Thread: Trying not to gloat...

But I'm really proud of myself. First time a question in the book had me stumped for a while, but I was able to figure it out. Check this baby out:

Code:

#include <stdio.h>
#include <stdlib.h>
int main()
{
    char user_letter;
    char pyramid_letter;
    char space_counter;
    int i=0;
    int j;

    printf("Enter a capital letter please.  Thank you very, very, much:    ");
    scanf("%c", &user_letter);
    space_counter=user_letter-1;
for(i='A'; i<=user_letter; i++)
{
    for(j='A'; j<=space_counter; j++)
        printf(" ");
    for(pyramid_letter='A'; pyramid_letter<=i; pyramid_letter++)
        printf("%c", pyramid_letter);

    for(pyramid_letter=pyramid_letter-2; pyramid_letter>='A'; --pyramid_letter)
        printf("%c", pyramid_letter);
    space_counter--;
    printf("\n");
}

    return 0;
}

Don't make too much fun of me LOL.

Originally Posted by gratiafide

First time a question in the book had me stumped for a while, but I was able to figure it out.

Congratulations!

Now, just think of the kicks you get when you get to solve a problem you or your coworkers/friends/anyone has, by writing your own program. It's more than a little addictive

The indentation could use a bit of fixing, but that's a readability issue for us humans; the compiler will not care.

Assuming you wish to progress to writing command-line tools at some point, you might rewrite the initial part of your code to take the character from the command line instead of scanning it from input:

Code:

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

int main(int argc, char *argv[])
{
    char user_letter;
    char pyramid_letter;
    char space_counter;
    int i=0;
    int j;

    if (argc != 2) {
        fprintf(stderr, "Usage: %s LETTER\n", argv[0]);
        return EXIT_FAILURE;
    }

    if (argv[1][0] == '\0') {
        fprintf(stderr, "No LETTER specified.\n");
        return EXIT_FAILURE;
    }

    if (argv[1][1] != '\0' || !(argv[1][0] >= 'A' && argv[1][0] <= 'Z')) {
        fprintf(stderr, "%s: Not a capital letter.\n", argv[1]);
        return EXIT_FAILURE;
    }

    user_letter = argv[1][0];
    space_counter = user_letter - 1;

Above, argv[0] is the name the program was run with, and argc is 1+the number of parameters, i.e. argc == the number of strings in the argv array.

If argc < 2 there is only the program name, no parameters.

If argc > 1, the first parameter is in argv[1]. It is then never a NULL pointer, but it may be an empty string (say, ./a.out ""), in which case argv[1][0] == '\0'.

In the above code, the first if statement checks that there is exactly one command-line parameter. (Program name and one parameter string means argc == 2.) Otherwise the program shows the usage and exits.

The second if statement checks if the first parameter is an empty string. I like to be careful, you see; it costs just a few lines of code, but saves a lot of trouble if you accidentally (say, in a script) supply an empty string.

The third if statement makes sure the first character is an ASCII uppercase letter A..Z, and only one-character long. Otherwise, it complains.

Thanks for the advice and encouragement to both of you! I really appreciate it and this forum.