Building QString out string

Question

With the following piece of code:

QString msg;
msg =   "Datalogging Item"          + ',' +
        QString::number(item)       + ',' +
        "Slave Index"               + ',' +
        QString::number(slaveIndex) + ',' +
        "Slave Name"                + ',' +
        slaveName                   + ',' +
        "Experiment Index"          + ',' +
        QString::number(expIndex)   + ',' +
        "Experiment Name"           + ',' +
        expName                     + ',' +
        "Aquisition Frequency "     + ',' +
        "33 Hz"                     + "\n";
qDebug() << msg;

I was getting the following debug output

"riment Index0,Slave Index,0,Slave Name,Execute 1,Experiment Index,0,Experiment Name,Read All (Barebone),Aquisition Frequency ,33 Hz\n"

Wich is not really what I was expected to get.

But with the code modified with the QString typecast:

QString msg;
msg =   QString("Datalogging Item")     + QString(',') +
        QString::number(item)           + QString(',') +
        QString("Slave Index")          + QString(',') +
        QString::number(slaveIndex)     + QString(',') +
        QString("Slave Name")           + QString(',') +
        slaveName                       + QString(',') +
        QString("Experiment Index")     + QString(',') +
        QString::number(expIndex)       + QString(',') +
        QString("Experiment Name")      + QString(',') +
        expName                         + QString(',') +
        QString("Aquisition Frequency ")+ QString(',') +
        QString("33 Hz")                + QString("\n");
qDebug() << msg;

I get what I expected to get:

"Datalogging Item,0,Slave Index,0,Slave Name,Execute 1,Experiment Index,0,Experiment Name,Read All (Barebone),Aquisition Frequency ,33 Hz\n"

Do you have any clue on what is going on? I just don't get what is going on and I'm a newbie to QT.

Thanks.

It's unclear what you're asking, please add the expected behaviour and the current behaviour in a clear cut manner to your question description. — HopefullyHelpful, Nov 21 '16 at 01:18
He's asking why the first example code is printing garbage characters in the output. Which I attempt to answer below. — selbie, Nov 21 '16 at 02:21
You may want to use QString::arg() instead. That is what I do. — drescherjm, Nov 21 '16 at 02:25

score 3 · Answer 1 · answered Nov 21 '16 at 02:18

You can repro similar results even simpler:

 msg =   "Datalogging Item" + ',' +  QString::number(item);

And now it occurs to be exactly what the issue is. The first part of that expression is simply:

"Datalogging Item" + ','

In the above expression, a raw string literal is having the integer value of 44 (ascii val for comma) added to it's pointer address. Hence, you are now in "undefined" territory since the length of "Datalogging Item" is far less than 44 characters. Most likely what's happening is that the above expression is now a pointer expression into the middle of one of your other string literals defined nearby.

The solution is simply to declare the first element to be a QString so that all subsequent "additions" will use the overloaded + operator of QString:

msg =   QString("Datalogging Item") + ',' +
        QString::number(item)       + ',' +
        ...

All you need to reproduce is `std::cout << "Datalogging Item" + ','` :) No need for Qt. Or also `printf("%s\n", "Datalogging Item" + ',');`. No need for C++ :) — Kuba hasn't forgotten Monica, Nov 21 '16 at 14:03

Kuba hasn't forgotten Monica · Accepted Answer · 2016-11-21T19:42:04.720

The expression

"Datalogging Item" + ','

has the type and meaning of

static_cast<const char *>("Datalogging Item" + 44)

You're taking the address of the beginning of the Datalogging string constant and adding 44 to it - advancing it by 44 characters. This is undefined behavior as you're moving a pointer past the object it points to. By coincidence the pointer happens to point to another C string constant.

A short program that reproduces the issue:

#include <iostream>
int main() {
  std::cout << "Datalogging Item Foo Bar Baz" + '\n' << std::endl;
}

Output:

g Item Foo Bar Baz

One way of rewriting your expression to make it valid and less verbose would be to use the arg-substitution mechanism of QString:

auto msg = QStringLiteral(
  "Datalogging Item,%1,Slave Index,%2,Slave Name,%3,Experiment Index,%4,"
  "Experiment Name,%5,Acquisition Frequency,%6\n")
   .arg(item).arg(slaveIndex).arg(slaveName).arg(expIndex)
   .arg(expName).arg(QStringLiteral("33 Hz"));

If you're using Qt 4, replace QStringLiteral with QString::fromUtf8. The QStringLiteral is a mechanism introduced in Qt 5 that builds up a constant QString instance at compile time, yielding smaller code and better runtime performance than the use of a QString constructor.

Thanks. I get it now. You answer was helpful! – Sebastien Nov 23 '16 at 01:58 — Sebastien, Nov 23 '16 at 01:58

score 0 · Answer 3 · answered Nov 21 '16 at 11:07

0

It is better to use QStringLiteral macro for static string in code.

msg = QStringLiteral("Datalogging Item") + ',' +

More info here: http://doc.qt.io/qt-5/qstring.html#QStringLiteral

answered Nov 21 '16 at 11:07

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Building QString out string

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