Create pointer to pointer in while statement issue - c++

Question

in Visual Studio 2010 i create a while statement in which i assign a pointer to pointer to a map.
Example:
std::map<int,std::tuple<int,std::string>** > dmap;
int i=0;
while (i<3){
std::tuple<int,std::string>* t = new std::tuple<int,std::string>(10+i,std::string("test"));
dmap[i] = &t;
dmap[i + 1 ] = &t;
i++;
}
.
.
.
for (auto it = d.begin();it!=d.end();++it)
{
if(*(it->second) != nullptr){
delete *(it->second);
*(it->second) = nullptr;
}
}
The problem is that the address of &t is always the same and so at the end the map always contains , for all keys that i entered, the last *t value.
What's the problem? (Resolved)
[Edit]
Now i modify the code beacause before it was incomplete, if i want to avoid to delete nullptr i need to have a pointer to pointer. Or not?

Answer 1

The problem is that you're putting a pointer to a local variable t into the map. After each loop, t is destroyed and the pointer is no longer valid.
I've no idea why you're using pointers at all, let alone pointers to pointers. You probably want to put the tuples themselves in the map:
std::map<int,std::tuple<int,std::string>> dmap;
for (int i = 0; i<3; ++i){
dmap[i] = {10+i, "test"};
}

Answer 2

i create a while statement in which i assign a pointer to pointer to a map
Sorry for saying this, but it sounds to me like you have bigger problems than the fact that t is the same (this looks like the xy problem).
Consider (in order) one of these alternatives:
store your tuples by value
store your tuples by single pointer (worse than "by value", better than "by pointer to pointer"). If you can do this, consider declaring your map over std::shared_ptr<std::tuple<...>>)
if you really need a map of pointers to pointers to tuples, consider creating a minimal proxy object that acts like a smart pointer to pointer internally (and manages the allocations for you in a safe manner) and like a regular type from the outside (and redeclare your map accordingly).
Either way, if you really need a map of pointers to pointers to tuples (for some reason), the allocation should be done like this:
std::map<int,std::tuple<int,std::string>**> dmap;
int i=0;
while (i<3) {
*dmap[ i ] = new std::tuple<int,std::string>{10 + i, "test"};
++i;
}
(The way you did it added the address of the same local (stack) variable to the map, which would lead to undefined behavior after you exit the local function).

Answer 3

Why are you interested in std::tuple<int,std::string>** ?
Wouldn't a std::tuple<int,std::string>* be sufficient ?
std::map<int,std::tuple<int,std::string>* > dmap;
int i=0;
while (i<3){
std::tuple<int,std::string>* t = new std::tuple<int,std::string>(10+i,std::string("test"));
dmap[i] = t;
i++;
}

Answer 4

Well, the address of t is always the same, because it is local variable that is stored on your stack. Each time you enter the block, t will be allocated on the same spot (as you're destroying t after you get out of your while body).
Instead, you need to allocate it on the heap (if this is really what you want to do).
std::tuple<int,std::string>** t = new std::tuple<int,std::string>*();
*t = new std::tuple<int,std::string>(10+i,std::string("test"));
dmap[i] = t;
I can't see what you're trying to accomplish, but this would be a better solution:
std::map<int,std::tuple<int,std::string>* > dmap;
int i=0;
while (i<3){
std::tuple<int,std::string>* t = new std::tuple<int,std::string>(10+i,std::string("test"));
dmap[i] = t;
i++;
}
Even better would be to use smart pointer instead raw one.
Even better would be to store objects by value (no pointers at all).