C ++ as the return value of the reference
By using references instead of pointers, C ++ makes programs easier to read and maintain. C ++ function can return a reference, and returns a pointer to a similar manner.
When the function returns a reference, it returns a pointer to the implicit return value. Thus, the function can be placed in the left side of an assignment statement. For example, consider the following simple procedure:
#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]; // 返回第 i 个元素的引用 } // 要调用上面定义函数的主函数 int main () { cout << "改变前的值" << endl; for ( int i = 0; i < 5; i++ ) { cout << "vals[" << i << "] = "; cout << vals[i] << endl; } setValues(1) = 20.23; // 改变第 2 个元素 setValues(3) = 70.8; // 改变第 4 个元素 cout << "改变后的值" << endl; for ( int i = 0; i < 5; i++ ) { cout << "vals[" << i << "] = "; cout << vals[i] << endl; } return 0; }
When the above code is compiled and executed, it produces the following results:
改变前的值 vals[0] = 10.1 vals[1] = 12.6 vals[2] = 33.1 vals[3] = 24.1 vals[4] = 50 改变后的值 vals[0] = 10.1 vals[1] = 20.23 vals[2] = 33.1 vals[3] = 70.8 vals[4] = 50
When returning a reference, pay attention to the referenced object can not go out of scope. It returns a reference to a local variable is not legal, however, you can return a reference to a static variable.
int& func() { int q; //! return q; // 在编译时发生错误 static int x; return x; // 安全,x 在函数作用域外依然是有效的 }