please, what is the best way to assign int's to a char array ?

Code:

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char alphabet_string[CHAR_COUNT][26] = { 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z', '\0' };

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Please, pseudo code, or is it possible inside a for loop, each letter of alphabet, 1 for a and 26 for z, inside a for loop, returning the int or a data type variable. Thanks.

Quote:

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each letter of alphabet, 1 for a and 26 for z, inside a for loop, returning the int or a data type variable.

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Can you clarify this, please? I can't understand exactly what you want to do.

Quote:

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Originally Posted by _jamie

Code:

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char alphabet_string[CHAR_COUNT][26] = { 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z', '\0' };

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Please, pseudo code, or is it possible inside a for loop, each letter of alphabet, 1 for a and 26 for z, inside a for loop, returning the int or a data type variable. Thanks.

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It's not quite clear what you want to do. But say we want 1 to represent 'a' and 26 to represent 'z', and we want to write to a string at a position, we can write a function like this

Code:

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void writeletter(char *str, int index, int letter)
{
    assert(index >= 0 && index < strlen(str));
    assert(letter >= 1 && letter <= 26);

    str[index] = 'a' + letter -1;
}

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looks good Malcom, I will study this code tonight. Great use of the use of assert() function, thanks.

Quote:

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each letter of alphabet, 1 for a and 26 for z, inside a for loop

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Code:

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#include <stdio.h>
#include <string.h>

char alphabet[26] = "abcdefghijklmnopqrstuvwxyz";

int main( int argc, char *argv[] ) {
 
    char userinput[256];
   
    printf( "enter string: " );
    fgets( userinput, 256, stdin );
    userinput[strlen(userinput)-1] = '\0';     
   
    for (int i=0; i<strlen(userinput); i++) {
      for (int j=0; j<strlen(alphabet); j++) {
       
        if (userinput[i] == alphabet[j]) {
          int converted = (j+1);
          printf( "%i,", converted );
        }

      }
    }
   
  printf( "\n" );
  return 0;
}

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> char alphabet[26] = "abcdefghijklmnopqrstuvwxyz";
There is no guarantee that you're getting a \0 in your string.
So strlen() on it is broken.

Quote:

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Originally Posted by Salem

> char alphabet[26] = "abcdefghijklmnopqrstuvwxyz";
There is no guarantee that you're getting a \0 in your string.
So strlen() on it is broken.

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Code:

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char alphabet[27] = "abcdefghijklmnopqrstuvwxyz\0";

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Make the compiler do the work.

Code:

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char alphabet[] = "abcdefghijklmnopqrstuvwxyz";

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>looks good Malcom, I will study this code tonight. Great use of the use of assert() function, thanks.

It's all fun and games until NDEBUG is defined :P