CODEWITHSUNDEEP
c++
user3498780
user3498780

Reputation: 189

error : expression must have class type

I have a problem with the class in C++; it's written like this.

in second source,

void idealtype::compare(idealtype T1)
{
if (height.size() > T1.height.size())
    cout << T1.getname() << " " << T1.getage() << "\t" << T1.getheight() << "\n";
else if (height.size() < T1.height.size())
    cout << getname() << " " << getage() << "\t" << getheight() << "\n";
else if (height.size() == T1.height.size())
{
    cout << T1.getname() << " " << T1.getage() << "\t" << T1.getheight() << "\n";
    cout << getname() << " " << getage() << "\t" << getheight() << "\n";
}
cout << "\n";
}

in header;

class idealtype
{public:
void compare(idealtype);

....

private:
int height;
}

in main source; ....

idealtype A(a,b,c) // c is "height"

....

idealtype B(a,b,c) // c is "height"
B.compare(A)

I think it's all done well, but Visual keeps showing me,

(in second source, on every if() state) error : expression must have class type

So, what's the KEY of this problem?

Plz help me, guys :)

Upvotes: 0

Views: 342

Answers (1)

πάντα ῥεῖ
πάντα ῥεῖ

Reputation: 1

So, what's the KEY of this problem?

In your code you say

if (height.size() > T1.height.size())

class idealtype {
    // ...
private:
    int height; // <<<<<<<<<<<<
};

since height is declared as int it doesn't have any class like methods. That's why the compiler complains.

Upvotes: 2

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