Full square

This is a discussion on Full square within the C++ Programming forums, part of the General Programming Boards category; I would like to make a program that reads in the side of a square (between 1 and 20) and ...

  1. #1
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    Post Full square

    I would like to make a program that reads in the side of a square (between 1 and 20) and prints it out of (+ or - or * ). The simplest way possible, because im not that good at C++ yet.
    Last edited by iLike; 09-26-2009 at 07:10 PM. Reason: to clarify

  2. #2
    Not stupid, just stupider yaya's Avatar
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    Well... what have you tried?

  3. #3
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    this is what i have now
    Code:
    #include <stdio.h>
    
    int main(void)
    {
       int row;
       int column;
       int x;
    
    
    
      printf("Enter a number of row to be printed\n" );
       scanf ( "%d", &x );
    
       row = x;
       
       while ( row > 0 ) 
       {
            printf("*");
            row--;
       }
       printf("\n");
    
       row = x;
       column = x;
    
    return 0;
    
    }

  4. #4
    Registered User C_ntua's Avatar
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    Filled square or empty square?

    If you are using C++ and not just C you should use cout/cin and not printf().

    Also this:
    Code:
       scanf ( "%d", &x );
       row = x;
    can be simply done like this:
    Code:
       scanf ( "%d", &row );
    Well, you print a line.Read also columns and reapeat "columns times" your program. And you will print eventually a square.

    Do some input checking if it is 1-20 if you want. Here is how your program should look (one way to do it)
    Code:
    while (columns > 0)
    {
       while (rows > 0)
       {
           //rest of code
       }
       //rest of code
    }

  5. #5
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    Is there way to just use printf and loop?

    Code:
    int main(void)
    {
       int row;
       int column;
       int x;
    
    
    
      printf("Enter a number of row to be printed\n" );
       scanf ( "%d", &row );
    
       row = x;
       while (column > 0){
       while ( row > 0 ) {
            printf("*");
            row--;
       }
       printf("\n");
    
       row = x - 2;
       column = x;
       }
    return 0;
    }
    is this what you told me?

  6. #6
    Malum in se abachler's Avatar
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    printf pwnz cout every day of the week, especially since cout is (or at least used to be) implemented using printf.

    plus its really easy to change a printf statement into a sprintf or fprintf, try doign a fcout or scout

    you can implement the whoel thing as a single for loop
    Until you can build a working general purpose reprogrammable computer out of basic components from radio shack, you are not fit to call yourself a programmer in my presence. This is cwhizard, signing off.

  7. #7
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    Question

    Can anyone tell me if its possible to do with printf and while and keep it simple( nothing like cout, cin, sprintf etc.), im just beginner. If it is possible can you tell me how should i modify my code?

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