我正在尝试制作一个模板函数并通过引用将两个变量传递给它,一切听起来都不错,但它从未编译过,错误消息是:-
error C4430: missing type specifier - int assumed. Note: C++ does not support default-int
我试了一小部分代码,它给了我同样的错误,请帮忙吗??
这部分代码,所有其他代码都是这样的:
int size , found = -1 ;
template<class type> Read_Data( type &key , type &arr)
{
cout << " please enter the size of your set \n " ;
cin >> size ;
arr = new type[size];
cout << " please enter the elements of your set : \n " ;
for (int i = 0 ; i <size ; i ++ )
{
cout << " enter element number " << i << ": " ;
cin >> arr[i] ;
}
cout << " please enter the key elemente you want to search for : \n " ;
cin >> key ;
}
void main(void)
{
int key , arr ;
Read_Data (key, arr);
/*after these function there is two other functions one to search for key
in array "arr" and other to print the key on the screen */
}
最佳答案
您只是缺少一些东西来编译代码(解决您看到的错误)。
基本上,在int main()
中,需要在实例化模板时指定类型,如:
Read_Data <int> (key, value);
这样编译器就知道你真正想用什么类型来实例化它。
另请注意,数组的类型应为 int *
。所以在模板函数中,签名必须更改为:
template<class type> Read_Data( type &key , type* &arr)
这是更正后的代码,不会显示您之前看到的错误:
#include<iostream>
using namespace std;
int size , found = -1 ;
template<class type> void Read_Data( type &key , type* &arr)
{
cout << " please enter the size of your set \n " ;
cin >> size ;
arr = new type[size];
cout << " please enter the elements of your set : \n " ;
for (int i = 0 ; i <size ; i ++ )
{
cout << " enter element number " << i << ": " ;
cin >> arr[i] ;
}
cout << " please enter the key elemente you want to search for : \n " ;
cin >> key;
}
int main()
{
int key;
int * arr;
Read_Data<int> (key, arr);
/* after these function there is two other functions one to search for key
* in array "arr" and other to print the key on the screen
*/
}
关于c++ - 通过模板函数 C++ 的引用传递,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/22434783/