Function Objects Set in C++
So basically I want to create a set of Function objects.
In python if we do:
def func():
print "a"
a = func
b = func
fset = set()
fset.insert(a)
fset.insert(b)
In this case fset will have only one function since both a and b are same in python.
But in C++, if I create function objects for same function both a and b will be two different objects of a set. Is there any way that two objects of same function be same?
In C++:
void func(){
cout << "a";
}
function<void()> a = bind(func);
function<void()> b = bind(func);
Now I want if a or its pointer is already present in the set, b should not be added.
c++
asked Nov 23 '18 at 5:20
250250
211212
|
show 4 more comments
So basically I want to create a set of Function objects.
In python if we do:
def func():
print "a"
a = func
b = func
fset = set()
fset.insert(a)
fset.insert(b)
In this case fset will have only one function since both a and b are same in python.
But in C++, if I create function objects for same function both a and b will be two different objects of a set. Is there any way that two objects of same function be same?
In C++:
void func(){
cout << "a";
}
function<void()> a = bind(func);
function<void()> b = bind(func);
Now I want if a or its pointer is already present in the set, b should not be added.
c++
asked Nov 23 '18 at 5:20
250250
211212
1
aandbare essentially two pointers to the same object, not two instances of the same class. You can havestd::set<std::function<void()>*>, or possibly a smart pointer - it'll behave in a similar manner.
– Igor Tandetnik
Nov 23 '18 at 5:22
But in this case also every time creating a pointer of function<void()> a = function(); function<void()> b = function(); Both a and b will still have different pointers right?
– 250
Nov 23 '18 at 5:23
2
A set, by definition, contains values distinct from one another. It's not clear what you're talking about here, when you suggest that you are putting the "same function" into a set as two different objects. Perhaps you should back up your question with some C++ code to illustrate your problem.
– paddy
Nov 23 '18 at 5:25
You say "creating a pointer", but your code doesn't actually create any pointers. It's also not clear what you mean byfunction()- that doesn't appear syntactically valid and likely won't compile.
– Igor Tandetnik
Nov 23 '18 at 5:25
@IgorTandetnik they are talking about your comment which storesstd::function<void()>*
– paddy
Nov 23 '18 at 5:26
|
show 4 more comments
So basically I want to create a set of Function objects.
In python if we do:
def func():
print "a"
a = func
b = func
fset = set()
fset.insert(a)
fset.insert(b)
In this case fset will have only one function since both a and b are same in python.
But in C++, if I create function objects for same function both a and b will be two different objects of a set. Is there any way that two objects of same function be same?
In C++:
void func(){
cout << "a";
}
function<void()> a = bind(func);
function<void()> b = bind(func);
Now I want if a or its pointer is already present in the set, b should not be added.
c++
asked Nov 23 '18 at 5:20
250250
211212
So basically I want to create a set of Function objects.
In python if we do:
def func():
print "a"
a = func
b = func
fset = set()
fset.insert(a)
fset.insert(b)
In this case fset will have only one function since both a and b are same in python.
But in C++, if I create function objects for same function both a and b will be two different objects of a set. Is there any way that two objects of same function be same?
In C++:
void func(){
cout << "a";
}
function<void()> a = bind(func);
function<void()> b = bind(func);
Now I want if a or its pointer is already present in the set, b should not be added.
c++
c++
asked Nov 23 '18 at 5:20
250250
211212
asked Nov 23 '18 at 5:20
250250
211212
edited Nov 23 '18 at 5:32
250
asked Nov 23 '18 at 5:20
250250
211212
asked Nov 23 '18 at 5:20
250250
211212
asked Nov 23 '18 at 5:20
250250
211212
211212
1
aandbare essentially two pointers to the same object, not two instances of the same class. You can havestd::set<std::function<void()>*>, or possibly a smart pointer - it'll behave in a similar manner.
– Igor Tandetnik
Nov 23 '18 at 5:22
But in this case also every time creating a pointer of function<void()> a = function(); function<void()> b = function(); Both a and b will still have different pointers right?
– 250
Nov 23 '18 at 5:23
2
A set, by definition, contains values distinct from one another. It's not clear what you're talking about here, when you suggest that you are putting the "same function" into a set as two different objects. Perhaps you should back up your question with some C++ code to illustrate your problem.
