My function goes over the length of string - c

Question

I am trying to make function that compares all the letters from alphabet to string I insert, and prints letters I didn't use. But when I print those letters it goes over and gives me random symbols at end. Here is link to function, how I call the function and result: http://imgur.com/WJRZvqD,U6Z861j,PXCQa4V#0
Here is code: (http://pastebin.com/fCyzFVAF)
void getAvailableLetters(char lettersGuessed[], char availableLetters[])
{
char alphabet[]={'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'};
int LG,LG2,LA=0;
for (LG=0;LG<=strlen(alphabet)-1;LG++)
{
for(LG2=0;LG2<=strlen(lettersGuessed)-1;LG2++)
{
if (alphabet[LG]==lettersGuessed[LG2])
{
break;
}
else if(alphabet[LG]!=lettersGuessed[LG2] &&LG2==strlen(lettersGuessed)-1)
{
availableLetters[LA]=alphabet[LG];
LA++;
}
}
}
}
Here is program to call the function:
#include <stdio.h>
#include <string.h>
#include "hangman.c"
int main()
{
int i = 0;
char result[30];
char text[30];
scanf("%s", text);
while(i != strlen(text))
{
i++;
}
getAvailableLetters(text, result);
printf("%s\n", result);
printf ("%d", i);
printf ("\n");
}
Here is result when I typed in abcd: efghijklmnopqrstuvwxyzUw▒ˉ

Answer 1

If you want to print result as a string, you need to include a terminating null at the end of it (that's how printf knows when to stop).

Answer 2

for %s printf stops printing when it reaches a null character '\0', because %s expects the string to be null terminated, but result not null terminated and that's why you get random symbols at the end
just add availableLetters[LA] = '\0' at the last line in the function getAvailableLetters
http://pastebin.com/fCyzFVAF

Answer 3

Make sure your string is NULL-terminated (e.g. has a '\0' character at the end). And that also implies ensuring the buffer that holds the string is large enough to contain the null terminator.
Sometimes one thinks they've got a null terminated string but the string has overflowed the boundary in memory and truncated away the null-terminator. That's a reason you always want to use the form of functions (not applicable in this case) that read data, like, for example, sprintf() which should be calling snprintf() instead, and any other functions that can write into a buffer to be the form that let's you explicitly limit the length, so you don't get seriously hacked with a virus or exploit.

Answer 4

char alphabet[]={'a','b','c', ... ,'x','y','z'}; is not a string. It is simply an "array 26 of char".
In C, "A string is a contiguous sequence of characters terminated by and including the first null character. ...". C11 §7.1.1 1
strlen(alphabet) expects a string. Since code did not provide a string, the result is undefined.
To fix, insure alphabet is a string.
char alphabet[]={'a','b','c', ... ,'x','y','z', 0};
// or
char alphabet[]={"abc...xyz"}; // compiler appends a \0
Now alphabet is "array 27 of char" and also a string.
2nd issue: for(LG2=0;LG2<=strlen(lettersGuessed)-1;LG2++) has 2 problems.
1) Each time through the loop, code recalculates the length of the string. Better to calculate the string length once since the string length does not change within the loop.
size_t len = strlen(lettersGuessed);
for (LG2 = 0; LG2 <= len - 1; LG2++)
2) strlen() returns the type size_t. This is some unsigned integer type. Should lettersGuessed have a length of 0 (it might have been ""), the string length - 1 is not -1, but some very large number as unsigned arithmetic "wraps around" and the loop may never stop. A simple solution follows. This solution would only fail is the length of the string exceeded INT_MAX.
int len = (int) strlen(lettersGuessed);
for (LG2 = 0; LG2 <= len - 1; LG2++)
A solution without this limitation would use size_t throughout.
size_t LG2;
size_t len = strlen(lettersGuessed);
for (LG2 = 0; LG2 < len; LG2++)