Does std::string const& parameter copies the passed const char*?

Question

sorry if this is a very noob question. I need to to know if c++ already supported the following behavior:

void sampleFunc(std::string const& p_sVal){
    
    Log("this is your string: " + p_sVal);

}

int main(){
    
    sampleFunc("my string");   //does the param allocates a copy of "my string"...?
    
}

does the param std::string const& p_sVal allocates a copy of the passed const char* or the compiler is smart enough to know that the passed is const char* and can be optimized that it don't need to allocate and copy?

If not, can you suggest example code that achieves this? (Note: aside for string_view solution.)

Answer 1

A temporary std::string is constructed, from the string literal you pass to it, for the std::string const& parameter to refer to.

As suggested in the comment, std::string owns its storage (otherwise it would be std::string_view , wouldn't it?). This observation alone, in my opinion, doesn't really tell you that the temporary std::string will not be able to steal resources from the passed argument.

The resource could be still moved, because you're passing an rvalue to sampleFun , right? No, wrong, because, as you can see here , a literal string, like "my string" in your main , is a lvalue, not an rvalue.

Oh, and even if a string literal was an rvalue, from the documentation on std::string 's (well, std::basic_string 's) constructor you see that there's no other constructor taking a && other then the "move constructor" (ie the moving version of the copy constructor, ie a constructor which takes the same type that it is constructing, ie std::string , not const char* ).

Notice that since you are doing "this is your string: " + p_sVal you do want a std::string to be constructed at some point because you need std::operator+ to work. You could have sampleFunc take by const char* , but then you would have to wrap either p_sVal or the string literal in std::string{…} for + to work.