Trouble \0 null terminating a string (C)

Question

I seem to have some trouble getting my string to terminate with a \0. I'm not sure if this the problem, so I decided to make a post.

First of all, I declared my strings as:

char *input2[5];

Later in the program, I added this line of code to convert all remaining unused slots to become \0, changing them all to become null terminators. Could've done with a for loop, but yea.

while (c != 4) {
    input2[c] = '\0';
    c++;
}

In Eclipse when in debug mode, I see that the empty slots now contain 0x0, not \0. Are these the same things? The other string where I declared it as

char input[15] = "";

shows \000 when in debug mode though.

My problem is that I am getting segmentation faults (on Debian VM. Works on my Linux 12.04 though). My GUESS is that because the string hasn't really been terminated, the compiler doesn't know when it stops and thus continues to try to access memory in the array when it is clearly already out of bound.

Edit: I will try to answer all other questions soon, but when I change my string declaration to the other suggested one, my program crashes. There is a strtok() function, used to chop my fgets input into strings and then putting them into my input2 array.

So,

input1[0] = 'l'
input1[1] = 's'
input1[2] = '\n'

input2[0] = "ls". 

This is a shell simulating program with fork and execvp. I will post more code soon.

Regarding the suggestion:

char *input2[5]; This is a perfectly legal declaration, but it defined input2 as an array of pointers. To contain a string, it needs to be an array of char.

I will try that change again. I did try that earlier, but I remember it giving me another run-time error (seg fault?). I think it is because of the way I implemented my strtok() function though. I will check it out again. Thanks!

EDIT 2: I added a response below to update my progress so far. Thanks for all the help!

It is here.

.

Answer 1

Response 1:

Thanks for all the answers!

I made some changes from the responses, but I reverted the char suggestion (or correct string declaration) because like someone pointed out, I have a strtok function. Strtok requires me to send in a char *, so I reverted back to what I originally had (char * input[5]). I posted my code up to strtok below. My problem is that the program works fine in my Ubuntu 12.04, but gives me a segfault error when I try to run it on the Debian VM.

I am pretty confused as I originally thought the error was because the compiler was trying to access an array index that is already out of bound. That doesn't seem like the problem because a lot of people mentioned that 0x0 is just another way of writing \000. I have posted my debug window's variable section below. Everything seems right though as far as I can see.. hmm..

Input2[0] and input[0], input[1 ] are the focus points.

Here is my code up to the strtok function. The rest is just fork and then execvp call:

int flag = 0;
int i = 0;
int status;
char *s; //for strchr, strtok
char input[15] = "";
char *input2[5];
//char input2[5];

//Prompt
printf("Please enter prompt:\n");

//Reads in input
fgets(input, 100, stdin);

//Remove \n
int len = strlen(input);
if (len > 0 && input[len-1] == '\n')
    input[len-1] = ' ';

//At end of string (numb of args), add \0

//Check for & via strchr
s = strchr (input, '&');
if (s != NULL) { //If there is a &
    printf("'&' detected. Program not waiting.\n");
    //printf ("'&' Found at %s\n", s);
    flag = 1;
}


//Now for strtok
input2[i] = strtok(input, " "); //strtok: returns a pointer to the last token found in string, so must declare
                                //input2 as char * I believe

while(input2[i] != NULL)
{
    input2[++i] = strtok( NULL, " ");
}

if (flag == 1) {
    i = i - 1; //Removes & from total number of arguments
}

//Sets null terminator for unused slots. (Is this step necessary? Does the C compiler know when to stop?)
int c = i;
while (c < 5) {
    input2[c] = '\0';
    c++;
}

Answer 2

You code should rather look like this:

char input2[5];

for (int c=0; c < 4; c++) {
    input2[c] = '\0';
}

0x0 and \0 are different representation of the same value 0;