory address is always going to be greater than zero
    static intptr_t initial_address=0;

    if (n!=0)
    {

        initial_address = initial_address > (intptr_t) &n ? initial_address : (intptr_t) &n;

        // pushing upto the stack (will grow down...)
        printf("%d...at address %p | offset: -%#zx\n", n, &n, initial_address - (intptr_t) &n); 
        countdown(n-1);

        // stack unwinds after function starts returning
        printf("%d...at address %p | offset: -%#zx\n", n, &n, initial_address - (intptr_t) &n); // pushing upto the stack (will grow down...)
    }
    else 
        printf("Liftoff!\n");
}

4…地址0x7ffe6944967c处偏移量：-0
3…地址0x7ffe6944965c处偏移量：-0x20
2…地址0x7ffe6944963c处偏移量：-0x40
1…地址为0x7ffe6944961c |偏移量：-0x60
发射
1…地址为0x7ffe6944961c |偏移量：-0x60
2…地址0x7ffe6944963c处偏移量：-0x40
3…地址0x7ffe6944965c处偏移量：-0x20
4…地址0x7ffe6944967c处|偏移量：-0

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…或者您可以将另一个参数传递给倒计时函数，例如，指向状态结构的指针。@因此，对于上述内容，您只需添加另一个参数，例如

void count（int n，state*state）{…}

？