通过地址，但它在C工作，如呼叫按值？ [英] Passing address, but it is working like call by value in C?

问题描述

您好，我的C语言编程初学者。最近我读到关于呼叫按值和呼叫按地址。我已经了解到，呼叫由被调用函数地址更改反映被调用者。而下面的code不起作用这样的。

Hello I am a beginner in C programming language. Recently I read about call by value and call by address. I have learned that in call by address changes in the called functions reflects the callee. However the following code does not work like that.

int x = 10,y = 20;
void change_by_add(int *ptr) {
    ptr = &y;
    printf("\n Inside change_by_add\t %d",*ptr);
    // here *ptr is printing 20
}

void main(){
    int *p;
    p = &x;
    change_by_add(p);
    printf("\nInside main\t %d", *p);
    // here *p is still pointing to address of x and printing 10
}

当我传递地址，然后按为何调用的函数所做的更改不会反映访问者？

When I am passing address then why the changes made by called function does not reflect caller?

推荐答案

该函数分配一个新的地址指针，但本身是由值传递的指针，因为所有的参数都在C.要改变的值指针变量的指针本身必须通过地址：

The function is assigning a new address to the pointer but the pointer itself is being passed by value, as all arguments are in C. To change the value of a pointer variable the address of the pointer itself must be passed:

void change_by_add(int **ptr)
{
    *ptr = &y;
}

change_by_add(&p);

请参阅Ç常见问题解答问题4.8 。

通过引用传递中不存在C，但可以通过使可变的谁的值被改变为一个函数的地址来实现。例如：

Passing by reference does not exist in C but can be achieved by passing the address of the variable who's value is to be changed to a function. For example:

void add_to_int(int* a_value, int a_increment)
{
    *a_value += a_increment;
}

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