How to resolve the "expression must have class type" error in C++
The “expression must have class type” is an error that is thrown when the . operator, which is typically used to access an object’s fields and methods, is used on pointers to objects.
When used on a pointer to an object, the . operator will try to find the pointer’s fields or methods, but it won’t, because they do not exist. These fields or methods are part of an object, rather than a part of the pointer to an object.
Wrong code
The code below tries to apply the . operator on a pointer to an object, rather than on an object itself. The code throws an "expression must have class type" error.
#include <iostream>using namespace std;class Myclass{public:void myfunc() {cout<<"Hello World"<<endl;}};int main(){// A pointer to our class is created.Myclass *a = new Myclass();// myfunc is part of the class object, rather than// part of the pointer to that class' object.a.myfunc();}
Correct code
The code below correctly applies the . operator on an object to access the object’s member functions. Since a is now a class object, the code does not throw any error.
#include <iostream>using namespace std;class Myclass{public:void myfunc() {cout<<"Hello World"<<endl;}};int main(){// A class object of Myclass is created.Myclass a ;// The . operator is used to access the class' methods.a.myfunc();}
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