[C++] Returning values by reference in C++

最新推荐文章于 2025-08-22 17:02:47 发布

转载最新推荐文章于 2025-08-22 17:02:47 发布 · 80 阅读

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原文链接：http://www.cnblogs.com/tianhangzhang/p/4983694.html

文章标签：

#c/c++

本文探讨了在AC++中通过使用引用而非指针来简化程序的方法，包括如何返回引用以及其优势，通过实例展示引用在修改数组元素方面的应用。

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A C++ program can be made easier to read and maintain by using references rather than pointers. A C++ function can return a reference in a similar way as it returns a pointer.

When a function returns a reference, it returns an implicit pointer to its return value. This way, a function can be used on the left side of an assignment statement. For example, consider this simple program:

#include <iostream>
#include <ctime>
 
using namespace std;
 
double vals[] = {10.1, 12.6, 33.1, 24.1, 50.0};
 
double& setValues( int i )
{
  return vals[i];   // return a reference to the ith element
}
 
// main function to call above defined function.
int main ()
{
 
   cout << "Value before change" << endl;
   for ( int i = 0; i < 5; i++ )
   {
       cout << "vals[" << i << "] = ";
       cout << vals[i] << endl;
   }
 
   setValues(1) = 20.23; // change 2nd element
   setValues(3) = 70.8;  // change 4th element
 
   cout << "Value after change" << endl;
   for ( int i = 0; i < 5; i++ )
   {
       cout << "vals[" << i << "] = ";
       cout << vals[i] << endl;
   }
   return 0;
}

When the above code is compiled together and executed, it produces the following result:

Value before change
vals[0] = 10.1
vals[1] = 12.6
vals[2] = 33.1
vals[3] = 24.1
vals[4] = 50
Value after change
vals[0] = 10.1
vals[1] = 20.23
vals[2] = 33.1
vals[3] = 70.8
vals[4] = 50

When returning a reference, be careful that the object being referred to does not go out of scope. So it is not legal to return a reference to local var. But you can always return a reference on a static variable.

int& func() {
   int q;
   //! return q; // Compile time error
   static int x;
   return x;     // Safe, x lives outside this scope
}

转载于:https://www.cnblogs.com/tianhangzhang/p/4983694.html