Increasing list in c

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

int main()
{
    int total = 10;
    char letra = '*';

    char *retorno;
    retorno = (char *) malloc(total + 1);

    for (int i=0; i<total; i++) {
      retorno[i] = letra;
      retorno[i+1] = '\0';

      printf("%s\n", retorno);
    }

    return 0;
}
c
  • 2

    Do you have the code snippet you made? Try to explain your question better

  • it is necessary to do two loop repetition. If the question is not answered, when I get home I send the answer

  • 1

    @Andrelacomski doesn’t need two ties to do this

1 answer

5

It is possible to solve this issue with only one loop, although it requires more variables.
Follows a possible solution:

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

int main()
{
    int total = 10;
    char letra = '*';

    char *retorno;
    retorno = (char *) malloc(total + 1);

    for (int i=0; i<total; i++) {
      retorno[i] = letra;
      retorno[i+1] = '\0';

      printf("%s\n", retorno);
    }

    return 0;
}

Already with two ties, maybe it’s easier to understand:

#include <stdio.h>

int main()
{
    int total = 10;
    char letra = '*';

    for (int i=0; i<total; i++) {
      for (int j=0; j<=i; j++){
        printf("%c", letra);
      }
      printf("\n");
    }

    return 0;
}

I apologize if I blew the question (since I myself voted to close it because it is not very clear), but the comments do not me allowed to pass..

  • 2

    Good! You can really do only with 1 loop. I hadn’t even thought of this creative solution, of concatenating only one char in the same string.

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