Mixing
Delayed instantiation in C++
I'm trying to create a variable without instantiating the object.
In Python it would look like this:
graph = mymap[c] if c in mymap else Graph()
So I basically check if I already have that graph otherwise I create it. My understanding is that if I declare the following in C++ it will call the constructor and be wasteful.
Graph g;
So I'm trying to use pointers to avoid this "waste":
Graph* g;
if (graphs.find(c) == graphs.end()){
g = new Graph();
graphs[c] = *g;
} else {
g = &(graphs[c]);
}
std::cout << g << std::endl;
std::cout << &(graphs[c]) << std::endl;
The problem is that the addresses printed in the end do not match. In fact, my tests show that something weird is happening like a new instance of Graph is being created every time.
What am I doing wrong?
c++
asked Nov 20 '18 at 22:33
RafaelRafael
345310
add a comment |
I'm trying to create a variable without instantiating the object.
In Python it would look like this:
graph = mymap[c] if c in mymap else Graph()
So I basically check if I already have that graph otherwise I create it. My understanding is that if I declare the following in C++ it will call the constructor and be wasteful.
Graph g;
So I'm trying to use pointers to avoid this "waste":
Graph* g;
if (graphs.find(c) == graphs.end()){
g = new Graph();
graphs[c] = *g;
} else {
g = &(graphs[c]);
}
std::cout << g << std::endl;
std::cout << &(graphs[c]) << std::endl;
The problem is that the addresses printed in the end do not match. In fact, my tests show that something weird is happening like a new instance of Graph is being created every time.
What am I doing wrong?
c++
asked Nov 20 '18 at 22:33
RafaelRafael
345310
3
Maps already do this. You are allocating an object, then dereferencing it and copying it inside your map, leaking the allocation. Basically not doing what you think it does.
– Matthieu Brucher
Nov 20 '18 at 22:36
1
Could you complete the example, showing the type ofcand ofgraphs(the latter presumably anstd::maporstd::unordered_map?
– jwimberley
Nov 20 '18 at 22:36
1
There are better solutions, as detailed in 1+ answers, but to clarify:graphs[c] = *gis not doing what you think it is.
– jwimberley
Nov 20 '18 at 22:39
add a comment |
I'm trying to create a variable without instantiating the object.
In Python it would look like this:
graph = mymap[c] if c in mymap else Graph()
So I basically check if I already have that graph otherwise I create it. My understanding is that if I declare the following in C++ it will call the constructor and be wasteful.
Graph g;
So I'm trying to use pointers to avoid this "waste":
Graph* g;
if (graphs.find(c) == graphs.end()){
g = new Graph();
graphs[c] = *g;
} else {
g = &(graphs[c]);
}
std::cout << g << std::endl;
std::cout << &(graphs[c]) << std::endl;
The problem is that the addresses printed in the end do not match. In fact, my tests show that something weird is happening like a new instance of Graph is being created every time.
What am I doing wrong?
c++
asked Nov 20 '18 at 22:33
RafaelRafael
345310
I'm trying to create a variable without instantiating the object.
In Python it would look like this:
graph = mymap[c] if c in mymap else Graph()
So I basically check if I already have that graph otherwise I create it. My understanding is that if I declare the following in C++ it will call the constructor and be wasteful.
Graph g;
So I'm trying to use pointers to avoid this "waste":
Graph* g;
if (graphs.find(c) == graphs.end()){
g = new Graph();
graphs[c] = *g;
} else {
g = &(graphs[c]);
}
std::cout << g << std::endl;
std::cout << &(graphs[c]) << std::endl;
The problem is that the addresses printed in the end do not match. In fact, my tests show that something weird is happening like a new instance of Graph is being created every time.
What am I doing wrong?
c++
c++
asked Nov 20 '18 at 22:33
RafaelRafael
345310
asked Nov 20 '18 at 22:33
RafaelRafael
345310
asked Nov 20 '18 at 22:33
RafaelRafael
345310
asked Nov 20 '18 at 22:33
RafaelRafael
345310
asked Nov 20 '18 at 22:33
RafaelRafael
345310
345310
3
Maps already do this. You are allocating an object, then dereferencing it and copying it inside your map, leaking the allocation. Basically not doing what you think it does.
– Matthieu Brucher
Nov 20 '18 at 22:36
1
Could you complete the example, showing the type ofcand ofgraphs(the latter presumably anstd::maporstd::unordered_map?
– jwimberley
Nov 20 '18 at 22:36
1
There are better solutions, as detailed in 1+ answers, but to clarify:graphs[c] = *gis not doing what you think it is.
– jwimberley
Nov 20 '18 at 22:39
add a comment |
3
Maps already do this. You are allocating an object, then dereferencing it and copying it inside your map, leaking the allocation. Basically not doing what you think it does.