– paddy
Nov 23 '18 at 5:25
You say "creating a pointer", but your code doesn't actually create any pointers. It's also not clear what you mean byfunction()- that doesn't appear syntactically valid and likely won't compile.
– Igor Tandetnik
Nov 23 '18 at 5:25
@IgorTandetnik they are talking about your comment which storesstd::function<void()>*
– paddy
Nov 23 '18 at 5:26
|
show 4 more comments
1
aandbare essentially two pointers to the same object, not two instances of the same class. You can havestd::set<std::function<void()>*>, or possibly a smart pointer - it'll behave in a similar manner.
– Igor Tandetnik
Nov 23 '18 at 5:22
But in this case also every time creating a pointer of function<void()> a = function(); function<void()> b = function(); Both a and b will still have different pointers right?
– 250
Nov 23 '18 at 5:23
2
A set, by definition, contains values distinct from one another. It's not clear what you're talking about here, when you suggest that you are putting the "same function" into a set as two different objects. Perhaps you should back up your question with some C++ code to illustrate your problem.
– paddy
Nov 23 '18 at 5:25
You say "creating a pointer", but your code doesn't actually create any pointers. It's also not clear what you mean byfunction()- that doesn't appear syntactically valid and likely won't compile.
– Igor Tandetnik
Nov 23 '18 at 5:25
@IgorTandetnik they are talking about your comment which storesstd::function<void()>*
– paddy
Nov 23 '18 at 5:26
1
1
a and b are essentially two pointers to the same object, not two instances of the same class. You can have std::set<std::function<void()>*>, or possibly a smart pointer - it'll behave in a similar manner.– Igor Tandetnik
Nov 23 '18 at 5:22
a and b are essentially two pointers to the same object, not two instances of the same class. You can have std::set<std::function<void()>*>, or possibly a smart pointer - it'll behave in a similar manner.– Igor Tandetnik
Nov 23 '18 at 5:22
But in this case also every time creating a pointer of function<void()> a = function(); function<void()> b = function(); Both a and b will still have different pointers right?
– 250
Nov 23 '18 at 5:23
But in this case also every time creating a pointer of function<void()> a = function(); function<void()> b = function(); Both a and b will still have different pointers right?
– 250
Nov 23 '18 at 5:23
2
2
A set, by definition, contains values distinct from one another. It's not clear what you're talking about here, when you suggest that you are putting the "same function" into a set as two different objects. Perhaps you should back up your question with some C++ code to illustrate your problem.
– paddy
Nov 23 '18 at 5:25
A set, by definition, contains values distinct from one another. It's not clear what you're talking about here, when you suggest that you are putting the "same function" into a set as two different objects. Perhaps you should back up your question with some C++ code to illustrate your problem.
– paddy
Nov 23 '18 at 5:25
You say "creating a pointer", but your code doesn't actually create any pointers. It's also not clear what you mean by
function() - that doesn't appear syntactically valid and likely won't compile.– Igor Tandetnik
Nov 23 '18 at 5:25
You say "creating a pointer", but your code doesn't actually create any pointers. It's also not clear what you mean by
function() - that doesn't appear syntactically valid and likely won't compile.– Igor Tandetnik
Nov 23 '18 at 5:25
@IgorTandetnik they are talking about your comment which stores
std::function<void()>*– paddy
Nov 23 '18 at 5:26
@IgorTandetnik they are talking about your comment which stores
std::function<void()>*– paddy
Nov 23 '18 at 5:26
|
show 4 more comments
1 Answer
1
active
oldest
votes
If you have only void functions(or all functions have the same signature), use simply C type function pointers as std::sets's template type.
This will work and as a plus no type erasure overheads of std::function.
void func() {}
void func2() {}
using fPtrType = void(*)(); // convenience type
int main()
{
std::set<fPtrType> fset;
fPtrType a = func;
fPtrType b = func;
fset.emplace(a);
fset.emplace(b);
fset.emplace(func2);
std::cout << fset.size(); // prints 2
return 0;
}
answered Nov 23 '18 at 5:46
JeJoJeJo
4,3573725
Will this also work if both a and b have a different scope?
– 250
Nov 23 '18 at 5:59
1
@250 Did you mean this.
– JeJo
Nov 23 '18 at 6:21
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If you have only void functions(or all functions have the same signature), use simply C type function pointers as std::sets's template type.
