Adding empty element to declared container without declaring type of element
When we use a complicated container in C++, like
std::vector<std::map<std::string, std::set<std::string>>> table;
The only way to add an empty map (which may represent a row or column) is to initialize a new element and push it back. For example with
table.push_back(std::map<std::string, std::set<std::string>>());
Is there any way to avoid redeclaring the type, and just adding the correct typed element?
c++ c++11
add a comment |
When we use a complicated container in C++, like
std::vector<std::map<std::string, std::set<std::string>>> table;
The only way to add an empty map (which may represent a row or column) is to initialize a new element and push it back. For example with
table.push_back(std::map<std::string, std::set<std::string>>());
Is there any way to avoid redeclaring the type, and just adding the correct typed element?
c++ c++11
add a comment |
When we use a complicated container in C++, like
std::vector<std::map<std::string, std::set<std::string>>> table;
The only way to add an empty map (which may represent a row or column) is to initialize a new element and push it back. For example with
table.push_back(std::map<std::string, std::set<std::string>>());
Is there any way to avoid redeclaring the type, and just adding the correct typed element?
c++ c++11
When we use a complicated container in C++, like
std::vector<std::map<std::string, std::set<std::string>>> table;
The only way to add an empty map (which may represent a row or column) is to initialize a new element and push it back. For example with
table.push_back(std::map<std::string, std::set<std::string>>());
Is there any way to avoid redeclaring the type, and just adding the correct typed element?
c++ c++11
c++ c++11
edited Mar 25 at 10:41
Mohammad Usman
21.4k134758
21.4k134758
asked Mar 25 at 6:35
VineetVineet
354411
354411
add a comment |
add a comment |
5 Answers
5
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From CLion's IntelliSense, I later found that one useful method is emplace_back()
. This constructs a new object of correct type and adds it to the end of the vector.
table.emplace_back();
add a comment |
You can take advantage of copy-list-initialization (since C++11) and just write
table.push_back({});
add a comment |
Before C++11 sometimes I use x.resize(x.size()+1)
, in C++11 or later you can use x.push_back({})
.
add a comment |
Though the other answers are correct, I will add that if you couldn't take that approach, you could have benefitted from declaring some type aliases to shorten that container type name.
I can of course only guess at the logical meaning of your containers, which is another thing that this fixes!
using PhilosopherNameType = std::string;
using NeighboursType = std::set<PhilosopherNameType>;
using NeighbourMapType = std::map<PhilosopherNameType, NeighboursType>;
std::vector<NeighbourMapType> table;
table.push_back(NeighbourMapType());
I mention this because you can likely still benefit from this in other places.
2
agreed. Along the same lines are the types that the STL gives you (vector::value_type iirc)
– sudo rm -rf slash
Mar 26 at 8:01
add a comment |
As a example:
vector<int> arr;
arr.push_back(1);
arr.push_back(2);
To use it, you should:
include
and also, either use std::vector
in your code or add
using std::vector;
or
using namespace std;
after the #include
line.
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
From CLion's IntelliSense, I later found that one useful method is emplace_back()
. This constructs a new object of correct type and adds it to the end of the vector.
table.emplace_back();
add a comment |
From CLion's IntelliSense, I later found that one useful method is emplace_back()
. This constructs a new object of correct type and adds it to the end of the vector.
table.emplace_back();
add a comment |
From CLion's IntelliSense, I later found that one useful method is emplace_back()
. This constructs a new object of correct type and adds it to the end of the vector.
table.emplace_back();
From CLion's IntelliSense, I later found that one useful method is emplace_back()
. This constructs a new object of correct type and adds it to the end of the vector.
table.emplace_back();
edited Mar 25 at 7:12
answered Mar 25 at 6:51
VineetVineet
354411
354411
add a comment |
add a comment |
You can take advantage of copy-list-initialization (since C++11) and just write
table.push_back({});
add a comment |
You can take advantage of copy-list-initialization (since C++11) and just write
table.push_back({});
add a comment |
You can take advantage of copy-list-initialization (since C++11) and just write
table.push_back({});
You can take advantage of copy-list-initialization (since C++11) and just write
table.push_back({});
edited Mar 25 at 6:46
answered Mar 25 at 6:37
songyuanyaosongyuanyao
93.8k11182249
93.8k11182249
add a comment |
add a comment |
Before C++11 sometimes I use x.resize(x.size()+1)
, in C++11 or later you can use x.push_back({})
.
add a comment |
Before C++11 sometimes I use x.resize(x.size()+1)
, in C++11 or later you can use x.push_back({})
.
add a comment |
Before C++11 sometimes I use x.resize(x.size()+1)
, in C++11 or later you can use x.push_back({})
.
