引用作为函数参数？ [英] References as function arguments?

问题描述

我有参考的麻烦。
考虑这个代码：

  void pseudo_increase（int a）{a ++;} 
 int main {
 int a = 0; 
 // .. 
 pseudo_increase（a）; 
 // .. 
} 
  

这里，变量 a

$ b现在让我们考虑另一个例子：

  void true_increase（int& a）{a ++;} 
 int main（）{
 int a = 0; 
 // .. 
 true_increase（a）; 
 // .. 
} 
  

这里说的是<$ true_increase（a）

c $ c>被调用，将传递 a 的副本。它将是一个不同的变量。因此，& a 将与 a 的真实地址不同。那么 a 的价值是如何增加的？

请考虑以下示例：

 <$ c $ <$> 




解决方案c> int a = 1;
int& b = a;
b = 2; // this will set a to 2
printf（a =％d\\\
，a）; // output：a = 2


这里 b 对于 a 可以像别名一样处理。无论您分配给 b ，也会分配给 a （因为 b 是对 a 的引用）。通过引用传递参数没有什么不同：

  void foo & b）
 {
b = 2; 
} 
 
 int main（）
 {
 int a = 1; 
 foo（a）; 
 printf（a =％d\\\
，a）; // output：a = 2 
 return 0; 
} 
  




I have a trouble with references. Consider this code:

void pseudo_increase(int a){a++;}  
int main(){  
    int a = 0;
    //..
    pseudo_increase(a);
    //..
}

Here, the value of variable a will not increase as a clone or copy of it is passed and not variable itself.
Now let us consider an another example:

void true_increase(int& a){a++;}
int main(){  
    int a = 0;
    //..
    true_increase(a);
    //..
}

Here it is said value of a will increase - but why?

When true_increase(a) is called, a copy of a will be passed. It will be a different variable. Hence &a will be different from true address of a. So how is the value of a increased?

Correct me where I am wrong.

解决方案

Consider the following example:

int a = 1;
int &b = a;
b = 2; // this will set a to 2
printf("a = %d\n", a); //output: a = 2

Here b can be treated like an alias for a. Whatever you assign to b, will be assigned to a as well (because b is a reference to a). Passing a parameter by reference is no different:

void foo(int &b)
{
   b = 2;
}

int main()
{
    int a = 1;
    foo(a);
    printf("a = %d\n", a); //output: a = 2
    return 0;
}

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