This will work and as a plus no type erasure overheads of std::function.
void func() {}
void func2() {}
using fPtrType = void(*)(); // convenience type
int main()
{
std::set<fPtrType> fset;
fPtrType a = func;
fPtrType b = func;
fset.emplace(a);
fset.emplace(b);
fset.emplace(func2);
std::cout << fset.size(); // prints 2
return 0;
}
answered Nov 23 '18 at 5:46
JeJoJeJo
4,3573725
Will this also work if both a and b have a different scope?
– 250
Nov 23 '18 at 5:59
1
@250 Did you mean this.
– JeJo
Nov 23 '18 at 6:21
add a comment |
If you have only void functions(or all functions have the same signature), use simply C type function pointers as std::sets's template type.
This will work and as a plus no type erasure overheads of std::function.
void func() {}
void func2() {}
using fPtrType = void(*)(); // convenience type
int main()
{
std::set<fPtrType> fset;
fPtrType a = func;
fPtrType b = func;
fset.emplace(a);
fset.emplace(b);
fset.emplace(func2);
std::cout << fset.size(); // prints 2
return 0;
}
answered Nov 23 '18 at 5:46
JeJoJeJo
4,3573725
Will this also work if both a and b have a different scope?
– 250
Nov 23 '18 at 5:59
1
@250 Did you mean this.
– JeJo
Nov 23 '18 at 6:21
add a comment |
If you have only void functions(or all functions have the same signature), use simply C type function pointers as std::sets's template type.
This will work and as a plus no type erasure overheads of std::function.
void func() {}
void func2() {}
using fPtrType = void(*)(); // convenience type
int main()
{
std::set<fPtrType> fset;
fPtrType a = func;
fPtrType b = func;
fset.emplace(a);
fset.emplace(b);
fset.emplace(func2);
std::cout << fset.size(); // prints 2
return 0;
}
answered Nov 23 '18 at 5:46
JeJoJeJo
4,3573725
If you have only void functions(or all functions have the same signature), use simply C type function pointers as std::sets's template type.
This will work and as a plus no type erasure overheads of std::function.
void func() {}
void func2() {}
using fPtrType = void(*)(); // convenience type
int main()
{
std::set<fPtrType> fset;
fPtrType a = func;
fPtrType b = func;
fset.emplace(a);
fset.emplace(b);
fset.emplace(func2);
std::cout << fset.size(); // prints 2
return 0;
}
answered Nov 23 '18 at 5:46
JeJoJeJo
4,3573725
answered Nov 23 '18 at 5:46
JeJoJeJo
4,3573725
answered Nov 23 '18 at 5:46
JeJoJeJo
4,3573725
answered Nov 23 '18 at 5:46
JeJoJeJo
4,3573725
4,3573725
Will this also work if both a and b have a different scope?
– 250
Nov 23 '18 at 5:59
1
@250 Did you mean this.
– JeJo
Nov 23 '18 at 6:21
add a comment |
Will this also work if both a and b have a different scope?
– 250
Nov 23 '18 at 5:59
1
@250 Did you mean this.
– JeJo
Nov 23 '18 at 6:21
Will this also work if both a and b have a different scope?
– 250
Nov 23 '18 at 5:59
Will this also work if both a and b have a different scope?
– 250
Nov 23 '18 at 5:59
1
1
@250 Did you mean this.
– JeJo
Nov 23 '18 at 6:21
@250 Did you mean this.
– JeJo
Nov 23 '18 at 6:21
add a comment |
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1
aandbare essentially two pointers to the same object, not two instances of the same class. You can havestd::set<std::function<void()>*>, or possibly a smart pointer - it'll behave in a similar manner.– Igor Tandetnik
Nov 23 '18 at 5:22
But in this case also every time creating a pointer of function<void()> a = function(); function<void()> b = function(); Both a and b will still have different pointers right?
– 250
Nov 23 '18 at 5:23
2
A set, by definition, contains values distinct from one another. It's not clear what you're talking about here, when you suggest that you are putting the "same function" into a set as two different objects. Perhaps you should back up your question with some C++ code to illustrate your problem.
– paddy
Nov 23 '18 at 5:25
You say "creating a pointer", but your code doesn't actually create any pointers. It's also not clear what you mean by
function()- that doesn't appear syntactically valid and likely won't compile.– Igor Tandetnik
Nov 23 '18 at 5:25
@IgorTandetnik they are talking about your comment which stores
std::function<void()>*– paddy
Nov 23 '18 at 5:26