C character array comparison - c

Question

Ok so I was doing a coding challenge on codewars.com and the challenge was to take a string as input and return a string where in place of the letters are instead the number of the alphabet that matches the letter.
Everything except letters are to be ignored.
ex: "aab" would return "1 1 2"
There should be a space between each number that represents a letter in the alphabet.
So, when I run this code on my IDE (which is xcode using c99) Everything looks good and the strcmp() function says the 2 strings are equal.
The website I'm on uses C11 I believe but I don't see that causing the error.
When I run this code on the challenge website it passes a couple of tests but then fails a couple also. It fails when the input string is "", and it also fails on the string that I have used in the code below, but again it does not fail when I run it on my ide.
My questions are:
1) Any idea what is causing this bug?
2) What would you have done differently as far as the code is concerned
Thanks
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
char *alphabet_position(char *text)
{
int i,letter_position;
unsigned long int size = strlen(text);
char *result = malloc(sizeof(int)*size + 1);
char int_string[10];
char temp = ' ';
//If String wasn't blank
if (strcmp(text, "")!=0)
{
for (i=0; i<size-1; i++)
{
//If it is a letter
if (isalpha(text[i]))
{
temp = tolower(text[i]);
if (temp == 'a')
strcat(result, "1");
else
{
letter_position = temp - 'a' + 1;
sprintf(int_string,"%d",letter_position);
strcat(result, int_string);
}
//Print space after letter until the last letter
if (i!=size-2)
strcat(result, " ");
}
}
strcat(result, "\0");
return result;
}
else
{
strcat(result, "\0");
return result;
}
}
int main(void)
{
char *string = alphabet_position("The narwhal bacons at midnight.");
char *expected_output = "20 8 5 14 1 18 23 8 1 12 2 1 3 15 14 19 1 20 13 9 4 14 9 7 8 20";
printf("Your output %s\n", alphabet_position("The narwhal bacons at midnight."));
printf("Expt output %s\n", "20 8 5 14 1 18 23 8 1 12 2 1 3 15 14 19 1 20 13 9 4 14 9 7 8 20");
printf("\n");
printf("your len %lu\n", strlen(alphabet_position("The narwhal bacons at midnight.")));
printf("Expt len %lu\n", strlen(expected_output));
if (strcmp(string, expected_output)==0)
printf("Equal\n");
else
printf("Not equal\n");
return 0;
}

Answer 1

You have two serious problems.
First, you're not initializing the contents of the result array to an empty string. After you call malloc(), add:
result[0] = '\0';
malloc() doesn't initialize the memory it allocates. There's another function calloc() that takes slightly different arguments and initializes the memory to zeroes. But you only need the first character to be zero, so there's no need for that.
Second, the for loop is not processing the last character of text. It should be:
for (i = 0; i < size; i++)
Similarly, the test for whether to add a space should be if (i != size-1). Did you think strlen() counts the null character at the end?
The amount of space you specify in malloc() is not correct, but in practice it won't cause a problem. sizeof(int) has nothing to do with the number of characters it takes to show the value of an integer. Since you're just printing the alphabetical position, it will be at most 26, so you need 3 characters for every input character. Therefore, it should be:
char *result = malloc(3 * size + 1);
Your allocation works because sizeof(int) is generally at least 4, so you're allocating more than enough space.
There are other minor problems that don't affect the correctness of the result:
You don't need the if statement that treats 'a' specially, the code you have in else will work for all letters.
You don't need strcat(result, "\0") at the end. result has to already be null-terminated in order for you to use it as an argument to strcat(), so there's no point in using strcat() to add a null terminator.