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c++qtqbytearray

Why size of QByteArray is `int` rather than `unsigned int`

I have such expressions in my code:

QByteArray idx0 = ...

unsigned short ushortIdx0;

if ( idx0.size() >= sizeof(ushortIdx0) ) {
    // Do something
}

But I'm getting the warning:

warning: comparison between signed and unsigned integer expressions [-Wsign-compare]

if ( idx0.size() >= sizeof(ushortIdx0) ) {
              ~~~~~~~~~~~~^~~~~~~~~~

Why size() of QByteArray is returned as int rather than unsigned int? How can I get rid of this warning safely?

Solution

  • Some folk feel that the introduction of unsigned types into C all those years ago was a bad idea. Such types found themselves introduced into C++, where they are deeply embedded in the C++ Standard Library and operator return types.

    Yes, sizeof must, by the standard, return an unsigned type.

    The Qt developers adopt the modern thinking that the unsigned types were a bad idea, and favour instead making the return type of size a signed type. Personally I find it idiosyncratic.

    To solve, you could (i) live with the warning, (ii) switch it off for the duration of the function, or (iii) write something like

    (std::size_t)idx0.size() >= sizeof(ushortIdx0)

    at the expense of clarity.