为什么call-by-value示例不修改输入参数？ [英] Why does call-by-value example not modify input parameter?

问题描述

在下面的call-by-value示例中，我不明白为什么这个代码不会将5的值更改为6.

In the following call-by-value example, I cannot understand why this code is not changing the value of the 5 to a 6.

第11行调用function changeValue的值为6，所以我会认为应该输出6，但仍然输出5。

Line 11 calls the function changeValue which has the value 6, so I would have thought 6 should be output, however 5 is still output?

#include <iostream>
using namespace std;

void changeValue(int value);

int main()
{
  int value = 5;
  changeValue(value);

  cout << "The value is : " << value << "." << endl;

  return 0;
}

void changeValue(int value)
{
  value = 6;
}

// This doesn't change the value from 5 to 6. 5 is output?

推荐答案

当传递函数参数 by value 将对象的副本传递给函数而不是原始对象。无论你明确指定函数的参数，总是通过C / C ++中的值传递。

When you pass a function argument by value a copy of the object gets passed to the function and not the original object.Unless you specify explicitly arguments to functions are always passed by value in C/C++.

您的函数：

void changeValue(int value)

通过值接收参数，简单地是<$ c $>中的 value $ c> main（）创建并传递给函数，该函数对该值操作，而不是中的值 > main（）。

receives the argument by value, in short a copy of value in main() is created and passed to the function, the function operates on that value and not the value in main().

如果您要修改原文，则需要使用 传递参照。

If you want to modify the original then you need to use pass by reference.

void changeValue(int &value)

b $ b

现在，将对原始值的引用（别名）传递给函数，函数对其进行操作，从而反映 main（）。

Now a reference(alias) to the original value is passed to the function and function operates on it, thus reflecting back the changes in main().

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