qDebug() not showing const std::string&

Question

I am trying to use some vector data's name with struct. I am trying to get see which name in qDebug()

To be more clear:

const std::string& testName = "asdfqwer";
qDebug() << testName;

It gives en error message in build:

Error: no match for 'operator<<' in 'qDebug()() << testName'

I don't have options to change const std::string& type. Could you please help me to solve this issue without changing type?

Solution 1

qDebug() does not know anything about std::string but it works with const char*. Appropriate operator you can find here. You can achieve this with data() or with c_str() which is better as Jiří Pospíšil said.

For example:

const std::string& testName = "asdfqwer";
qDebug() << testName.data() << testName.c_str();

Aslo you can convert std::string to QString with QString::fromStdString.

Solution 2

If you need writing std::string to qDebug() often in your code, you can implement this function globally (for example in you main.cpp):

#include <QDebug>
#include <string>

QDebug& operator<<(QDebug& out, const std::string& str)
{
    out << QString::fromStdString(str);
    return out;
}

int main()
{
    std::string jau = "jau";
    qDebug() << jau;
    return 0;
}

Comments

[johngull]

same you can use convertion to the QString with the QString::fromStdString

[goGud]

Amazing.. I didn't know that behaviours of qDebug()

[Jiří Pospíšil]

It'd be better to use c_str() instead of data() as the latter is not required to be null-terminated (until C++11).

[Jablonski]

@johngull Yes, you are right, but I thought that OP use here C++ only types, but I added this to answer, it correct and OP can choose what he wants.

[Jablonski]

@JiříPospíšil I added c_str() to my answer. Thank you.

[johngull]

@Chernobyl I think it is natural to use Qt types if you use qDebug. Impossible to use qDebug without using QtCore, so QString is available for him. Anyway, the main point here that convertion needed.

[smac89]

This makes a copy of the QDebug object. Shouldn't you be returning a reference instead?