temporary of type `int'

 

 執行下面的語句你會得到錯誤：

 invalid initialization of non-const reference of type ‘int&’ from a temporary of type ‘int’

 

 

int &z = 12;

 

12 這個值是沒有名字的，所以它是臨時的（temporary），不能將一個temporary的變量賦值給一個引用（&修飾的）類型。

 

但是可以將temporary量賦值給const修飾的引用類型。 這個是爲什麼呢？

 

國外網站對此有激烈討論：

http://stackoverflow.com/questions/8293426/error-invalid-initialization-of-non-const-reference-of-type-int-from-an-rval?rq=1

 

 

#include <stdio.h>
template <class T>
T returnSelf(T &v){
        return v;
}

template <class T>
int compare (const T& v1,const T& v2){
        if(v1 < v2) return -1; 
        if(v1 > v2) return 1;
        return 0;
}
int main(){
        int i = 1;
        float j = 2.0f;
        double k = 3.0;
        printf("i is :%d\n",returnSelf(i));
        printf("j is :%f\n",returnSelf(j));
        printf("k is :%f\n",returnSelf(k));
        compare("hi","88");//"hi" and "hi1" are different styles.
        int ret = compare(i, 10);//can't set temporary value to reference.
        printf("compare i with 10, the result is :%d",ret);
}
~