CODEWITHSUNDEEP
c++templatespointersiostreamcout
ranieri
ranieri

Reputation: 2078

cout char* printing address instead of value

In a template class, I overloaded the ostream<< operator with the following:

friend std::ostream& operator<<(std::ostream &out,
        const DataItem<T> &cDataItem) {
    out << cDataItem.m_value;
    return out;
}

And m_value is declared as a pointer to the generic type

T *m_value;

But for some reason, when I cout the object, it prints an hexadecimal value, which I guess is the address and NOT the value, as I expect:

Value: 0x7fff418f9d40

I learned that cout with a char* should print the content of the char*, not the address. Why then is this happening? Is the template a problem?

Upvotes: 3

Views: 2230

Answers (2)

Michał G&#243;rny
Michał G&#243;rny

Reputation: 19233

The problem is that in your case, T is char[40] which is a pointer already (equivalent to char*). This means that if you use T*, it becomes char** and that's what you are trying to print. Since it's a pointer to the pointer, generic template for pointers get applied and the address is printed.

For a test, you may try to print:

friend std::ostream& operator<<(std::ostream &out,
        const DataItem<T> &cDataItem) {
    out << *cDataItem.m_value;
    return out;
}

In this case, you will dereference the pointer to the original value, that is char[40] and you should get the expected output.

Also, I think your operator= is wrong. Since you pass T to it, and assign it to T* m_value, you implicitly store a pointer to a local value. You probably want to make it take T* likewise in the constructor. Or a T& reference in both cases.

Upvotes: -1

David Schwartz
David Schwartz

Reputation: 182761

char * is a special case. For any other pointer, unless you have some specialization you implemented yourself, outputting a pointer outputs that pointer, not what it points to.

Upvotes: 3

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