c++中临时变量不能作为非const的引用参数的方法

#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); //编译错误，把临时变量作为非const的引用参数传递了 g(a + b); //OK，把临时变量作为const&传递是允许的}

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定义的参数是const int&，编译通过。  问题是为什么临时变量作为引用参数传递时，必须是常量引用呢？很多人对此的解释是临时变量是常量，不允许赋值，改动，所以当作为非常量引用传递时，编译器就会报错。这个解释在关于理解临时变量不能作为非const引用参数这个问题上是可以的，但不够准确。事实上，临时变量是可以被作为左值(LValue)并被赋值的，请看下面的代码：#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;  //临时对象作为左值 return 0; }