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I am writing some code and have a question.
What is the difference between auto x = vector<int>(); and vector<int> x;? Are they all the same or there's some difference with the complexity?
c++ vector stl initialization
edited 1 hour ago
songyuanyao
94.5k11182250
asked 1 hour ago
AutoratchAutoratch
555
add a comment |
I am writing some code and have a question.
What is the difference between auto x = vector<int>(); and vector<int> x;? Are they all the same or there's some difference with the complexity?
c++ vector stl initialization
edited 1 hour ago
songyuanyao
94.5k11182250
asked 1 hour ago
AutoratchAutoratch
555
2
The first results in a call to a default constructor and a call to a move constructor. The second results in a call to a default constructor. Even if the compiler optimizes both to result in the same assembly, the second one is the one to go for readability.
– DeiDei
1 hour ago
add a comment |
I am writing some code and have a question.
What is the difference between auto x = vector<int>(); and vector<int> x;? Are they all the same or there's some difference with the complexity?
c++ vector stl initialization
edited 1 hour ago
songyuanyao
94.5k11182250
asked 1 hour ago
AutoratchAutoratch
555
I am writing some code and have a question.
What is the difference between auto x = vector<int>(); and vector<int> x;? Are they all the same or there's some difference with the complexity?
c++ vector stl initialization
c++ vector stl initialization
edited 1 hour ago
songyuanyao
94.5k11182250
asked 1 hour ago
AutoratchAutoratch
555
edited 1 hour ago
songyuanyao
94.5k11182250
asked 1 hour ago
AutoratchAutoratch
555
edited 1 hour ago
songyuanyao
94.5k11182250
edited 1 hour ago
songyuanyao
94.5k11182250
edited 1 hour ago
songyuanyao
94.5k11182250
94.5k11182250
asked 1 hour ago
AutoratchAutoratch
555
asked 1 hour ago
AutoratchAutoratch
555
asked 1 hour ago
AutoratchAutoratch
555
555
2
The first results in a call to a default constructor and a call to a move constructor. The second results in a call to a default constructor. Even if the compiler optimizes both to result in the same assembly, the second one is the one to go for readability.
– DeiDei
1 hour ago
add a comment |
2
The first results in a call to a default constructor and a call to a move constructor. The second results in a call to a default constructor. Even if the compiler optimizes both to result in the same assembly, the second one is the one to go for readability.
– DeiDei
1 hour ago
2
2
The first results in a call to a default constructor and a call to a move constructor. The second results in a call to a default constructor. Even if the compiler optimizes both to result in the same assembly, the second one is the one to go for readability.
– DeiDei
1 hour ago
The first results in a call to a default constructor and a call to a move constructor. The second results in a call to a default constructor. Even if the compiler optimizes both to result in the same assembly, the second one is the one to go for readability.
– DeiDei
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
They have the same effect since C++17. Both construct an object named x with type std::vector<int>, which is initialized by the default constructor of std::vector.
Precisely the 1st one is copy initialization, x is copy-initialized from a value-initialized temporary. From C++17 this kind of copy elision is guaranteed, as the result x is initialized by the default constructor of std::vector directly. Before C++17, copy elision is an optimization:
even when it takes place and the copy/move (since C++11) constructor is not called, it still must be present and accessible (as if no optimization happened at all), otherwise the program is ill-formed:
The 2nd one is default initialization, as a class type x is initialized by the default constructor of std::vector.
answered 1 hour ago
songyuanyaosongyuanyao
94.5k11182250
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
They have the same effect since C++17. Both construct an object named x with type std::vector<int>, which is initialized by the default constructor of std::vector.
Precisely the 1st one is copy initialization, x is copy-initialized from a value-initialized temporary. From C++17 this kind of copy elision is guaranteed, as the result x is initialized by the default constructor of std::vector directly. Before C++17, copy elision is an optimization:
even when it takes place and the copy/move (since C++11) constructor is not called, it still must be present and accessible (as if no optimization happened at all), otherwise the program is ill-formed:
The 2nd one is default initialization, as a class type x is initialized by the default constructor of std::vector.
answered 1 hour ago
songyuanyaosongyuanyao
94.5k11182250
add a comment |
They have the same effect since C++17. Both construct an object named x with type std::vector<int>, which is initialized by the default constructor of std::vector.
Precisely the 1st one is copy initialization, x is copy-initialized from a value-initialized temporary. From C++17 this kind of copy elision is guaranteed, as the result x is initialized by the default constructor of std::vector directly. Before C++17, copy elision is an optimization:
even when it takes place and the copy/move (since C++11) constructor is not called, it still must be present and accessible (as if no optimization happened at all), otherwise the program is ill-formed:
The 2nd one is default initialization, as a class type x is initialized by the default constructor of std::vector.
answered 1 hour ago
songyuanyaosongyuanyao
94.5k11182250
add a comment |
They have the same effect since C++17. Both construct an object named x with type std::vector<int>, which is initialized by the default constructor of std::vector.
Precisely the 1st one is copy initialization, x is copy-initialized from a value-initialized temporary. From C++17 this kind of copy elision is guaranteed, as the result x is initialized by the default constructor of std::vector directly. Before C++17, copy elision is an optimization:
even when it takes place and the copy/move (since C++11) constructor is not called, it still must be present and accessible (as if no optimization happened at all), otherwise the program is ill-formed:
The 2nd one is default initialization, as a class type x is initialized by the default constructor of std::vector.
answered 1 hour ago
songyuanyaosongyuanyao
94.5k11182250
They have the same effect since C++17. Both construct an object named x with type std::vector<int>, which is initialized by the default constructor of std::vector.
Precisely the 1st one is copy initialization, x is copy-initialized from a value-initialized temporary. From C++17 this kind of copy elision is guaranteed, as the result x is initialized by the default constructor of std::vector directly. Before C++17, copy elision is an optimization:
even when it takes place and the copy/move (since C++11) constructor is not called, it still must be present and accessible (as if no optimization happened at all), otherwise the program is ill-formed:
The 2nd one is default initialization, as a class type x is initialized by the default constructor of std::vector.
answered 1 hour ago
songyuanyaosongyuanyao
94.5k11182250
edited 1 hour ago
answered 1 hour ago
songyuanyaosongyuanyao
94.5k11182250
answered 1 hour ago
songyuanyaosongyuanyao
94.5k11182250
answered 1 hour ago
songyuanyaosongyuanyao
94.5k11182250
94.5k11182250
add a comment |
add a comment |
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2
The first results in a call to a default constructor and a call to a move constructor. The second results in a call to a default constructor. Even if the compiler optimizes both to result in the same assembly, the second one is the one to go for readability.
– DeiDei
1 hour ago