为什么我们不能用C按引用传递指针？ [英] why can't we pass the pointer by reference in C?

问题描述

我学习C.我写code，当我遇到一个分段故障来创建一个链表。我发现，在<一个解决我的问题href=\"http://stackoverflow.com/questions/2293033/why-is-this-c-linked-list-program-giving-segmentation-fault\">this题。
我试图通过引用传递一个指针。该解决方案说，我们不能这样做。我们有一个指针传递给一个指针。该解决方案为我工作。
不过，我不明白为什么会这样。谁能告诉原因？

I am learning C. I was writing code to create a linked list when i came across a segmentation fault. I found a solution to my problem in this question. I was trying to pass a pointer by reference. The solution says that we can't do so. We have to pass a pointer to a pointer. This solution worked for me. However, I don't understand why is it so. Can anyone tell the reason?

推荐答案

从的 C程序设计语言 - 第二版的（K＆安培; R 2）：

From The C Programming Language - Second Edition (K&R 2):

5.2指针和函数参数

由于C按值传递函数的参数，也没有直接的方法
  对被调用的功能，以改变在调用函数中的变量。

Since C passes arguments to functions by value, there is no direct way for the called function to alter a variable in the calling function.

...

指针参数使函数来访问和更改对象
  调用它的功能。

Pointer arguments enable a function to access and change objects in the function that called it.

如果你明白：

void fn1(int x) {
    x = 5; /* a in main is not affected */
}
void fn2(int *x) {
    *x = 5; /* a in main is affected */
}
int main(void) {
    int a;

    fn1(a);
    fn2(&a);
    return 0;
}

出于同样的原因：

for the same reason:

void fn1(element *x) {
    x = malloc(sizeof(element)); /* a in main is not affected */
}
void fn2(element **x) {
    *x = malloc(sizeof(element)); /* a in main is affected */
}
int main(void) {
    element *a;

    fn1(a);
    fn2(&a);
    return 0;
}

正如你所看到的，有一个与 INT 和一个指向元素没有什么区别，你需要一个指针传递给 INT ，在第二个你需要传递一个指针的指针的元素。

As you can see, there is no difference between an int and a pointer to element, in the first example you need to pass a pointer to int, in the second one you need to pass a pointer to pointer to element.

这篇关于为什么我们不能用C按引用传递指针？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！