– Matthieu Brucher
Nov 20 '18 at 22:36
1
Could you complete the example, showing the type ofcand ofgraphs(the latter presumably anstd::maporstd::unordered_map?
– jwimberley
Nov 20 '18 at 22:36
1
There are better solutions, as detailed in 1+ answers, but to clarify:graphs[c] = *gis not doing what you think it is.
– jwimberley
Nov 20 '18 at 22:39
3
3
Maps already do this. You are allocating an object, then dereferencing it and copying it inside your map, leaking the allocation. Basically not doing what you think it does.
– Matthieu Brucher
Nov 20 '18 at 22:36
Maps already do this. You are allocating an object, then dereferencing it and copying it inside your map, leaking the allocation. Basically not doing what you think it does.
– Matthieu Brucher
Nov 20 '18 at 22:36
1
1
Could you complete the example, showing the type of
c and of graphs (the latter presumably an std::map or std::unordered_map?– jwimberley
Nov 20 '18 at 22:36
Could you complete the example, showing the type of
c and of graphs (the latter presumably an std::map or std::unordered_map?– jwimberley
Nov 20 '18 at 22:36
1
1
There are better solutions, as detailed in 1+ answers, but to clarify:
graphs[c] = *g is not doing what you think it is.– jwimberley
Nov 20 '18 at 22:39
There are better solutions, as detailed in 1+ answers, but to clarify:
graphs[c] = *g is not doing what you think it is.– jwimberley
Nov 20 '18 at 22:39
add a comment |
2 Answers
2
active
oldest
votes
You actually don't have to do anything fancy here. You can simply do
Graph* g = &graphs[c];
The reason is that map::operator has the following behavior:
Returns a reference to the value that is mapped to a key equivalent to key, performing an insertion if such key does not already exist.
So if the key has an associated value already, you'll get a reference to that object. If that key does not yet exist, a value will be default constructed and inserted for you, and a reference to that new object will be returned.
answered Nov 20 '18 at 22:36
CoryKramerCoryKramer
74k1188141
Even better, it could just be Graph& g = graphs[c], where graphs callsGraph's default constructor, and no memory management is needed.
– jwimberley
Nov 20 '18 at 22:37
1
That is exactly what my answer is stating, the only difference is whether you return a reference or a pointer, the allocation and construction is identical to what you demonstrated in your comment
– CoryKramer
Nov 20 '18 at 22:38
Yeah, I get that, I was just making a comment to point out the extra niceties your suggestion provides, since the code in the OP's question is leaky...
– jwimberley
Nov 20 '18 at 22:41
Can I just doGraph g = graphs[c];? Would that be equivalent to the Python code? I started messing with pointers because I thought it was the only way.
– Rafael
Nov 20 '18 at 22:45
2
Graph g = graphs[c];would copy. Probably not what you're after.
– user4581301
Nov 20 '18 at 22:46
add a comment |
Another answer already says the right way to do what you're asking, but I'll address (ha!) why the addresses don't match:
On this line:
graphs[c] = *g;
You're creating a COPY of the graph you just made with new that g points to, and that copy has a different address. You're then printing the address of the original and the address of the copy, so obviously they're at different addresses.
graphs[c] = ...
This code already does the allocation and construction of a (default) Graph object to be stored inside the map when there isn't one mapped to c already (which you already checked for).. the assignment after that just makes the Graph object it just created look like the one that you had previously created with the call to new.
answered Nov 20 '18 at 22:42
xaxxonxaxxon
14.4k43060
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You actually don't have to do anything fancy here. You can simply do
Graph* g = &graphs[c];
The reason is that map::operator has the following behavior:
Returns a reference to the value that is mapped to a key equivalent to key, performing an insertion if such key does not already exist.
So if the key has an associated value already, you'll get a reference to that object. If that key does not yet exist, a value will be default constructed and inserted for you, and a reference to that new object will be returned.
answered Nov 20 '18 at 22:36
CoryKramerCoryKramer
74k1188141
Even better, it could just be Graph& g = graphs[c], where graphs callsGraph's default constructor, and no memory management is needed.
– jwimberley
Nov 20 '18 at 22:37
1
That is exactly what my answer is stating, the only difference is whether you return a reference or a pointer, the allocation and construction is identical to what you demonstrated in your comment
– CoryKramer
Nov 20 '18 at 22:38
Yeah, I get that, I was just making a comment to point out the extra niceties your suggestion provides, since the code in the OP's question is leaky...
– jwimberley
Nov 20 '18 at 22:41
Can I just doGraph g = graphs[c];? Would that be equivalent to the Python code? I started messing with pointers because I thought it was the only way.
