CODEWITHSUNDEEP
cmemorypointersfree
tpar44
tpar44

Reputation: 1451

Memory Allocation C

I have a memory allocation question. If I declare a char* within an else block that means the char* is destroyed when the else block is done executing. The else block is located in a while loop so it will iterate many times. However, what if the char* declared in the else block is aliased to a malloc'd variable as seen in the below example. My question is how do I fee something like this? I feel as though if I free the temp char* variable, I will cause a segmentation fault because I will free the variable I want to keep as well. If that is the case, I'm at a loss for where the free statement goes.

char* print_path = NULL;

(snip)

  (while)

 else{
        char* temp_path = print_path;
        int temp_size = strlen(temp_path)+strlen(file_name(child->fts_path))+1;
        print_path = (char*)malloc(temp_size);
        strcpy(print_path, temp_path);
        strncat(print_path, file_name(child->fts_path), strlen(file_name(child->fts_path)));
        printf("%s:\n\n", print_path);
        }

(snip)

I would like to point out that I do free print_path at the end of the program after I know it will not be executed again. However, it is the intermediary executions of the loop that I would like to free. Any help would be appreciated. Thanks!

Upvotes: 0

Views: 322

Answers (3)

Alejandro García
Alejandro García

Reputation: 1

You have 2 pointers:

print_paths: Points to an array like ['h','e','l','l','o'], this pointer contains the address for the first element of the array, in this case it points to h, lets say that the address of this h is 1, so, the address of e is 2 and so on... hence, print_path contains the number 1.

temp_path: this pointer points to nothing or to a random place in the memory.

When you call malloc, malloc ask for some memory and gives you the address of the new allocated memory, but this memory contains garbage.

So after 

print_path = (char*)malloc(temp_size);

print_path contains the address of the new allocated memory, a new array of size temp_size which has garbage, lets say that this address is 40, so the value of print_path is 40.

finally, when you call:

strcpy(print_path, temp_path);

you copy the values of the array temp_path (['h','e','l','l','o']) to the array pointed by print_path, in other words, the address 40 instead of containing garbage now contains h, 41 contains e and so on.

It's important to note that, even if both address (1 and 40) contains the value h, modifying one doesn't affect the other.

The best place for the free(temp_path) is just before the end of the else, just check that temp_path is allocated memory, if you try to free memory that it wasn't obtained with malloc your'e gonna have a bad time.

Upvotes: 0

tpar44
tpar44

Reputation: 1451

I believe that since, the gist of what I want to do is reallocate memory to the same variable, I should have looked into realloc instead of malloc.

Upvotes: 0

Kerrek SB
Kerrek SB

Reputation: 477700

It looks like free(temp_path) is the right thing. It should go like this:

char * print_path = malloc(...);    // "NULL" is also possible

while (condition)
{
    if (...)
    {
         // ...
    }
    else
    {
        char * temp_path = print_path;

        print_path = malloc(...);

        free(temp_path);
    }
}

free(print_path);

The invariant in your algorithm is (or should be) that print_path always points to dynamically mallo­cated memory. Note how we have precisely one free for every malloc.

Upvotes: 2

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