Before C++11 sometimes I use x.resize(x.size()+1)
, in C++11 or later you can use x.push_back({})
.
answered Mar 25 at 6:49
65026502
87.6k13115218
87.6k13115218
add a comment |
add a comment |
Though the other answers are correct, I will add that if you couldn't take that approach, you could have benefitted from declaring some type aliases to shorten that container type name.
I can of course only guess at the logical meaning of your containers, which is another thing that this fixes!
using PhilosopherNameType = std::string;
using NeighboursType = std::set<PhilosopherNameType>;
using NeighbourMapType = std::map<PhilosopherNameType, NeighboursType>;
std::vector<NeighbourMapType> table;
table.push_back(NeighbourMapType());
I mention this because you can likely still benefit from this in other places.
2
agreed. Along the same lines are the types that the STL gives you (vector::value_type iirc)
– sudo rm -rf slash
Mar 26 at 8:01
add a comment |
Though the other answers are correct, I will add that if you couldn't take that approach, you could have benefitted from declaring some type aliases to shorten that container type name.
I can of course only guess at the logical meaning of your containers, which is another thing that this fixes!
using PhilosopherNameType = std::string;
using NeighboursType = std::set<PhilosopherNameType>;
using NeighbourMapType = std::map<PhilosopherNameType, NeighboursType>;
std::vector<NeighbourMapType> table;
table.push_back(NeighbourMapType());
I mention this because you can likely still benefit from this in other places.
2
agreed. Along the same lines are the types that the STL gives you (vector::value_type iirc)
– sudo rm -rf slash
Mar 26 at 8:01
add a comment |
Though the other answers are correct, I will add that if you couldn't take that approach, you could have benefitted from declaring some type aliases to shorten that container type name.
I can of course only guess at the logical meaning of your containers, which is another thing that this fixes!
using PhilosopherNameType = std::string;
using NeighboursType = std::set<PhilosopherNameType>;
using NeighbourMapType = std::map<PhilosopherNameType, NeighboursType>;
std::vector<NeighbourMapType> table;
table.push_back(NeighbourMapType());
I mention this because you can likely still benefit from this in other places.
Though the other answers are correct, I will add that if you couldn't take that approach, you could have benefitted from declaring some type aliases to shorten that container type name.
I can of course only guess at the logical meaning of your containers, which is another thing that this fixes!
using PhilosopherNameType = std::string;
using NeighboursType = std::set<PhilosopherNameType>;
using NeighbourMapType = std::map<PhilosopherNameType, NeighboursType>;
std::vector<NeighbourMapType> table;
table.push_back(NeighbourMapType());
I mention this because you can likely still benefit from this in other places.
answered Mar 25 at 17:12
Lightness Races in OrbitLightness Races in Orbit
294k54477811
294k54477811
2
agreed. Along the same lines are the types that the STL gives you (vector::value_type iirc)
– sudo rm -rf slash
Mar 26 at 8:01
add a comment |
2
agreed. Along the same lines are the types that the STL gives you (vector::value_type iirc)
– sudo rm -rf slash
Mar 26 at 8:01
2
2
agreed. Along the same lines are the types that the STL gives you (vector::value_type iirc)
– sudo rm -rf slash
Mar 26 at 8:01
agreed. Along the same lines are the types that the STL gives you (vector::value_type iirc)
– sudo rm -rf slash
Mar 26 at 8:01
add a comment |
As a example:
vector<int> arr;
arr.push_back(1);
arr.push_back(2);
To use it, you should:
include
and also, either use std::vector
in your code or add
using std::vector;
or
using namespace std;
after the #include
line.
add a comment |
As a example:
vector<int> arr;
arr.push_back(1);
arr.push_back(2);
To use it, you should:
include
and also, either use std::vector
in your code or add
using std::vector;
or
using namespace std;
after the #include
line.
add a comment |
As a example:
vector<int> arr;
arr.push_back(1);
arr.push_back(2);
To use it, you should:
include
and also, either use std::vector
in your code or add
using std::vector;
or
using namespace std;
after the #include
line.
As a example:
vector<int> arr;
arr.push_back(1);
arr.push_back(2);
To use it, you should:
include
and also, either use std::vector
in your code or add
using std::vector;
or
using namespace std;
after the #include
line.
answered yesterday
Mr. SemicolonMr. Semicolon
5301422
5301422
add a comment |
add a comment |
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