Interesting effect with tr().arg()
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I discussed in the past this and also the season why you should only use
arg(QString)and not the other overloads -
@ChrisVT Hi,Friend,Welcome.
You can write some test code, to find why. It is interesting.
Do not omit Intermediate results.
You will clear well from following snippet:
#include <QCoreApplication> #include <QtCore/QDebug> int main(int argc, char *argv[]) { QCoreApplication a(argc, argv); qDebug() << QObject::tr("%1%2%3"); /// %1%2%3 qDebug() << QObject::tr("%1%2%3").arg("A"); /// A%2%3 qDebug() << QObject::tr("%2%3").arg("A"); ///A%3 qDebug() << QObject::tr("%1%2%3").arg("A").arg("%2A"); /// A%2A%3 qDebug() << QObject::tr("%1%2%3").arg("A").arg("%2A").arg("B"); /// ABA%3 // To replace First %, is %1=>A, result is tr(A%2%3).arg(%2A).arg(B) // In Result To replace First %, is %2=>%2, result is (A%2A%3).arg(B) // In Result To replace First %, is %2=>B, result is ABA%3 return a.exec(); }@joeQ
I still don't get how this behaviour is anything to do with usingQObject::tr(), in your examples and this thread title? I can't test because I don't use C++, but wouldn't your examples behave just the same if you usedQStringin place ofQObject::tr? Thearg()s are evaluated after thetr(), not before, aren't they? -
Further to what I have written above, and to @VRonin 's post:
ISTM that you should never use
qString.arg().arg(). The chainedargs, which work on the substituted result from the previousarg()calls, are far too dangerous. [You wouldn't use C'ssprint(format)with a non-literal format string for the same kind of reason, would you?]Like with
sprintf, or repeatedreplaces or regular expression substitutions, you can only afford to use a substituter which works from a single, original string specifying what to do, not one affected by a previous substitution. You need a function which does multiple substitutions at the same time, not one after the other, to be safe.So I don't see/where is Qt's
QString::argList()function, so you can go like:QString("%1%2%3")::argList(arg1, arg2, arg3), which is what you need to do? -
@ChrisVT
Firstly, I can't see how this can have anything to do withtr()function. As per http://doc.qt.io/qt-5/qobject.html#tr,QObject::tr()simply returns aQString, only then does it evaluate the.arg()s, so the whole thing is simplyQString::arg()calls, forget abouttr().- Check what plain
tr("%1%2%3")returns? - Check what
QString("%1%2%3").arg("A").arg("%2A").arg("B")returns?
- Check what plain
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Further to what I have written above, and to @VRonin 's post:
ISTM that you should never use
qString.arg().arg(). The chainedargs, which work on the substituted result from the previousarg()calls, are far too dangerous. [You wouldn't use C'ssprint(format)with a non-literal format string for the same kind of reason, would you?]Like with
sprintf, or repeatedreplaces or regular expression substitutions, you can only afford to use a substituter which works from a single, original string specifying what to do, not one affected by a previous substitution. You need a function which does multiple substitutions at the same time, not one after the other, to be safe.So I don't see/where is Qt's
QString::argList()function, so you can go like:QString("%1%2%3")::argList(arg1, arg2, arg3), which is what you need to do?@JonB said in Interesting effect with tr().arg():
You need a function which does multiple substitutions at the same time, not one after the other, to be safe.
ππππππππ
@JonB said in Interesting effect with tr().arg():
which is what you need to do?
Actually it's very easy to solve the problem at hand:
tr("%1%2%3").arg(QStringLiteral("A"),QStringLiteral("%2A"),QStringLiteral("B"))"La mort n'est rien, mais vivre vaincu et sans gloire, c'est mourir tous les jours"
~Napoleon BonaparteOn a crusade to banish setIndexWidget() from the holy land of Qt
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@JonB said in Interesting effect with tr().arg():
You need a function which does multiple substitutions at the same time, not one after the other, to be safe.
ππππππππ
@JonB said in Interesting effect with tr().arg():
which is what you need to do?
Actually it's very easy to solve the problem at hand:
tr("%1%2%3").arg(QStringLiteral("A"),QStringLiteral("%2A"),QStringLiteral("B")) -
Yes, up to 9: http://doc.qt.io/qt-5/qstring.html#arg-21
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I discussed in the past this and also the season why you should only use
arg(QString)and not the other overloadsThanks for all the responses - very insightful!
@VRonin This also happens with QStrings:
qDebug() << QString("%1%2%3").arg("A").arg("%2A").arg("B"); -> "ABA%3"
qDebug() << QString("%1%2%3").arg(QString("A")).arg(QString("%2A")).arg(QString("B")); -> "ABA%3"
But yes, implicit type conversion can be very hard to figure out. Been there!@joeB Good advice about using the arglist instead, but
(a) this only works for QStrings, not for numerical formats, and
(b) Qt offers the ability to use chained args which MOSTLY work.
I was incorrectly assuming that arg(a).arg(b) should be the same as .arg(a,b) for constant QStrings a and b. I'll still say it's not all that intuitive that it isn't. -
Thanks for all the responses - very insightful!
@VRonin This also happens with QStrings:
qDebug() << QString("%1%2%3").arg("A").arg("%2A").arg("B"); -> "ABA%3"
qDebug() << QString("%1%2%3").arg(QString("A")).arg(QString("%2A")).arg(QString("B")); -> "ABA%3"
But yes, implicit type conversion can be very hard to figure out. Been there!@joeB Good advice about using the arglist instead, but
(a) this only works for QStrings, not for numerical formats, and
(b) Qt offers the ability to use chained args which MOSTLY work.
I was incorrectly assuming that arg(a).arg(b) should be the same as .arg(a,b) for constant QStrings a and b. I'll still say it's not all that intuitive that it isn't.@ChrisVT
I love things which "MOSTLY" work :)Be aware, if you are going to use "numerical formats" as you say then, as @VRonin pointed out in the post he referred to, these will not be localed. If you care about that, you must format them to localed strings in the first place, then you can pass them to the list-
arg()because they are now strings.
