C can someone tell me what is going on here?

Question

I can not figure out what the heck is happening here. What I expect is that the output should say that there is only 1 element in keys, it's saying there are 7 when I have allocated only the 0 position with 120 bytes.

void add2(char **b, char *i) {
    if (!i) {
        b[0] = (char*) malloc(120);
        sprintf(b[0], "%s", "hithere");
    } else {
        strcat(b[0], "\\\\");
        strcat(b[0], i);
    }
}

void add1(char **b) {
    add2(b, NULL);
    add2(b, "one");
    add2(b, "two");
    add2(b, "three");
}

void add() {
    char *keys[2];
    int i;
    add1(keys);
    fprintf(stderr, "%s\n", keys[0]);
    for (i = 0; keys[i]; i++)
        fprintf(stderr, "%d\n", i);
    free(keys[0]);
}

int main (int argc, char** argv)
{   
     add();
     add();
     return 255;
}

outputs: hithere\one\two\three 0 1 2 3 4 5 6 7 hithere\one\two\three 0 1 2 3 4 5 6 7

the strings are as I expect them to be however i thought only 0 should be printed out after since 0 is the only element I added to. I need to be able to free each spot instead of just free[0] but when i put the free[i] in the for loop that prints out i it stack dumps.

with regard to initialization from response below, if i need an array that is like 1000 instead of 2 then how do i init them all to 0 without typing out 1,000 0's

Answer 1

You haven't initialized the keys array, so it contains whatever happened to be in memory. keys[1] and on up to 7 weren't zero in that instance.

Answer 2

/* ... */
void add() {
    char *keys[2];
/* ... */

keys is an array of 2 pointers to char, but it is not initialized.
Try this

/* ... */
void add() {
    char *keys[2] = {0, 0}; /* explicit initialization of the two elements */
/* ... */

In the absence of explicit initializers for all members, the uninitialized ones are initialized to zero (the right zero for their type).

/* ... */
void add() {
    char *keys[1000] = {42}; /* explicit initialize the first element to 42 */
                             /* and implicitly initialize all other elements to 0 */
                             /* **ATTENTION** */
                             /* initializing a pointer to 42 is wrong */
                             /* I only wrote that as an example */
/* ... */

---

Edit, quote from the Standard

6.7.8 Initialization
Syntax
1

initializer:

assignment-expression
{ initializer-list }
{ initializer-list , }

initializer-list:

designation_opt initializer
initializer-list , designation_opt initializer

There is no syntax for an empty initializer-list.

Answer 3

For the second part of question,

If u need to initialize an array that is like 1000 instead of 2 then use the following

char *keys[1000] = {0};

If u initialize one element of array with a value, then all other members of array will be automatically initialized to zero.

ie, if u use char *keys[1000] = {42}; then first member of array will be initialized to 42 and all other members of array will be automatically initialized to zero.

Answer 4

What's the difference between implicitly initializing array elements to 0 and explicitly setting them to 0 with memset()?

char *data[1000] = {0};
memset(data, 0, sizeof data);

The first option (implicit initialization) sets every element of the array to zero of the proper type; the second option sets all bits of all elements (plus any padding) to 0. Usually (99.99% of all current computers) there is no difference between a typed 0 and all bits 0.

Imagine a computer with segmented memory ... where a pointer is composed of two parts. When you set it to 0 (the right type of 0) the compiler can make that 0 different than all bits 0. If you specifically set all bits to 0, you might end up with an invalid pointer.

void *test = 0; /* can make test something like "0xC0DE:0x0000" */
memset(test, 0, sizeof test); /* will make test as "0x0000:0x0000" */