c++ User defined conversion - implicit conversion

Question

I have a question about user defined conversion.

class String {
    char* m_data;

public:
    String(): m_data(NULL) {}
    String(const char* cstr): m_data(new char[strlen(cstr)+1]) {
        strcpy(m_data, cstr);
    }
    ~String() {
        delete[] m_data;
    }
    String& operator=(const char* cstr) {
        delete[] m_data;
        m_data = new char[strlen(cstr)+1];
        strcpy(m_data, cstr);
        return *this;
    }
    operator const char*() const { 
        return m_data;
    }
};

While this works:

int main(int argc, char** argv) {
    String a;
    String b;
    a = "aaa";
    b = (const char *)a;
    return 0;
}

This does not:

int main(int argc, char** argv) {
    String a;
    String b;
    a = "aaa";
    b = a;
    return 0;
}

I get double free or corruption runtime error. Valgrind says something about invalid delete.

Why do I have to explicitly typecast it? I thought it would work this way with explicit operator const char*(). Am I doing something wrong?

http://stackoverflow.com/questions/4172722/what-is-the-rule-of-three — David G, Apr 16 '15 at 21:12
Long story short: assignment operator `String& operator=(const String& other)` is missing. — emvee, Apr 16 '15 at 21:14

score 7 · Accepted Answer · edited Apr 16 '15 at 21:26

7

You forgot to define copy assignment operator:

String& operator=(const String& other) {
    if(this != &other) {
        char* new_data = new char[strlen(other.m_data)+1];
        strcpy(new_data, other.m_data);
        delete[] m_data;
        m_data = new_data;
    }
    return *this;
}

Because of that your compiler had to define "default" copy assignment operator, which simple assigns all fields of "other" to a current object:

String& operator=(const String& other) {
    m_data = other.m_data;
    return *this;
}

So you have two pointers to same m_data in a and b and on the exit from main, delete[] will be called twice.

edited Apr 16 '15 at 21:26

T.C.

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answered Apr 16 '15 at 21:14

myaut

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3

Please add the checks to take care of self assignment, which should be a no op. – R Sahu Apr 16 '15 at 21:16
Have you checked that a copy assignment operator is generated when an assignment operator (not copy assignment) is defined. – Cheers and hth. - Alf Apr 16 '15 at 21:16

c++ User defined conversion - implicit conversion

1 Answers1