I am passing a pointer a function that updates it. However when the function returns the pointer it returns to the value it had prior to the function call.
Here is my code:
#include <stdio.h>
#include <stdlib.h>
static void func(char *pSrc) {
int x;
for ( x = 0; x < 10; x++ ) {
*pSrc++;
}
printf("Pointer Within Function: %p\n", pSrc );
}
int main(void) {
char *pSrc = "Good morning Dr. Chandra. This is Hal. I am ready for my first lesson.";
printf("Pointer Value Before Function: %p\n", pSrc );
func(pSrc);
printf("Pointer Value After Function: %p\n", pSrc );
return EXIT_SUCCESS;
}
Here is the output
Pointer Value Before Function: 0x100403050
Pointer Within Function: 0x10040305a
Pointer Value After Function: 0x100403050
What I was expecting was the value after the function to match the one from within the function.
I tried switching to char **pSrc
but that did not have the desired affect.
I am sure the answer is fairly simple, but I am a recovering hardware engineer and can't seem to figure it out :-)
The pointer inside the function is a copy of the passed pointer.
They both hold the same address but have different addresses, so changing the address held by one of them doesn't affect the other.
If you want to increment the pointer inside the function pass it's address instead, like this
static void func(char **pSrc) {
int x;
for ( x = 0; x < 10; x++ ) {
(*pSrc)++;
}
printf("Pointer Within Function: %p\n", pSrc );
}
and
func(&pSrc);
Also, be careful not to modify the contents, because your pointer points to a string literal, and string literals cannot be modified.
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