Pointers as function arguments

Hello,

I am having some difficulty in understanding pointers when passed to functions. When I pass a pointer to a function and assign some dynamic memory to its argument in the other function, should that dynamic memory be reflected in the invoking block?

I have some code below. As you can see from executing it, the string and memory assigned in the function is not reflected in main. Why is this? If the value of testThis is an address and is being assigned to ttArg, shouldn't testThis and ttArg point to the same memory location?

Code:

---

#include <stdio.h>

void testPointer(char *);

main()
{
        char * testThis;
        testPointer(testThis);
       
        printf("testThis in main is %s\n", testThis);               

}        /* end main */

/* begint the testPointer */
void testPointer(char * ttArg)
{

        ttArg = (char *)malloc(10);
       
        strncpy(ttArg, "hello", 6);

        printf("ttArg in testPointer() is %s\n", ttArg);       


}        /* end testPointer */

---

Not really. You're still passing by value. You pass the value (the address that the pointer points to, which is totally random at this point) of the pointer to the function. You aren't passing the address of the pointer. If you wanted to actually update what the original pointer points at, you'd pass a pointer to the pointer.

Try this on for size:

Code:

---

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

void testPointer(char **);

main()
{
        char * testThis;
        testPointer(&testThis);
       
        printf("testThis in main is %s\n", testThis);               

}        /* end main */

/* begint the testPointer */
void testPointer(char ** ttArg)
{

        *ttArg = (char *)malloc(10);
       
        strncpy(*ttArg, "hello", 6);

        printf("ttArg in testPointer() is %s\n", *ttArg);       


}        /* end testPointer */

---

That should work. OTP.

Quzah.