Beginner at C++ Stuck on Homework

OK I have a 2dimensional array. I need to be able to search through the array and tell the user how many times each number appears in the array. My hack job code is as follows:

Code:

---

//this program counts the number of times A number occurs In an array

#include <iostream.h>

void main()
{
        short numbers[5][3] =  {{2, 3, 5},
                                                        {6, 3, 5},
                                                        {7, 2, 1},
                                                        {4, 5, 9},
                                                        {3, 5, 1}};
 
       
        short tmpNum = 1; //Number being seeked will increase by 1 Until done
        short tmpCount[] = {0}; //Number of times it has been found (counter)
        short xCt = 0;



        for(tmpNum = 1; tmpNum <10; tmpNum  = tmpNum +1) //Each Number               
//I need A way To look into the array For Each number from 1-9





//Display Results

 cout << "Number " << tmpNum << " appears " << tmpCount[xCt] << " times. " << endl;


} //end of main Function

'Code improved by vBulletin Tool (Save as...)

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Please Help, I'm way confused!

Use 2 for loops (one within the other) to loop through each dimension of the array.

Sorry, I am REAL NEW....Can you give me a push in the right direction?

Well err... hmm... I really don't know how to make it clearer without just doing the whole thing...

You know how to do VB right? I get the impression from your profile and the fact that you have 120 posts yet you rarely post in the C++ forum that you must do. Well how would you do it in VB? It's not a difficult problem, and I don't want to just give you the answer because that's not the point. If you're having problems converting VB code to C++ then I can help you with syntax and the like. Why don't you write it in VB first and then convert it?

Alright how's this:

dim number(5,3)
dim Find as integer
dim foundCount as integer
dim Y as Integer, X as Integer

find = 1
do until find = 10
call Go See Y
find = find + 1
Loop


______________________________________
Frivate Function Go See Y()
Do until X = 3

If number(Y,X) = Find then
findCount = FindCount +1
Y = Y +1
End if
If Y = 5 then
x = x + 1
Y = 1
end if
Loop
End Sub
__________________________________________

OK now let's have it!

Something like this maybe:

Code:

---

#include <iostream.h>

int main(int argc, char* argv[])
{
        short numbers[5][3] =  {{2, 3, 5},
                                                        {6, 3, 5},
                                                        {7, 2, 1},
                                                        {4, 5, 9},
                                                        {3, 5, 1}};
int count = 0; //counter
for(int i=0;i<5;i++)
{
        for(int y=0;y<3;y++)
        {
                if (numbers[i][y] == 3 //the number that is searched in the array)
                        count++;
        }
}
cout<<count<<endl;
        return 0;
}

---

Here is the code you need:

Code:

---

#include <iostream.h>

int main(int argc, char* argv[])
{
        short numbers[5][3] =  {{2, 3, 5},
                                                        {6, 3, 5},
                                                        {7, 2, 1},
                                                        {4, 5, 9},
                                                        {3, 5, 1}};
int count[9];

for(int nro = 0;nro<10;nro++)
{
        count[nro] = 0;
        for(int i=0;i<5;i++)
        {
                for(int y=0;y<3;y++)
                {               
                       
                        if (numbers[i][y] == nro)
                        {
                                ++count[nro];
                        }
                }
        }
}

for(int i = 0;i<10;i++)
cout<<i<<"--"<<count[i]<<endl;
//prints info how many times each number appears in the array
        return 0;
}

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Well it looks like the question is answered then :rolleyes:

Ooh...Harry looks bitter :eek:

Bitter, me? nahh, I'm listening to Slayer at the moment, how could I be bitter? ;)

Thanks guys

I don't know why C++ confuses the daylights out of me.

But it does.


Is Slayer still around?

Yep, still alive and very much kicking :) They toured with Sepultura and System of a Down fairly recently I think... like within the last year or two.