how C++ Implicitly convert c style string to a string object?

Question

string s = "abc";

The above statement will first invoke string ( const char * s ) constructor, then invoke copy constructor according to What are the differences in string initialization in C++? . Here is the question: how C++ know it should invoke string ( const char * s ) to convert literal string "abc" to a temporary string object?

Note: copy constructor won't be invoked in copy initialization.

It runs overload resolution on all of the converting constructors of the destination type std::string (and any non-explicit conversion functions of the source type, if it's a class type). — T.C.
– T.C., Commented Feb 23, 2016 at 4:16

R Sahu · Accepted Answer · 2016-02-23 04:18:02Z

3

Initializing an object by using the syntax

string s = "abc";

is called copy initialization.

There are several scenarios where that is legal initialization. In all cases the RHS must be convertible to a string for it to work.

One way a string literal can be converted to a string is through the constructor of string that takes a char const*. That is called a user defined conversion.

answered Feb 23, 2016 at 4:18

R Sahu

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