C - dynamical allocation/rellocation inside the function

Question

I have written this function in C (function should receive a char*, allocate space necessary, and insert character given a pointer to the index of character behind the pointer)

void add_to_str(char *character, char ** string, int* index)
{
    //we dont expect  *string to be NULL !! no condition done
    if (*index == 0)    //if str is empty we will alloc AVGLEN(32) characters space
       *string=malloc(AVGLEN*sizeof(char));

    if (*string == NULL)   //malloc fails?
    {
        fprintf(stderr,errors[MALLOC]);
        exit(99);
    }
    //string length exceeds 32 characters we will allocate more space
    if ( *index > (AVGLEN-1) || character== NULL )// or string buffering ended, we will free redundant space
    {
        *string=realloc(*string,sizeof(char)*((*index)+2));//+1 == '\0' & +1 bcs we index from 0
        if (*string==NULL)   //realloc fails?
        {
            fprintf(stderr,errors[REALLOC]);
            exit(99);
        }
    }
    *string[(*index)++]=*character;
}

When *index > 0, it gives me a segmentation fault on the line

*string[(*index)++]=*character;

A variant of this function (just malloc behind char* and then assign characters to string[i++]) works perfectly.

I suggest moving any `++` and such inside any non-trivial expressions to their own statements. Compiler should still produce same output after optimizations, and you may avoid some unexpected behavior and have clearer code. — hyde, Oct 22 '13 at 20:16

Filipe Gonçalves · Accepted Answer · 2013-10-22T19:56:17.043

You have to be careful with this statement:

*string[(*index)++]=*character;

Because array indexing has higher precedence than pointer dereferencing. Thus, this is the same as

*(string[(*index)++]) = *character;

Which is not what you want. You want this:

(*string)[(*index)++] = *character;

The faulty code works for *index == 0 because in that case the statement is equivalent to **string, which is still valid, but when index > 0, string is going to be dereferenced in an invalid position: string+index.

score 0 · Answer 2 · answered Oct 22 '13 at 20:12

0

Note that doing something like this:

ptr = realloc(ptr, ...);

is a really bad pattern. When realloc() fails your old allocated area is no longer accessible to the program and becomes leaked. The correct pattern would be:

char* str_new = realloc(string, ...);

if (str_new != NULL)
    string = str_new;
else
    /* handle allocation error */

answered Oct 22 '13 at 20:12

Alexander L. Belikoff

5,698
1
25
31

yeah i see now, btw in this program certainly its true that i dont need the rest of a string when reallocation fails, but at least i can free() it carefully, thanks :) – Smarty77 Oct 22 '13 at 21:38

score 0 · Answer 3 · answered Oct 22 '13 at 20:18

0

For a start the function fprintf requires a format string.

I.e.

 fprintf(stderr,errors[MALLOC]);

is probably invalid.

Also what is the purpose of this function. Seems to have no narrative,

answered Oct 22 '13 at 20:18

Ed Heal

59,252
17
87
127

btw it works errors* is a form of associative string array (smt like this)const char* const errors[] = { [MALLOC] = "ERROR: Malloc()'s dynamic memory alocation failed\n" } – Smarty77 Oct 22 '13 at 21:40

C - dynamical allocation/rellocation inside the function

3 Answers3