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Free pointer to dynamic array in c

I am creating a dynamic array in c with malloc. eg:.

myCharArray = (char *) malloc(16);

Now, if I create such a function and pass it myCharArray:

reset(char * myCharArrayp)
{
    free(myCharArrayp);
}

will work, or am I somehow freeing only a copy of the pointer (myCharArrayp), and not the actual myCharArray?

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3 answers

It will be perfect and free up memory as you expect.

I would like to write a function like

 void reset(char** myPointer) {
     if (myPointer) {
         free(*myPointer);
         *myPointer = NULL;
     }
 }

Thus, after releasing the pointer, it is set to NULL. Reusing previously freed pointers is a common source of errors.

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, - , . , . . , , (, int), .

void myFunction()
{
    char *myPointer;     // <- the function stack frame is set up with space for...
    int myOtherVariable; // <- ... these two variables

    myPointer = malloc(123); // <- some memory is allocated on the heap and your pointer points to it

    free(myPointer); // <- the memory on the heap is deallocated

} // <- the two local variables myPointer and myOtherVariable are freed as the function returns.
+14

Yes, that will work.

Although a copy of your pointer variable will be sent, it will still refer to the correct memory location, which will actually be released with a free call.

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