修改函数中的指针值 [英] Modify pointer value in a function

问题描述

我正在研究指针，并试图理解这段代码.

I'm studying pointers and i'm trying to understand this code.

int main()
{
    int i=66 ;
    int *x;
    x=&i;
    f(&x);
    printf("%d",*x);

    return 0;

}
void f(int *x)
{
    int j=7;
    x=&j;
    printf("%d-",*x);

} 

我期望输出: 7-7 ，但我得到 7-66.我想当我写

I'm expectin as output : 7-7 , but i get 7-66. I suppose that when i write

x=&j;

在函数中，原始指针现在被设置为 j 值的方向，这是可以的，因为 printf 返回 7- 但是为什么当我回到 main 时， printf 返回 66?指针现在不是设置为函数的 j 值吗?我知道关闭函数时函数中的值会丢失，但是为什么我的指针在运行函数后仍然设置为主函数中的 i 值?

in the function the original pointer now is setted to the direction of the j value and this is ok because the printf return 7- but why when i go back to the main the printf return 66? isn't the pointer now setted to the j value of the function ? i know the values in the function are lost when i close the function but why my pointer is still setted to the i value in the main after i run the function?

推荐答案

在您的函数 f(int* x) 中，您将变量 x 分配给地址您的本地临时变量 j.您真正想要做的是将 x 指向的 value 分配给 j 的 value.

In your function f(int* x), you are assigning the variable x to the address of your local, temporary variable j. What you really want to do is assign the value of what is pointed to by x to the value of j.

#include <stdio.h>
#include <stdlib.h>

void f(int *x)
{
    int j=7;
    *x=j;
    printf("%d-",*x);

} 

int main()
{
    int i=66 ;
    int *x;
    x=&i;
    f(x);
    printf("%d",*x);

    return 0;

}

这可以在这里运行.

第二个问题是，您在主函数中获取变量 x 的地址，并将其作为参数传递给 f(int*)，后者变为指向 int 指针的指针(int** 而不是 int*).请记住，x 已经是一个 int* 并且可以直接传递.

A second issue is the fact that you take the address of your variable x in the main function and pass that as a parameter to f(int*) which becomes a pointer to a pointer to an int (an int** not an int*). Remember that x is already an int* and can be passed directly.

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