指针不更新它在 void 函数内部指向的值 [英] pointer not updating the value it is pointing to inside void function

问题描述

我制作了一个简单的函数，它使用指针对参数之间的加法和绝对差异进行了简单的处理，但是当我尝试更新指针时，指针仍然具有旧值.为什么会这样，或者我做错了什么:

I've made a simple function that simple does the addition and absolute difference between the parameters using pointers, but when I try to update the pointer, the pointer still has the old value. why is this so, or what am I doing wrong:

#include <stdio.h>
#include <cstdlib> 

void update(int *a,int *b) {
int temp = *a;
int temp2 = *b;
int temp3 =0;
temp3 = temp + temp2;
printf("%d",temp3);
*b = abs(*a - *b);
 a = &temp3; // it is not updating
}

int main() {
    int a, b;
    int *pa = &a, *pb = &b;

    scanf("%d %d", &a, &b);
    update(pa, pb);
    printf("%d\n%d", a, b);

    return 0;
} 

指针 a 没有更新并且仍然保留它在函数 update 中的旧值

the pointer a is not updating and still retains its old value inside the function update

推荐答案

a 是传递的指针的副本.update 结束时，a 丢失.当你这样做时:

a is a copy of the pointer that was passed. At the end of update, a is lost. When you do this:

a = &temp3;

您更改了 a 的值，但这并不重要，因为无论如何 a 已经消失了.相反，将值分配给它指向的位置，就像您对 b 所做的一样:

You change the value of a, but that doesn't matter, because a is gone after that anyway. Instead, assign the value to where it's pointing at, much like you did with b:

*a = temp3;

你也可以使用引用代替指针:

You could also use references instead of pointers:

void update(int &a, int &b) {
    int temp = a;
    int temp2 = b;
    int temp3 = temp + temp2;
    printf("%d ", temp3);
    b = abs(a - b);
    a = temp3;
}

int main() {
    int a, b;
    scanf("%d %d", &a, &b);
    update(a, b);
    printf("%d\n%d", a, b);
    return 0;
}

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