使用指针模拟“按值传递"函数(C和C ++)中的“按引用传递" [英] Using pointers to emulate Pass-by-Reference in a Pass-by-Value function (C and C++)

问题描述

因此，我试图理解一段简单的代码，该代码使用按值传递"函数中的指针来模拟"按引用传递.该示例是C惯例，因为C中没有通过引用的传递.但是我也很好奇它对C ++的影响.

So I am trying to understand a simple piece of code that 'emulates' Pass-by-reference using pointers in a Pass by value function. The example is given as a C practice since there is no Pass by reference in C. but I am also curious about its effects on C++.

因此，运行此代码后，将交换值:

so after running this code the values are swapped:

void swap(int* a, int* b)
{
  int temp;
  temp = *a;
  *a = *b;
  *b = temp;
}

我试图理解为什么将指针作为参数传递会产生按引用传递效果?这是否意味着通过指针执行的所有操作都具有按引用传递"效果?因此，一旦退出函数，交换的内存位置将不会返回?这在C ++中也有效吗?

I am trying to understand why passing pointers as arguments create a pass-by-reference effect? Does it mean all actions that are performed through pointers have a pass-by-reference effect? So the swapped memory location will not go back once we quit the function? Would this also be valid in C++?

编码新手，所以我希望我能说清楚.

New to coding so I hope I could make myself clear.

推荐答案

基本上，您的代码正在将内存地址复制到函数中，而如果省略了*运算符，则函数将复制 函数的内存地址.这就是每行本质上的意思.

Basically, your code is copying a memory address to the function, while if you omitted the * operator, the function would be copying what's in a memory address to the function. Here's what each line is essentially saying.

// Give me the memory address of a and b.
void swap(int* a, int* b)
{
  int temp;
  // Whatever value is at memory address 'a', copy it to temp.
  temp = *a; 
  // Whatever value is at memory address 'b', copy it to memory address 'a'.
  *a = *b;
  // Copy temp over the value at memory address 'b'.
  *b = temp;
}

是的，在C ++中同样有效.

And yes, it would be equally valid in C++.

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