#include <iostream>
using namespace std;

void f(int &a)

 {
 cout << "f(" << a << ") is being called" << endl;
}

void g(const int &a)

{
 cout << "g(" << a << ") is being called" << endl;
}

 int main()

 {
 int a = 3, b = 4;
 f(a + b); //編譯錯(cuò)誤，把臨時(shí)變量作為非const的引用參數(shù)傳遞了
 g(a + b); //OK，把臨時(shí)變量作為const&傳遞是允許的
}

const_ref.cpp: In function `int main()':
const_ref.cpp:14: error: invalid initialization of non-const reference of type '
  int&' from a temporary of type 'int'
const_ref.cpp:4: error: in passing argument 1 of `void f(int&)' 而在g(a+b)中，由于g定義的參數(shù)是const int&，編譯通過。  問題是為什么臨時(shí)變量作為引用參數(shù)傳遞時(shí)，必須是常量引用呢？很多人對(duì)此的解釋是臨時(shí)變量是常量，不允許賦值，改動(dòng)，所以當(dāng)作為非常量引用傳遞時(shí)，編譯器就會(huì)報(bào)錯(cuò)。這個(gè)解釋在關(guān)于理解臨時(shí)變量不能作為非const引用參數(shù)這個(gè)問題上是可以的，但不夠準(zhǔn)確。事實(shí)上，臨時(shí)變量是可以被作為左值(LValue)并被賦值的，請(qǐng)看下面的代碼：
#include  <iostream> 
using namespace std;

class CComplex {  
friend CComplex operator+(const CComplex &cp1, const CComplex &cp2);
friend ostream& operator<<(ostream &os, const CComplex &cp);
private: 
 int x; 
public: 
 CComplex(){}
  
 CComplex(int x1) { 
 x = x1; 
 }
};
 
CComplex operator+(const CComplex &cp1, const CComplex &cp2)

{ 
 CComplex cp3; 
 cp3.x = cp1.x + cp2.x; 
 return cp3; 
} ostream& operator<<(ostream &os, const CComplex &cp)

{
 os << cp.x;
 return os;
}

int main()

{ 
 CComplex a(2), b(3), c(4); 
 cout << (a + b) << endl;
 cout << ((a + b) = c) << endl;  //臨時(shí)對(duì)象作為左值
 return 0; 
}