Thread: multiplication tables using for statements

I'm having severe difficulty creating a multiplication table. I need to create a multiplication table that looks like this by multiplying i*j

1*1= 1 1* 2=2 1* 3=3 1* 4=4 1* 5=5
2*1= 2 2* 2=4 2* 3=6 2* 4=8 2* 5=10
3*1= 3 3* 2=6 3* 3=9 3 * 4=12 3* 5=15
4*1= 4 4* 2=8 4*3=12 4* 4=16 4* 5=20
5 1= 5 5* 2=10 5*3=15 5* 4=20 5* 5=25
This is what I have. I can get a table of i*j I just need to figure out how to put a for statement that could increment itself 5 times for j and start over and then have i increment by 1.

Code:

#include <stdio.h>
#include "conio.h"

const int num_rows = 5;
const int num_columns = 5;

int main(void)
{
	int row, column, i, j;
	
	printf("i*j\t", '*');
	for (i= 1; i<=5; i++)
		if (i <= 5)
			break;
	printf( "%d", i);
	
	for (column = 1; column <num_columns; column++)
		printf("i*j\t", column);
	putchar('\n');
	for (i= 1; i<=5; i++)
		if (i <= 5)
			break;,md
	
	
	
	for (row =1; row <num_rows; row++)
	{
		printf("i*j\t", row);

				for(column =1; column < num_columns; column++)
		{
			printf("i*j\t", row *column);
		}
		printf("\n");
	}
	_getch();
	return 0;
}

Thanks everyone.

wow your code is very long, and not vary readable, here is what I would do:

Code:

    int i;
    for(i=0; i<25; i++){
        printf(" %-1d*%-1d=%-2d %c", (i/5)+1, (i%5)+1, ((i/5)+1)*((i%5)+1), (i%5-4)?' ':'\n');
    }

Code:

const int num_rows = 5;

it is better practice to use #define 5 N

Code:

printf("i*j\t", column);
printf("i*j\t", row *column);

I don't know what u expect this to do.

Code:

break;,md

or this

Originally Posted by Salem

It's a lot more readable than that one-line cryptic horror.

hhhhhhhhh cryptic horror, just wanted to show that it doesn't have to be so long.

So should I use the const int for number of rows and columns or can I do this with declaring i and j as int? I'm kinda lost on how I can do that though, I know that I should be able to use two "for" loops but I have no idea how to go about doing this. If I could program this right it would be due to serendipity because I don't really understand these for statements, especially in this application. Thanks everyone!

This is what I have so far

Code:

#include <stdio.h>
#include "conio.h"



const int num_rows = 5;
const int num_columns = 5;

int main(void)
{
	int row, column, i,j result;
	
	result = i * j ;
	for ( i=1; row < num_rows; row++) //for (row =1; row <num_rows; row++)
	
	{
	
		printf("%d*%d=%d\t", i,j, result);
		printf("\n");
		for (j =1; j < num_columns; j++)
		
		{
			printf("%d*%d=%d\t", i, j, result);
			printf("\n");
		}
		printf("\n");
	}

I was hoping changing the for statements and the printf statements around would fix the problem but it hasn't. It keeps telling me i and j are being used without being initialized but I can't figure out why Thanks everyone.

Yay I sort of got it to work but now the formatting is off and I can't fix it with \n.

Code:

#include <stdio.h>
#include "conio.h"

int main(void)
{
int i, j;

for (i = 1; i < 6; i++)
	{	printf("\t");
	
for (j = 1; j < 6; j++)	
	printf("\t");

printf("%d*%d = %d\t", i, j, (i * j));

}
}

_getch(); 
return 0;
}

or \t

You are basically there. The loops are still a little overworked. Here is one possible solution:

Code:

#include <stdio.h>
#include "conio.h"

#define NUM_ROWS 5
#define NUM_COLUMNS 5

int main(void)
{
	for (int column = 1; column < NUM_COLUMNS+1; column++){
		for (int row = 1; row < NUM_ROWS+1; row++){	
			printf("%d*%d = %d \t", row, column, (row * column)); 
                                                //A space is between %d and \t
		}
		printf("\n");
	}

	_getch();	
	return 0;
}