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QString to char conversion

I have a function (fopen in STL) that gives a char* argument as a path in my computer, but I must use QString in that place so it doesn't work.

How can I convert QString to char* to solve this problem?

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amiref
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2 Answers2

52

See here at How can I convert a QString to char* and vice versa?

In order to convert a QString to a char*, then you first need to get a latin1 representation of the string by calling toLatin1() on it which will return a QByteArray. Then call data() on the QByteArray to get a pointer to the data stored in the byte array. See the documentation:

https://doc.qt.io/qt-5/qstring.html#toLatin1 https://doc.qt.io/qt-5/qbytearray.html#data

See the following example for a demonstration:

int main(int argc, char **argv)
{
 QApplication app(argc, argv);
  QString str1 = "Test";
  QByteArray ba = str1.toLatin1();
  const char *c_str2 = ba.data();
  printf("str2: %s", c_str2);
  return app.exec();
}

Note that it is necessary to store the bytearray before you call data() on it, a call like the following

const char *c_str2 = str2.toLatin1().data();

will make the application crash as the QByteArray has not been stored and hence no longer exists

To convert a char* to a QString you can use the QString constructor that takes a QLatin1String, e.g:

QString string = QString(QLatin1String(c_str2)) ;

See the documentation:

https://doc.qt.io/qt-5/qlatin1string.html

Of course, I discovered there is another way from this previous SO answer:

QString qs;

// Either this if you use UTF-8 anywhere
std::string utf8_text = qs.toUtf8().constData();

// or this if you on Windows :-)
std::string current_locale_text = qs.toLocal8Bit().constData();
feedc0de
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    I think the wording needs to change. The statement `const char *c_str2 = str2.toLatin1().data();` should work fine. Unfortunately after the ';' the temporary QByteArray created by toLatin1() has been destroyed so `c_str2` now has an invalid pointer. Conversely you could use it in a call `doStuff(str2.toLatin1().data());` as the QByteArray is not destroyed until the ';' Thus: `printf("str2: %s", str2.toLatin1().data());` Should be OK. – Martin York Mar 31 '11 at 20:46
  • @Martin: I am just quoting Qt. –  Mar 31 '11 at 20:49
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    This. I have spent half of today debugging a problem caused by the underlying QByteArray being destroyed. It's a real pain to have to store the byte array, but it seems to be necessary. – mpenkov Nov 14 '11 at 11:24
  • Loki's comment seems spot-on. I use someFunction(myQString.toUtf8().constData()); all the time and it works fine, but if you try to use char *myCString = myQString.toUtf8().constData(); and then use myCString, you'll end up with a C string that sometimes works and sometimes doesn't, depending on whether the QByteArray happens to still be valid or not when you access its contents via a C pointer. Definitely need to be careful here! – Vern Jensen Jun 24 '14 at 19:48
  • i was hopping for char* not const char * – user889030 Jul 10 '19 at 12:37
  • Note: [toLocal8Bit](https://doc.qt.io/qt-5/qstring.html#toLocal8Bit) *If this string contains any characters that cannot be encoded in the locale, the returned byte array is undefined. Those characters may be suppressed or replaced by another.* -- it sounds like this is only valid if you know ahead of time that the `QString` contains only valid Latin1 characters. – jrh May 01 '20 at 13:58
3

You could use QFile rather than std::fstream.

QFile           file(qString);

Alternatively convert the QString into a char* as follows:

std::ifstream   file(qString.toLatin1().data());

The QString is in UTF-16 so it is converted toLatin1() here but QString has a couple of different conversions including toUtf8() (check your file-system it may use UTF-8).

As noted by @0A0D above: don't store the char* in a variable without also getting a local copy of the QByteArray.

char const*      fileName = qString.toLatin1().data();
std::ifstream    file(fileName);  // fileName not valid here.

This is because toLatin1() returns an object of QByteArray. As it is not actually bound to a variable it is a temporary that is destroyed at the end of the expression. Thus the call to data() here returns a pointer to an internal structure that no longer exists after the ';'.

Martin York
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