– Rafael
Nov 20 '18 at 22:45
2
Graph g = graphs[c];would copy. Probably not what you're after.
– user4581301
Nov 20 '18 at 22:46
add a comment |
You actually don't have to do anything fancy here. You can simply do
Graph* g = &graphs[c];
The reason is that map::operator has the following behavior:
Returns a reference to the value that is mapped to a key equivalent to key, performing an insertion if such key does not already exist.
So if the key has an associated value already, you'll get a reference to that object. If that key does not yet exist, a value will be default constructed and inserted for you, and a reference to that new object will be returned.
answered Nov 20 '18 at 22:36
CoryKramerCoryKramer
74k1188141
Even better, it could just be Graph& g = graphs[c], where graphs callsGraph's default constructor, and no memory management is needed.
– jwimberley
Nov 20 '18 at 22:37
1
That is exactly what my answer is stating, the only difference is whether you return a reference or a pointer, the allocation and construction is identical to what you demonstrated in your comment
– CoryKramer
Nov 20 '18 at 22:38
Yeah, I get that, I was just making a comment to point out the extra niceties your suggestion provides, since the code in the OP's question is leaky...
– jwimberley
Nov 20 '18 at 22:41
Can I just doGraph g = graphs[c];? Would that be equivalent to the Python code? I started messing with pointers because I thought it was the only way.
– Rafael
Nov 20 '18 at 22:45
2
Graph g = graphs[c];would copy. Probably not what you're after.
– user4581301
Nov 20 '18 at 22:46
add a comment |
You actually don't have to do anything fancy here. You can simply do
Graph* g = &graphs[c];
The reason is that map::operator has the following behavior:
Returns a reference to the value that is mapped to a key equivalent to key, performing an insertion if such key does not already exist.
So if the key has an associated value already, you'll get a reference to that object. If that key does not yet exist, a value will be default constructed and inserted for you, and a reference to that new object will be returned.
answered Nov 20 '18 at 22:36
CoryKramerCoryKramer
74k1188141
You actually don't have to do anything fancy here. You can simply do
Graph* g = &graphs[c];
The reason is that map::operator has the following behavior:
Returns a reference to the value that is mapped to a key equivalent to key, performing an insertion if such key does not already exist.
So if the key has an associated value already, you'll get a reference to that object. If that key does not yet exist, a value will be default constructed and inserted for you, and a reference to that new object will be returned.
answered Nov 20 '18 at 22:36
CoryKramerCoryKramer
74k1188141
answered Nov 20 '18 at 22:36
CoryKramerCoryKramer
74k1188141
answered Nov 20 '18 at 22:36
CoryKramerCoryKramer
74k1188141
answered Nov 20 '18 at 22:36
CoryKramerCoryKramer
74k1188141
74k1188141
Even better, it could just be Graph& g = graphs[c], where graphs callsGraph's default constructor, and no memory management is needed.
– jwimberley
Nov 20 '18 at 22:37
1
That is exactly what my answer is stating, the only difference is whether you return a reference or a pointer, the allocation and construction is identical to what you demonstrated in your comment
– CoryKramer
Nov 20 '18 at 22:38
Yeah, I get that, I was just making a comment to point out the extra niceties your suggestion provides, since the code in the OP's question is leaky...
– jwimberley
Nov 20 '18 at 22:41
Can I just doGraph g = graphs[c];? Would that be equivalent to the Python code? I started messing with pointers because I thought it was the only way.
– Rafael
Nov 20 '18 at 22:45
2
Graph g = graphs[c];would copy. Probably not what you're after.
– user4581301
Nov 20 '18 at 22:46
add a comment |
Even better, it could just be Graph& g = graphs[c], where graphs callsGraph's default constructor, and no memory management is needed.
– jwimberley
Nov 20 '18 at 22:37
1
That is exactly what my answer is stating, the only difference is whether you return a reference or a pointer, the allocation and construction is identical to what you demonstrated in your comment
– CoryKramer
Nov 20 '18 at 22:38
Yeah, I get that, I was just making a comment to point out the extra niceties your suggestion provides, since the code in the OP's question is leaky...
– jwimberley
Nov 20 '18 at 22:41
Can I just doGraph g = graphs[c];? Would that be equivalent to the Python code? I started messing with pointers because I thought it was the only way.
– Rafael
Nov 20 '18 at 22:45
2
Graph g = graphs[c];would copy. Probably not what you're after.
