为什么这个函数在第一次调用后没有返回指针？

int* intptr ()
{
    int i;
    i=rand();
    printf("%d_____",i);
    return(&i);
}

int main()
{
    int* j,k,l;
    j=intptr();
    k=intptr();
    l=intptr();
    printf("%d/n",j);printf("%d/n",k);printf("%d/n",l);
}

Here the intptr function returns a pointer in the first call and after that all the subsequent calls return the int 2752220

这里intptr函数在第一次调用中返回一个指针,之后所有后续调用都返回int 2752220

3 个解决方案

#1

The other answers address undefined behavior, which is the poor coding practice your code sample exhibits. However, as shown here, the real issue is that when you define

其他答案解决了未定义的行为,这是您的代码示例所展示的不良编码实践。但是,如此处所示,真正的问题是当您定义时

int* j,k,l;

only the first variable j is an int pointer. The others are regular integers. In order to make them all pointers, you have to define them like this:

只有第一个变量j是一个int指针。其他是常规整数。为了使它们成为所有指针,你必须像这样定义它们:

int* j; 
int* k;
int* l;

By running the code by chance each pointer may point to the same deleted value from your function, but at least they'll be pointers and not integers.

通过偶然运行代码,每个指针可能指向函数中相同的已删除值,但至少它们将是指针而不是整数。

#2

In your code, int i; is local to the intptr () function. You cannot return the address of i and use that in the caller. it invokes undefined behaviour.

在你的代码中,int i;是intptr()函数的本地。您不能返回i的地址并在调用者中使用它。它调用未定义的行为。

To elaborate, the lifetime of i has ended when the intptr () function has returned. The address you're returned (or tried to return) has become invalid. Using the return value is thus, UB.

详细说明,当intptr()函数返回时,i的生命周期已经结束。您返回(或尝试返回)的地址已失效。因此,使用返回值UB。

After that, please note

之后,请注意

k and l are of type int, and you're trying to store an int *. Wrong. The result is implementation-defined (refer §6.3.2.3, C11) and most likely that is not something you wanted in your code.

k和l的类型为int,并且您正在尝试存储int *。错误。结果是实现定义的(参见§6.3.2.3,C11),并且很可能在您的代码中不是您想要的。

j being a pointer, the correct way to print the pointer address is to use

j是一个指针,打印指针地址的正确方法是使用
```
printf("%p/n",(void *)j)
```
using wrong type of argument (%d expects int, not int *) is again UB.

使用错误类型的参数(%d期望int,而不是int *)再次是UB。

#3

You are trying to return something, which is local to the function intptr(), hence whatever output you are getting even for the first call, it is not valid. This is undefined behavior. Because after the stack unwinding happens(control returning from inptr() to main(), the idisappears and any value at the address of it could not be found.

您正在尝试返回函数intptr()的本地函数,因此即使是第一次调用,您获得的输出也是无效的。这是未定义的行为。因为在堆栈展开发生后(控制从inptr()返回到main(),idis出现并且无法找到它的地址处的任何值。

You can make it static to work.

你可以让它静止工作。

int* intptr ()
{
    static int i = 0;
    i=rand();
    printf("%d_____",i);
    return(&i);

 }

#1