In this code (using C++17), will "abc" be used to create a std::string temporary object and then copied into letters by push_back, or will it be moved ?
Is it more efficient to use push_back or emplace_back in this circumstance?

#include <iostream>
#include <string>
#include <vector>
 
int main()
{
    std::vector<std::string> letters;
    letters.push_back("abc");
}

In this code (using C++17), will "abc" be used to create a std::string temporary object and then copied into letters by push_back, or will it be moved ?
Is it more efficient to use push_back or emplace_back in this circumstance?

#include <iostream>
#include <string>
#include <vector>
 
int main()
{
    std::vector<std::string> letters;
    letters.push_back("abc");
}

• c++
• string
• vector
• copy
• move

Share Improve this question Follow edited Nov 18, 2024 at 11:56 wohlstad 30.4k1717 gold badges6161 silver badges9494 bronze badges asked Nov 15, 2024 at 19:32 codeologycodeology 35733 silver badges1212 bronze badges 3

• 1 Literal constant strings are really arrays of (constant) characters, and usually stored in memory that is not writeable. This array itself can't be moved. But the temporary std::string object that is created in the call can be moved, but that string is independent of the literal string you pass as an argument. With that said, emplace_back is probably better anyway. – Some programmer dude Commented Nov 15, 2024 at 19:33
• The foundation issue here is that a char-string literal needs to be converted to std::string before it can be put into the std::vector, because you declared a std::vector<std::string>. This conversion will take place first, then the new std::string will be placed into the std::vector. – Thomas Matthews Commented Nov 15, 2024 at 21:08
• Beware sloppiness with the word "it". In your title, you ask "will [the const char*] be copied or moved?" while the question itself asks "will [the temporary object] be moved?" Those are two rather different questions. (For the former, moving a pointer is the same as copying it, so you probably meant the pointed-to data instead of the pointer. However, the copy of the data occurs when the temporary is created, so still before what is asked in the question's body.) – JaMiT Commented Nov 16, 2024 at 6:31

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1 Answer 1

Sorted by: Reset to default 5

vector::push_back() has an overload that accepts an rvalue, so the temporary std::string will be moved and not copied.

If you use vector::emplace_back() instead, the temporary std::string will be avoided altogether.