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c++ 模板参数列表中参数 1 的类型/值不匹配

转载 作者:塔克拉玛干 更新时间:2023-11-03 01:19:23 25 4
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#include <iostream>
using namespace std;

template<class T>
class people{
public:
virtual void insert(T item)=0;
virtual T show(T info)=0;
};

template<class T>
class name
{
private:
T fname;
T lname;
public:
name(T first, T last);
// bool operator== (name & p1, name &p2)
};
template <class T>
name<T>::name(T first, T last){
fname = first;
lname = last;
}
template <class T>
class person : public people<T>
{
private:
T a[1];
int size;
public:
person();
virtual void insert(T info);
virtual T show();
};
template<class T>
person<T>::person(){
size = 0;
}
template<class T>
void person<T>::insert(T info){
a[0] =info;
}
template<class T>
T person<T>::show(){
return a[0];
}
int main(){
string first("Julia"), last("Robert");
name<string> temp(first,last);
people<name>* aPerson = new person();
aPerson-> insert(temp);
aPerson->show();
return 0;
}

这些是我不断遇到的错误,我无法确定真正的问题所在:

test.cpp:52: error: type/value mismatch at argument 1 in template parameter list for 'template<class T> class people'
test.cpp:52: error: expected a type, got 'name'
test.cpp:52: error: invalid type in declaration before '=' token
test.cpp:52: error: expected type-specifier before 'person'
test.cpp:52: error: expected ',' or ';' before 'person'
test.cpp:53: error: request for member 'insert' in '* aPerson', which is of non-class type 'int'
test.cpp:54: error: request for member 'show' in '* aPerson', which is of non-class type 'int'

最佳答案

name 是一个模板类,所以你必须指定模板:

people<name<string>>* aPerson = new person<name<string>>();

关于c++ 模板参数列表中参数 1 的类型/值不匹配,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/19877896/

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