Beginner at C++ Stuck on Homework

[Lee M.]

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:

//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...)

Please Help, I'm way confused!

[HarryW]

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

[Lee M.]

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

[HarryW]

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?

[Lee M.]

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!

[Vlatko]

Something like this maybe:

#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;
}

[Vlatko]

Here is the code you need:

#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;
}

[HarryW]

Well it looks like the question is answered then

[parksie]

Ooh...Harry looks bitter

[HarryW]

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

[Lee M.]

Thanks guys

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

But it does.


Is Slayer still around?

[HarryW]

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.