与 C 指针混淆

Question

 #include<stdio.h>

// Function to swap two values but does not work
void swapDoesNotWork (int *ptrX, int *ptrY);

// Function to swap two values and works fine
void swap (int *ptrX, int *ptrY);

void swap (int *px, int *py) {
 int temp;
 temp = *px;
 *px = *py;
 *py = temp;
}

void swapDoesNotWork (int *px, int *py) {
 printf("\n\n");
 int temp;
 temp = *px;
 px = py; 
 py = &temp;
}

int main() {
 int x = 5;
 int y = 10; 
 swapDoesNotWork(&x, &y);
 printf("++++++++++++++++++\n");
 printf("pre x:%d\n", x); 
 printf("pre y:%d\n", y); 
 printf("\n");
 printf("After calling swapDoesNotWork(&x, &y)...\n");
 printf("post x:%d\n", x); 
 printf("post y:%d\n", y); 
 printf("++++++++++++++++++\n\n");
 x = 5;
 y = 10; 
 printf("= = = = = = = = =\n\n");
 printf("pre x:%d\n", x); 
 printf("pre y:%d\n", y); 
 swap(&x, &y);

 printf("\n");
 printf("After calling swap(&x, &y)...\n");
 printf("post x:%d\n", x); 
 printf("post y:%d\n", y); 
 printf("= = = = = = = = =\n\n");
 return 0;

以上程序编译执行后的输出为:

infi@linux% ./swap_test.o 


++++++++++++++++++
pre x:5
pre y:10

After calling swapDoesNotWork(&x, &y)...
post x:5
post y:10
++++++++++++++++++

= = = = = = = = =
= = = = = = = = =    
pre x:5
pre y:10

After calling swap(&x, &y)...
post x:10
post y:5

可以看出，swapDoesNotWork 函数似乎没有像 swap 函数那样更改值。

我是 C 语言的新手，主要来自脚本背景。有人能帮我解释为什么 swapDoesNotWork 函数没有改变值吗？

Answer 1

以下是 swapDoesNotWork 执行的所有修改:

 temp = ...
 px = ... 
 py = ...

所有这些都是对局部变量的赋值。局部变量在函数返回时被销毁，因此 swapDoesNotWork 没有持久效果。它只会更改即将停止存在的变量。

另一方面，swap 包含这些行:

 *px = ...
 *py = ...

这些是对分别由 px 和 py 指向的位置的赋值。尽管 px 和 py 本身是局部变量，但它们可以指向当前函数之外的变量(在这种情况下，该函数被称为 swap(&x, &y)，因此它们指向 main 中的 x 和 y 变量。