C++ const char* to std::string &

Question

I have a code that compiles, and it looks like:

    void test1(const std::string &name)................
    test1("myName");

But, if I remove the const from the declaration as:

    void test2(std::string &name)................
    test2("myName");

The code just doesn't compile.

Why the const std::string makes it compile? As far as I understand the same implicit constructor will be called for the char * to be converted to a std::string.

I am using Visual Studio 2013, in case this might be related to the answer.

@AhmedAkhtar yeah, sorry, I am adapting my code to be brief here and I missed this detail, thanks! — Michel Feinstein, Feb 10 '16 at 04:16
Another way to solve the compiler error is to make the function call in the second case as `test1(std::string("myName"));`. — ubuntugod, Feb 10 '16 at 04:45

score 5 · Accepted Answer · answered Feb 10 '16 at 04:10

5

When the implicit conversion happens, you get a temporary std::string (technically, the important thing is that it is an xvalue). You cannot bind a temporary to a lvalue reference std::string & but you can bind to a const lvalue reference const std::string & or an rvalue reference std::string &&.

The same thing happens here:

const int &x = 5;
int &x = 5; // error

answered Feb 10 '16 at 04:10

Dietrich Epp

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I have heard about lvalue before...but not xvalue or rvalue...would you mind to elaborate more on the difference between them? – Michel Feinstein Feb 10 '16 at 04:19
@mFeinstein: http://stackoverflow.com/questions/3601602/what-are-rvalues-lvalues-xvalues-glvalues-and-prvalues – Dietrich Epp Feb 10 '16 at 05:30

score 2 · Answer 2 · answered Feb 10 '16 at 04:09

2

You get an error in second case because a temporary can not bind to a non-const reference.

test("myName"); tries to bind temporary constructed object to a lvalue in second case, hence an error

answered Feb 10 '16 at 04:09

P0W

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score 2 · Answer 3 · answered Feb 10 '16 at 04:14

"myName" is a c-style string and it has the type of const char[]. Since it is not a std::string it needs to be converted to one. When that happens a temporary string is created. That temporary string cannot be bound to a reference but it can be bound to a const reference as it will extend its lifetime to the end of the expression.

You could also pass the string by rvalue reference like void foo(std::string && bar). This will move the temporary string into the function and give you the same effect.

A third option is to pass the string by value void foo(std::string bar). This works with temporaries and non temporaries and will wither make a copy or a move depending on the source of the string.

@mfeinstein you should read this: http://herbsutter.com/2008/01/01/gotw-88-a-candidate-for-the-most-important-const/ — NathanOliver, Feb 10 '16 at 04:27

score 0 · Answer 4 · answered Feb 10 '16 at 04:09

0

The problem is with the & you are using for reference. When calling the function with a string literal you cannot pass a non-const reference.

answered Feb 10 '16 at 04:09

Ahmed Akhtar

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C++ const char* to std::string &

4 Answers4