Simple C question

Question

I just started to learn C, and one question in book that I'm using is:
Use nested loops to produce the following pattern:

$
$$
$$$
$$$$
$$$$$

And of course I got stuck.

#include <stdio.h>
int main (void)
{
    int i, j;
    char ch = '$';
    for(i = 1; i < 5; i++)
    {
        for(j = 1; j <=5; j++)
           printf("%c", ch);
        printf("\n");
    }
    return 0;
}

Answer 1

Please try this code, If you want to generalized ( means user can give any number for printing the pattern you can take a input from user). like n then replace for(row = 1; row <= 5; row++) to for(row = 1; row <= n; row++).

#include<stdio.h>
    int main()
    {
       int row, col;
       for(row = 1; row <=5; row++)
       {
           for(col = 0; col < row; col++)
              printf("$");
           printf("\n");
       }
       return 0;
    }

Answer 2

#include <stdio.h>
int main (void)
{
    for (int i = 1; i <= 6; i++)
    {
        for (int j = 1; j <= 6; j++)
        {
            if (i - j >= 0)
                printf("$");
        }
        printf("\n");
    }
 return 0;
}

Answer 3

First some basic about loops is that - When outer loop execute one time then inner loop will complete its whole iteration. in your case - for(i = 1; i < 5; i++) // outer loop for each changing value of i such as i= 1,2,3,4 Inner loop for(j = 1; j <=5; j++) // will complete its whole iteration (i.e 5 times because you are using j=1 to j<=5.

Now the come to your problem with your question is that-

for(i = 1; i < 5; i++) //here is problem this will run only 4 time because i<5, and you require 5 times as according to your output given,replace it with i<=5 
{
    for(j = 1; j <=5; j++) //here is also because you are using j<=5, as I mention above it will run 5 times for each value of i (in outer loop),so replace j<=5 by j<=i, because for each change value of i, you require same time execution of inner loop to print the value of "ch" variable) 
       printf("%c", ch);
    printf("\n");
}

so here is modified code

int main() {

// your code goes here
int i, j;
char ch = '$';
for(i = 1; i < =5; i++) // outer loop
{
    for(j = 1; j <=i; j++) // inner loop
       printf("%c", ch);
    printf("\n"); // to move on next line after the complition of inner loop 
}
return 0;

}

May be this is helpful for you.

Answer 4

#include <stdio.h>
int main (void)
{
int i, j;
char ch = '$';

for(i = 1; i < 5; i++)
{
   for(j = 1; j <=i; j++)
      printf("%c", ch);
   printf("\n");
}
 return 0;

}

description: first for loop is for printing the row.. and nested for loop is for no of '$' have to print

Answer 5

it was simple,why because,in it you just want to print in the form of rows and columns and that should be in increasing order. in the first for loop,you are going to print rows and with included for loop you need to do it for columns.

Answer 6

• Your outer loop needs to run from 1 to 5, it only runs from 1 to 4 as written. for( i = 1; i <= 5; i++ )
• The inner loop needs to run from 0 to i-1 (or 1 to i if you prefer). for( j = 0; j < i; i++ )
• The variable ch is redundant since it never changes, you can print the character directly using a literal string in the printf() call, or better a character constant with putchar().
• The best way to solve such simple problems is to step the code in your debugger and observe the code flow and how the variables change with each step.

Answer 7

Since answering this question with code is cheating, here are some hints:

• For each line you're printing, you want to print a number of $s equal to the line number.
• printf doesn't add a newline character unless you tell it to, so successive calls to printf can put characters on the same line.

If you have code that doesn't work, post it. We'll be happy to help you fix it.

Edit: Based on your sample code, you have some very small problems.

First, in your outer loop, you want a <= instead of a <. That gets you up to 5. Second, in your inner loop, j <= 5 should be j <= i. Though I would have written the inner loop with j starting at 0 and < i, that's just a stylistic preference.

The printf("%c", ch) is equivalent to printf("$") too, in case you weren't sure.

For reference, here's my first crack at an answer. It's very similar to yours:

#include <stdio.h>

int main()
{
    int line, dollar;
    for (line=1; line <= 5; line++)
    {
        for (dollar = 0; dollar < line; dollar++)
        {
            printf ("$");
        }
        printf ("\n");
    }
    return 0;
}

Answer 8

The logic of what you need to do is pretty simple: you need to print 5 rows, where the i-th row has got i '$'.

Pseudo code would look like this:

for any i from 1 to 5:
  print '$' times i
  print newline

print '$' times i could look like this:

for any j from 1 to i:
  print '$'

It shouldn't be too hard to rewrite this using C syntax.