c - dereferencing issue - c

Question

I have simplified an issue that I've been having trying to isolate the problem, but it is not helping.
I have a 2 dimensional char array to represent memory. I want to pass a reference to that simulation of memory to a function. In the function to test the contents of the memory I just want to iterate through the memory and print out the contents on each row.
The program prints out the first row and then I get seg fault.
My program is as follows:
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <string.h>
void test_memory(char*** memory_ref) {
int i;
for(i = 0; i < 3; i++) {
printf("%s\n", *memory_ref[i]);
}
}
int main() {
char** memory;
int i;
memory = calloc(sizeof(char*), 20);
for(i = 0; i < 20; i++) {
memory[i] = calloc(sizeof(char), 33);
}
memory[0] = "Mem 0";
memory[1] = "Mem 1";
memory[2] = "Mem 2";
printf("memory[1] = %s\n", memory[1]);
test_memory(&memory);
return 0;
}
This gives me the output:
memory[1] = Mem 1
Mem 0
Segmentation fault
If I change the program and create a local version of the memory in the function by dereferencing the memory_ref, then I get the right output:
So:
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <string.h>
void test_memory(char*** memory_ref) {
char** memory = *memory_ref;
int i;
for(i = 0; i < 3; i++) {
printf("%s\n", memory[i]);
}
}
int main() {
char** memory;
int i;
memory = calloc(sizeof(char*), 20);
for(i = 0; i < 20; i++) {
memory[i] = calloc(sizeof(char), 33);
}
memory[0] = "Mem 0";
memory[1] = "Mem 1";
memory[2] = "Mem 2";
printf("memory[1] = %s\n", memory[1]);
test_memory(&memory);
return 0;
}
gives me the following output:
memory[1] = Mem 1
Mem 0
Mem 1
Mem 2
which is what I want, but making a local version of the memory is useless because I need to be able to change the values of the original memory from the function which I can only do by dereferencing the pointer to the original 2d char array.
I don't understand why I should get a seg fault on the second time round, and I'd be grateful for any advice.
Many thanks
Joe

Answer 1

Try:
printf("%s\n", (*memory_ref)[i]);
The current version is equivalent to
*(*(memory_ref + i));
This is because the [] operator has higher precedence than the dereference *. Which means that when i is larger than 0 you try to read memory after the char*** temporary memory_ref
The second version is equal to (*memory_ref)[i] which means that you will be indexing the correct memory.
EDIT: The reason it works on the first iteration is because:
*(*(memory_ref + 0)) == *((*memory_ref) + 0)

Answer 2

This looks like a precedence issue. The [] is getting evaluated first, and the * second. What you would need is something like the following for your first code example ...
printf("%s\n", (*memory_ref)[i]);

Answer 3

The quick fix is to use
printf("%s\n", (*memory_ref)[i]);
But I suspect you have an problem in the lines
memory[0] = "Mem 0";
These do not copy the string "Mem 0" into your memory array. They make memory[i] point to the string.
You need to explicitly copy the data using strncpy
e.g.
char s = "Mem 0";
strncpy( memory1, s, max( strlen(s) + 1, 29) ) // max of 30= (29 char + '\0' as that is the row length you allocated - better to #define this as a constant