how to access elements of std::vector<int>*

markhod · 08-02-2005, 08:17 AM

Hello,

What I want is to make a pointer to an std::vector, to put something in it and then retrieve the value. All I can find that will compile is:

std::vector<Int_t>* test = new std::vector<Int_t>;
test->push_back(5);
Int_t test5 = test[5000][0];

I dont understand why though or what the first [5000] means (5000 can be any number to make it compile). I cant find any help from google or previous posts. Does anyone know?

I am using gcc 3.2.3 on linux.

Thanks,

Mark

biertrinken · 08-02-2005, 09:00 AM

you have to dereference the test pointer
before using the index-operator

Int_t test5 = (*test)[0];

a std::vector is not the same as a 2 dimensional array !

tobi

markhod · 08-02-2005, 09:03 AM

Quote:

Originally posted by biertrinken
you have to dereference the test pointer
before using the index-operator

Int_t test5 = (*test)[0];

a std::vector is not the same as a 2 dimensional array !

tobi

oh, I see. I had been trying *test[0] and it wouldnt work. I guess brackets will help then!

I still dont see why test[5000][0][ compiles given it isnt a 2D thing though.

Thanks,

Mark

deiussum · 08-02-2005, 10:17 AM

It compiles, but l'm guessing you will segfault when the first element is anything other than 0. Pointers can be dereferenced as though they were an array as well. For instance:

Code:

int *p1 = new int[10];
int *p2 = new int;

*p1 = 10; // OK, same as p1[0] = 10;
p2[0] = 10;  // OK, same as *p2 = 10;

p1[9] = 10; // Ok, since there are 10 elements
p2[9] = 10; // NOT ok, it only points to 1 int not an array

That last line is a segfault waiting to happen...