传递C函数的参数[重复]

Question

这个问题在这里已有答案：

• What (actually) happens, when a function with the warning “control reaches end of non-void function” is called? 7个答案

我是新手。所以请耐心等待，

#include<stdio.h>

    int abc(int k)
    {
    k++;

    }
    int main()
    {
    int a=1;
    printf("%d",abc(a));

    return 0;
    }

以上程序的输出是：1

我的问题是输出不应该是'2'，因为实际参数将'1'的值传递给形参，并且必须通过函数abc递增。

当我将函数调用更改为

printf("%d",abc(1));

输出是一些垃圾值。

参数传递如何在这里工作？请解释。

Answer 1

您获得的意外结果不是由“参数传递”产生的，而是来自abc函数不返回任何值的事实。您应该使用return k;语句来获得您期望的输出。但是对于参数传递，它们是通过值传递的，即传递的值被复制到临时位置k（仅在函数中可见），而不是在它之外修改。

Answer 2

您已经通过值传递a的代码示例。你可以把它想象成a的副本。您使用注释修改了代码：

#include<stdio.h>

int abc(int k)
{
    // k is a copy of a, it is not a, since k is a copy, it has the 
    // value of a at the point of the copy.  So, k is 1

    k++;  // k is now 2
    return k; // return the computed value to the caller and destroy k
}

int main()
{
    int a=1;

    // as previously written, without the return statement in abc()
    // this function returned nothing.  So, the compiler just arranges
    // for something to be used from the stack where the return would 
    // have placed 2.  (I'm not terribly familiar
    // with assembly and so I'm not sure which register it would use).
    // That's why you get non-nonsensical data, whatever is in memory is
    // what you get and without the return statement, there's nothing 
    // meaningful there.

    // Also, as I commented above, abc() takes a **copy** of a.  Thus,
    // the contents of a are unmodified.  See how the printf() is
    // changed.  What does it print? 
    printf("%d %d",abc(a), a);

    return 0;
}