Thread: exercise: Print out the sign 'X' with the size depending on an input-value.

Hi!

I want to print out a 'X' with as many rows, as an input-value would have.
This picture will describe hopefully better, what I want to do:

Screenshot - 0781ad21c51043868573c8ecc3b4b448 - Gyazo

I have broken this task down in
a.) at first I only consider for n being an odd integer value.
b.) At first I do row 1 - 4 (basically a 'V' sign, we could say).

My code so far:

Code:

#include <stdio.h>

int main() {
    int i, j, n;
    scanf("%d", &n);


    for (i = 0, j = n - 2 ; i <= (n/2), j >= -1; i++, j = j - 2) {
        printf("%*s", i, "");
        printf("*");
        printf("%*s", j, "");
        printf("*\n");
    }


     return 0;
}

Also I have another fundamental question:
When I would nest a for loop in another one, then the outer loop makes one iteration and arrives in the inner loop.
The inner loop now finishes all its iterations (until cancel condition is reached) and goes back to the outer loop which now would start its second iteration.

But what tool we use, instead of nested for loops, when we want a loop with lets say five iterations iterate only one and then want a following for loop also iterate only one-time, before it goes back to the first loop for its second iteration, after from there to second loop for its second iteration and so on...?

Also, can someone point me out what is happening in this code:

Code:

#include <stdio.h>

int main() {
    int i, j, n;
    scanf("%d", &n);


    for (i = 0; i <= (n/2); i++) {
        printf("%*s", i, "");
        printf("*");
        for(j = n-2; j >= 0; j = j - 2) {
            printf("%*s", j, "");
            printf("*\n");
        }
    }


     return 0;
}

Regards,
Placebo

Hi!

Thanks for your initial help.
Before I try again to print out an X, I want to understand and finally execute your exercise (printing out the choosen sequence).

I post here my first attempt and will afterwards describe my problems, so that you can see them:

Attachment 15735

Code:

#include <stdio.h>#include <stdlib.h>


int main() {
int row, n = 7, initialIndent, spacesBetween = n;


    for ( row = 0 ; row < n ; row++ ) {
    int initialIndent = row;
    int spacesBetween = spacesBetween - 2;
    printf("%d %d\n", initialIndent, spacesBetween);
    }
}

1.) After declaration, I have provided the variable n the value of 7.
Now I have provided to the variable spacesBetween the same value of n, so it should be 7.
My calculation for spacesBetween is per loop:
spacesBetween = spacesBetween - 2;

Why is the first output 6 and not 5?

2.) With your code-hint?
How can I reverse both sequences after the first 4 loops?
For instance, when we look at initialIndent:
I can do 0, 1, 2, 3;
And the in a separate loop 2, 1, 0;
But I cannot do this switch in one loop and its one calculation - is this possible?

Although, even with separating the increasing sequence form the decreasing sequence and by using "spaghetti" code, I cannot execute your sequence
What is happening in regards to the red-highlighted output?

Originally Posted by Salem

It would help if you could put your code as text we can copy/paste.

But you might want to compile with
-Wshadow
Using the GNU Compiler Collection (GCC): Warning Options

Hey, I did put the relevant code in text using the CODE-tags - is this what you mean?
The second code can be ignored.

Code:

int spacesBetween = spacesBetween - 2;

That line doesn't use the spacesBetween variable that was defined 5 lines above it. Rather, it defines a new variable called spacesBetween and initializes it with its own value minus 2. Since that spacesBetween was not initialized before its value is used, it it set to a garbage value.

Start at the first and last elements of the array and work your way to the middle swapping them over.

As an exercise, open a spreadsheet and make a1=1, a2=2,... a5=5.

Use b1 like a temp variable

Using 'copy and paste' Swap the values of a1 and a5, making sure that you store the value of one in b1 so you don't lose it.

Next a2 and a4, and then a3 doesn't need to be swapped. However, if there is an even amount of elements, there will be no centre one.

This is the algorithm that you will need to use in your code.

When you are struggling to conceptualise an algorithm, it helps to go through it manually. Spreadsheets are great for this, notepad for pseudo code, or just the humble pen and paper

Somehow this thread does not bring me closer on topic tbh - I will try your exercise advice and the last one (pen and paper is anyways my default start), but can someone please concretely post here his code-solution for:

"Make this output the sequence
0 5
1 3
2 1
3 0
2 1
1 3
0 5
"

So that I can see, where I was wrong?

I mean, it's not that I have not been trying to solve the given exercise (number-sequences) myself - then I have posted my results where I can see what the code does different than what I want it to do, posted some follow up questions and did not hear back.
I just do not want to jump right away to your exercise while the first one is not done yet.
Anyways, thanks for the hint in regards to style and codeblocks...


btw., lol - I somehow managed to finish the very original exercise (printing out an X-sign - mentioned it in my first post), but the code looks terrible and it feels that the learning-effect was low, since it feels that I have missed core strategies there:

Code:

#include <stdio.h>#include <stdlib.h>


int main() {
    int z, s, leer_li, leer_re, n;
    printf("Please enter the number of rows: ");
    scanf("%d", &n);


    if (n % 2 != 0) {
            for (z = 1; z <= n/2; z++) {
                printf("%*s", z - 1, "");
                printf("*");
                for (s = 1; s <=1; s++) {
                    printf("%*s" , n - 2*z, "");
                    printf("*\n");
                }
            }
                printf("%*s", z - 1, "");
                printf("*\n");


                for (z = n/2; z >= 1; z--) {
                    printf("%*s", z - 1, "");
                    printf("*");
                    for (s = 1; s <=1; s++) {
                        printf("%*s" , n - 2*z, "");
                        printf("*\n");
                    }
                }
            } else {
                for (z = 1; z < n/2; z++) {
                    for (leer_li = 1; leer_li <= 1; leer_li++) {
                        printf("%*s", z - 1, "");
                        printf("*");
                    }
                    for (leer_re = 1; leer_re <= 1; leer_re++) {
                        printf("%*s", n - 2*z, "");
                        printf("*\n");
                    }
                }


                for (leer_li = 1; leer_li <= 2; leer_li++) {
                    printf("%*s", z - 1, "");
                    printf("**\n");
                }


                for (leer_li = n/2 - 2; leer_li >= 0; leer_li--) {
                    printf("%*s", leer_li, "");
                    printf("*");
                    for (leer_re = 1; leer_re <= 1; leer_re++) {
                        printf("%*s", (n - 2) - 2 * leer_li, "");
                        printf("*\n");
                    }


                }


            }


   return 0;
}

It somehow works, but feels still off.
Would still like to know how this one works (in dependence of a number-variable n) :
"Make this output the sequence
0 5
1 3
2 1
3 0
2 1
1 3
0 5
"

I feel that being able to put out this sequence would have currently a large learning impact.
After that, I might wanna try your exercise, but need to reread and try to understand it and/or get familiar with spreadsheets at first...