how to conver QString to const char*

[ChrisW67]

I'll quote the Qt docs again with emphasis:

The pointer remains valid as long as the byte array isn't reallocated or destroyed.

and add some from the C++ standards:

$12.2/3- "Temporary objects are destroyed as the last step in evaluating the full-expression (1.9) that (lexically) contains the point where they were created. This is true even if that evaluation ends in throwing an exception."

So, at the statement boundary of:

char *str = s.toAscii().data();

char *str = s.toAscii().data();

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(s is QString) the temporary QByteArray created by the toAscii() method is destroyed. Most of the time immediately subsequent access through str will appear to work because the memory has not yet been allocated elsewhere and modified. You can be sure when it does fail intermittently you will be scratching your head. If you are going to store a pointer to the char data for later use then you must control the scope of the QByteArray by making it explicit in some form (which is what was suggested).

That does not, in any way, prohibit this immediate usage:

printf("%s\n", s.toAscii().constData());

printf("%s\n", s.toAscii().constData());

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which is safe because the scope of the temporary QByteArray is the end of the complete statement.

[high_flyer]

y QByteArray created by the toAscii() method is destroyed.

True, however it is implicitly shared, so the pointer it was pointing at still lives, as long as the original 's' is alive.(unless you have done something to alter the content of the pointer, which is why const is a better choice).

At any rate, my suggestion was wrongly understood.
If you read my original post, you will see the syntax is not correct too.
I just meant that one could use toAscii() by getting the QByteArray from the QString, that is all.
So when you do it, make sure you are not working with the temp instance of it.
But the reaction "YOU SHOULD NEVER..." assertion starts to annoy me, (it gets more and more to be seen around here).
If you should never call a specific function, it should not exist.
If it does, you should call it, in constraints of its usage and when its appropriate - that was my point.
If it is, or not appropriate in a specific case, is another issue.

[ChrisW67]

True, however it is implicitly shared, so the pointer it was pointing at still lives, as long as the original 's' is alive.(unless you have done something to alter the content of the pointer, which is why const is a better choice).

You are retrieving a pointer to a data buffer inside the temporary QByteArray not a pointer to the buffer underlying the QString. The data buffer in the temporary QByteArray is a transformed copy of the data buffer that is in the string's buffer. That buffer is not shared with the string and goes away with the temporary QByteArray.

[high_flyer]

Just to be clear: my point was NOT about using a temp objects.

Back to our discussion:

The data buffer in the temporary QByteArray is a transformed copy of the data buffer that is in the string's buffer.

I just checked and you are right:
toAscii() (via toLatin1()) indeed transforms the data:

QByteArray QString::toLatin1() const
{
    QByteArray ba;
    if (d->size) {
        ba.resize(d->size);
        const ushort *i = d->data;
        const ushort *e = d->data + d->size;
        uchar *s = (uchar*) ba.data();
        while (i != e) {
            *s++ = (*i>0xff) ? '?' : (uchar) *i;
            ++i;
        }
    }
    return ba;
}

QByteArray QString::toLatin1() const
{
    QByteArray ba;
    if (d->size) {
        ba.resize(d->size);
        const ushort *i = d->data;
        const ushort *e = d->data + d->size;
        uchar *s = (uchar*) ba.data();
        while (i != e) {
            *s++ = (*i>0xff) ? '?' : (uchar) *i;
            ++i;
        }
    }
    return ba;
}

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One of the main disadvantages answering such post during work is that I can often just skim the posts. :-\