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What's the difference between `auto x = vector()` and `vector x`?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
The Ask Question Wizard is Live!
Data science time! April 2019 and salary with experience
Should we burninate the [wrap] tag?What is the difference between #include <filename> and #include “filename”?Regular cast vs. static_cast vs. dynamic_castWhat are the differences between a pointer variable and a reference variable in C++?The Definitive C++ Book Guide and ListDifference between private, public, and protected inheritanceWhat is the difference between const int*, const int * const, and int const *?Why is “using namespace std” considered bad practice?Why are elementwise additions much faster in separate loops than in a combined loop?What is the difference between 'typedef' and 'using' in C++11?Why is it faster to process a sorted array than an unsorted array?
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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
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
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
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
94.5k11182250
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
.
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
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
.
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
.
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
.
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
.
edited 1 hour ago
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