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Home/ Questions/Q 296325
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Editorial Team
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Editorial Team
Asked: May 12, 20262026-05-12T06:32:19+00:00 2026-05-12T06:32:19+00:00

When I compiled my C++ code with GCC 4.3 for the first time, (after

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When I compiled my C++ code with GCC 4.3 for the first time, (after having compiled it successfully with no warnings on 4.1, 4.0, 3.4 with the -Wall -Wextra options) I suddenly got a bunch of errors of the form warning: type qualifiers ignored on function return type.

Consider temp.cpp:

class Something
{
public:
    const int getConstThing() const {
        return _cMyInt;
    }
    const int getNonconstThing() const {
        return _myInt;
    }

    const int& getConstReference() const {
        return _myInt;
    }
    int& getNonconstReference() {
        return _myInt;
    }

    void setInt(const int newValue) {
        _myInt = newValue;
    }

    Something() : _cMyInt( 3 ) {
        _myInt = 2;
    }
private:
    const int _cMyInt;
    int _myInt;
};

Running g++ temp.cpp -Wextra -c -o blah.o:

temp.cpp:4: warning: type qualifiers ignored on function return type
temp.cpp:7: warning: type qualifiers ignored on function return type

Can someone tell me what I am doing wrong that violates the C++ standard? I suppose that when returning by value, the leading const is superfluous, but I’m having trouble understanding why it’s necessary to generate a warning with it. Are there other places where I should leave off the const?

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  1. Editorial Team
    Editorial Team
    2026-05-12T06:32:19+00:00Added an answer on May 12, 2026 at 6:32 am

    It doesn’t violate the standard. That’s why they’re warnings and not errors.

    And indeed you’re right — the leading const is superfluous. The compiler warns you because you’ve added code that in other circumstances might mean something, but in this circumstance means nothing, and it wants to make sure you won’t be disappointed later when your return values turn out to be modifiable after all.

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