Segmentation fault, I don't know why

Question

I have a seg fault but I don't know why. I know it should be working but it kee telling me there is a seg fault someone has a solution? Please someone help I need to know for my BA.

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

int split_allocate(const char* s, char*** word_array){
    char c;
    char tmp[100];
    int id_s,i=0, j=0,k;
    for(id_s =0 ; (c=s[id_s])!='\0' ; id_s++){
        printf("%c\n", c);
        if(c!=' ' && c!='\t' && c!='\n'){
            tmp[i]=c;
            i++;
            printf("i if : %d\n", i);
            continue;
        }
        tmp[i]='\0';
        printf("i else : %d & tmp : %s\n", i, tmp);
        (*word_array)[j] = (char*)malloc(sizeof(char)*(i+1));
        printf("666\n");
        if(NULL== (*word_array)[j]){
            return -1;
        }
        for(k=0 ; tmp[k] != '\0' ; k++){
            (*word_array)[j][k]=tmp[k];
        }
        (*word_array)[j][k+1]='\0';
        j++;
        printf("j for : %d\n", j);
        i=0;
    }
    return j;
}

int main(void) {
    char* s = "Salut,   cet examen\n a l'air long...";
    char*** word_array;
    printf("number of words :%d\n", split_allocate(s,word_array));
    return 0;
}

Undefined behavior for using the value of an object with automatic storage duration while it is indeterminate. — EOF, Aug 16 '17 at 12:23
Also, being a [three-start programmer](http://wiki.c2.com/?ThreeStarProgrammer) is *not* something to strive for. — Some programmer dude, Aug 16 '17 at 12:24
I have to use the three star cause it's in the prototype my professor gave us — Sam45505, Aug 16 '17 at 12:28
Then you should probably pass the address of a pointer-to-pointer to `char`, i.e. have `char **word_array; split_allocate(..., &word_array);` *And* allocate memory in the function, like `*word_array = malloc(...)` — Some programmer dude, Aug 16 '17 at 12:31
Also Your logic is wrong. E.g 1) You need to skip the spaces. 2) This will not get the last word. — BLUEPIXY, Aug 16 '17 at 12:33
What steps have you taken in attempting to debug the problem? — faintsignal, Aug 16 '17 at 12:37
`(*word_array)[j][k+1] ` I bet that it does not do what you @Sam45505 think it does — 0___________, Aug 16 '17 at 12:43

score 1 · Answer 1 · answered Aug 16 '17 at 12:58

I would recommend you to replace

if(c!=' ' && c!='\t' && c!='\n')

with

if(!isspace(c))  // Iam easier and more readable

from ctype.h. Which detects all these characters

' '      space 
'\t'     horizontal tab 
'\n'     newline
'\v'     vertical tab 
'\f'     feed 
'\r'     carriage return

Also you should change char*** word_array; (three star general) to pointer to pointer which is enough. Then alloc memory on heap (dynamic storage duration). You didnt do it and it caused segmentaion fault (dereferencing uninitialzed pointer).

char ** word_array = malloc (sizeof(char *) * ROWS);
for (int i = 0; i < ROWS; ++i)
{
    word_array[i] = malloc (sizeof(char) * ROW_LEN);
}

You shouldn't cast malloc()'s return value because it may lead to problems.

Also you should check number of rows, and if needed use realloc to get more rows, because accessing out of bounds leads to undefined behavior.

Instead of copying character by character use strcpy, since you know there is enough space. Its more readable and easier.

for(k=0 ; tmp[k] != '\0' ; k++){
    (*word_array)[j][k]=tmp[k];
}

to

strcpy(word_array[j], tmp); // In case word_array is char **

I can see spaces in your string but you arent skiping them, this may help

while ((c=s[id_s++]) && isspace(c))
    ;

Segmentation fault, I don't know why

1 Answers1