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formating output to new lines

thebirdman26 (5)
The program i am working on is taking an array of 50 numbers and i need the output to display 10 numbers and then start a new line. What do i need to add to the cout statement to get it to print that way?


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#include <iostream>
#include <conio.h>
#include <iomanip>

using namespace std;
void ArrayTest (int, double[]);
double numbers[50];
int i;

void ArrayTest (int i, double numbers[])
{
	for (i = 0; i < 50; i++)
	{
		if (i <= 24)
			numbers[i] = i * i;
			
		if (i >= 25)
			numbers[i] = i / 2;	
	}
}

int main()
{

	ArrayTest (i, numbers);

	for (i = 0; i < 50; i++)
		cout << numbers[i] << " " << '\n';

	getch();
	return 0;
}

tmoney91 (13)
Are you trying to display 5 rows with 10 numbers on each row? If so, you just have to alter your cout statement. I'm sure there is a way to do this more efficiently but this works. 
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#include <iostream>
#include <conio.h>
#include <iomanip>

using namespace std;
void ArrayTest ();
double numbers[50];
int i;

void ArrayTest ()
{
	for (i = 0; i < 50; i++)
	{
		if (i <= 24)
			numbers[i] = i * i;

		if (i >= 25)
			numbers[i] = i / 2;
	}
}

int main()
{

	ArrayTest ();

	for (i = 0; i < 10; i++)
		cout << numbers[i] << " ";

    cout<< endl;

    for (i = 10; i < 20; i++)
		cout << numbers[i] << " ";

    cout<< endl;

    for (i = 20; i < 30; i++)
		cout << numbers[i] << " ";

    cout<< endl;

    for (i = 30; i < 40; i++)
		cout << numbers[i] << " ";

    cout<< endl;

    for (i = 40; i < 50; i++)
		cout << numbers[i] << " ";

	return 0;
}
thebirdman26 (5)
Thanks! This solves the problem for now. I know there has to be a way to do this more efficiently, but this gets the job done.
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