better ways to use a for loop:

[jocdrew21]

#include <iostream>
#include <ctime>
#include <vector>
#include <random>

struct MyvecSort {
    bool operator () (int x, int y) {return x<y;};
}SortingVector;

int main(int argc, const char * argv[])
{
    std::vector<int>MyVector;
    
    for (int i=0; i<3600; ++i)
    {
        
    std::random_device rd;
        //asking as if they are two dice
    std::uniform_int_distribution<int> uniform_dist (2,12); //producing random #'s 2-12
    std::mt19937 e1(rd());
    
    int num = uniform_dist(e1);
        
        MyVector.push_back(num);  //pushing 3600 random #'s to vector
        
    }
    
    std::sort(MyVector.begin(), MyVector.end(), SortingVector); //sorting 3600 random numbers
    size_t limit = MyVector.size();
    
    for (int i=0; i<limit; ++i) {
        int count =1;
        while (i<limit && MyVector[i]==MyVector[i+1] ) {
        count++;
        i++;
    }
        std::cout<<MyVector.at(i)<<"\t"<<count<<"\n";
        
    }

    
    return 0;
}

This code is going to look strikingly familiar to the one I just made on counting numbers from a file. However now I am just producing random one each time.

I am just having a hard time thinking of different ways to do this. White Flags made me see that a for loop can be used is several other ways when he wrote:

void AddressBook::cleanup()
{
    nodePtr next;
    for (nodePtr walk = head; walk != NULL; walk = next) {
        next = walk->next;
        delete walk;
    }

}

Am I facing the same situation here? Am I able to ditch the while loop and just use a for loop? Now I know I have been told place i++ into my while is bad programming practice. However I am wrestling with the idea of how to check the number and send the count to the proper number. Therefore I have this above.

[whiteflags]

I really wasn't using a for loop in a "better" or special way. I am simply used to writing for loops where I can. I assigned a variable called next to walk in the update step of a for loop header. When you delete a linked list, you need to save the next link and assign it as the current node eventually, anyway. In a while loop, of course this would look a little different, but only slightly, since the exact same statement moved from the body to the header of a for loop.

It is the same for any other algorithm. I might write your code one way, you will write it a different way, and we both ultimately express the same idea.

set count to 0
take the first element of the vector, call it value
for each item in the vector:
  if item == value:
    increase count by 1
  else:
    print count
    set count to 1
    set value to item
end for
print count

These are probably the basic steps any implementation will take. I would go as far to say that any logically equivalent implementation is as good as something that I would do.

There are also alternatives to counting everything up after you sort, but I didn't want to be the one to spoil your fun. Still, in the interest of sharing knowledge: there is a solution that works for a range of values and requires space proportional to the size of that range.

[jocdrew21]

There are also alternatives to counting everything up after you sort, but I didn't want to be the one to spoil your fun. Still, in the interest of sharing knowledge: there is a solution that works for a range of values and requires space proportional to the size of that range.

If I would heading in the direction of hash maps would I be wrong? I am about to start learning about them soon, however I have read a little about it.

I really wasn't using a for loop in a "better" or special way.

I just never seen it done like that and a little light bulb went off when I seen it. I suppose I was looking for another one.

[whiteflags]

If I would heading in the direction of hash maps would I be wrong? I am about to start learning about them soon, however I have read a little about it.

A hash map would be overkill: a normal array would work.

[jocdrew21]

Ahhh yes an array I see it now.... Compare the array against the sorted numbers, if they are the, same count the location in the array.

[tabstop]

Ahhh yes an array I see it now.... Compare the array against the sorted numbers, if they are the, same count the location in the array.

If I'm thinking what whiteflags is thinking, then "sorted" is already doing too much work.

You're doing just the same as all my statistics students: they insist on going through the list one time for every class and counting as they go, and if I'm lucky they sort it first so they only have to go through the list twice. But there's not really a point, when you're creating a frequency list, to go through the list more than once (which leaves sorting right out, since you have to go through the list once (at least!) to sort and then again to count). Just count the numbers as you come across them! In fact, for what you're doing here, storing the numbers at all is just a waste of space. "count[num]++" in place of the push_back and you're good to go.