CODEWITHSUNDEEP
c++listpointers
Arefe
Arefe

Reputation: 12397

size of a list using pointer in C++

How to get size of a list using pointer ? I have the following piece of code. I put few integers inside and later, tried to get the size of the list. 

#include <iostream>
#include <list>

using namespace std;

int main ()

{


list<int>* t;

for (int i=1; i<10; ++i)
{
    t->push_back(i*10);


}

cout<<"size:\t"<<t->size()<<endl;

return 0;
}

Upvotes: 0

Views: 776

Answers (2)

Lightness Races in Orbit
Lightness Races in Orbit

Reputation: 385104

You need to initialise the pointer to actually point somewhere. That could be a dynamically-allocated list using new (in which case you should be using a smart pointer of some description), or it could be some other object that already exists (in which case this is "point"less, isn't it?).

Better yet, don't use a pointer:

list<int> l;
for (int i = 1; i < 10; ++i)
    l.push_back(i*10);
cout << "size:\t" << l.size() << endl;

Upvotes: 7

Alex Telishev
Alex Telishev

Reputation: 2274

You haven't allocated t. Write list<int>* t = new list<int>; or better unique_ptr<list<int>> t (new list<int>); if you are in c++11 . But in this case allocating list on the stack will work better anyway.

Upvotes: 2

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