– user4581301
Nov 20 '18 at 22:46
Even better, it could just be Graph& g = graphs[c], where graphs calls
Graph's default constructor, and no memory management is needed.– jwimberley
Nov 20 '18 at 22:37
Even better, it could just be Graph& g = graphs[c], where graphs calls
Graph's default constructor, and no memory management is needed.– jwimberley
Nov 20 '18 at 22:37
1
1
That is exactly what my answer is stating, the only difference is whether you return a reference or a pointer, the allocation and construction is identical to what you demonstrated in your comment
– CoryKramer
Nov 20 '18 at 22:38
That is exactly what my answer is stating, the only difference is whether you return a reference or a pointer, the allocation and construction is identical to what you demonstrated in your comment
– CoryKramer
Nov 20 '18 at 22:38
Yeah, I get that, I was just making a comment to point out the extra niceties your suggestion provides, since the code in the OP's question is leaky...
– jwimberley
Nov 20 '18 at 22:41
Yeah, I get that, I was just making a comment to point out the extra niceties your suggestion provides, since the code in the OP's question is leaky...
– jwimberley
Nov 20 '18 at 22:41
Can I just do
Graph g = graphs[c];? Would that be equivalent to the Python code? I started messing with pointers because I thought it was the only way.– Rafael
Nov 20 '18 at 22:45
Can I just do
Graph g = graphs[c];? Would that be equivalent to the Python code? I started messing with pointers because I thought it was the only way.– Rafael
Nov 20 '18 at 22:45
2
2
Graph g = graphs[c]; would copy. Probably not what you're after.– user4581301
Nov 20 '18 at 22:46
Graph g = graphs[c]; would copy. Probably not what you're after.– user4581301
Nov 20 '18 at 22:46
add a comment |
Another answer already says the right way to do what you're asking, but I'll address (ha!) why the addresses don't match:
On this line:
graphs[c] = *g;
You're creating a COPY of the graph you just made with new that g points to, and that copy has a different address. You're then printing the address of the original and the address of the copy, so obviously they're at different addresses.
graphs[c] = ...
This code already does the allocation and construction of a (default) Graph object to be stored inside the map when there isn't one mapped to c already (which you already checked for).. the assignment after that just makes the Graph object it just created look like the one that you had previously created with the call to new.
answered Nov 20 '18 at 22:42
xaxxonxaxxon
14.4k43060
add a comment |
Another answer already says the right way to do what you're asking, but I'll address (ha!) why the addresses don't match:
On this line:
graphs[c] = *g;
You're creating a COPY of the graph you just made with new that g points to, and that copy has a different address. You're then printing the address of the original and the address of the copy, so obviously they're at different addresses.
graphs[c] = ...
This code already does the allocation and construction of a (default) Graph object to be stored inside the map when there isn't one mapped to c already (which you already checked for).. the assignment after that just makes the Graph object it just created look like the one that you had previously created with the call to new.
answered Nov 20 '18 at 22:42
xaxxonxaxxon
14.4k43060
add a comment |
Another answer already says the right way to do what you're asking, but I'll address (ha!) why the addresses don't match:
On this line:
graphs[c] = *g;
You're creating a COPY of the graph you just made with new that g points to, and that copy has a different address. You're then printing the address of the original and the address of the copy, so obviously they're at different addresses.
graphs[c] = ...
This code already does the allocation and construction of a (default) Graph object to be stored inside the map when there isn't one mapped to c already (which you already checked for).. the assignment after that just makes the Graph object it just created look like the one that you had previously created with the call to new.
answered Nov 20 '18 at 22:42
xaxxonxaxxon
14.4k43060
Another answer already says the right way to do what you're asking, but I'll address (ha!) why the addresses don't match:
On this line:
graphs[c] = *g;
You're creating a COPY of the graph you just made with new that g points to, and that copy has a different address. You're then printing the address of the original and the address of the copy, so obviously they're at different addresses.
graphs[c] = ...
This code already does the allocation and construction of a (default) Graph object to be stored inside the map when there isn't one mapped to c already (which you already checked for).. the assignment after that just makes the Graph object it just created look like the one that you had previously created with the call to new.
answered Nov 20 '18 at 22:42
xaxxonxaxxon
14.4k43060
edited Nov 20 '18 at 22:57
answered Nov 20 '18 at 22:42
xaxxonxaxxon
14.4k43060
answered Nov 20 '18 at 22:42
xaxxonxaxxon
14.4k43060
answered Nov 20 '18 at 22:42
xaxxonxaxxon
14.4k43060
14.4k43060
add a comment |
add a comment |
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3
Maps already do this. You are allocating an object, then dereferencing it and copying it inside your map, leaking the allocation. Basically not doing what you think it does.
– Matthieu Brucher
Nov 20 '18 at 22:36
1
Could you complete the example, showing the type of
cand ofgraphs(the latter presumably anstd::maporstd::unordered_map?– jwimberley
Nov 20 '18 at 22:36
1
There are better solutions, as detailed in 1+ answers, but to clarify:
graphs[c] = *gis not doing what you think it is.– jwimberley
Nov 20 '18 at 22:39