last number in a function array - c

Question

I want to write a function where I have a given array and number N. The last occurrence of this number I want to return address. If said number cannot be found I want to use a NULL-pointer
Start of the code I've made:
int main(void) {
int n = 3;
int ary[6] = { 1,3,7,8,3,9 };
for (int i = 0; i <= 6; i++) {
if (ary[i] == 3) {
printf("%u\n", ary[i]);
}
}
return 0;
}
result in command prompt:
3
3
The biggest trouble I'm having is:
it prints all occurrences, but not the last occurrence as I want
I haven't used pointers much, so I don't understand how to use the NULL-pointer

Answer 1

I see many minor problems in your program:
If you want to make a function, make a function so your parameters and return types are explicit, instead of coding directly in the main.
C arrays, like in most languages, start the indexing at 0 so if there are N element the first has index 0, then the second has 1, etc... So the very last element (the Nth) has index N-1, so in your for loops, always have condition "i < size", not "i <= size" or ( "i <= size-1" if y'r a weirdo)
If you want to act only on the last occurence of something, don't act on every. Just save every new occurence to the same variable and then, when you're sure it was the last, act on it.
A final version of the function you describe would be:
int* lastOccurence(int n, int* arr, int size){
int* pos = NULL;
for(int i = 0; i < size; i++){
if(arr[i] == n){
pos = &arr[i]; //Should be equal to arr + i*sizeof(int)
}
}
return pos;
}
int main(void){
int n = 3;
int ary[6] = { 1,3,7,8,3,9 };
printf("%p\n", lastOccurence(3, ary, 6);
return 0;
}
Then I'll add that the NULL pointer is just 0, I mean there is literally the line "#define NULL 0" inside the runtime headers. It is just a convention that the memory address 0 doesn't exist and we use NULL instead of 0 for clarity, but it's exactly the same.

Answer 2

Bugs:
i <= 6 accesses the array out of bounds, change to i < 6.
printf("%u\n", ary[i]); prints the value, not the index.
You don't actually compare the value against n but against a hard-coded 3.
I think that you are looking for something like this:
#include <stdio.h>
int main(void)
{
int n = 3;
int ary[6] = { 1,3,7,8,3,9 };
int* last_index = NULL;
for (int i = 0; i < 6; i++) {
if (ary[i] == n) {
last_index = &ary[i];
}
}
if(last_index == NULL) {
printf("Number not found\n");
}
else {
printf("Last index: %d\n", (int)(last_index - ary));
}
return 0;
}
The pointer last_index points at the last found item, if any. By subtracting the array's base address last_index - ary we do pointer arithmetic and get the array item.
The cast to int is necessary to avoid a quirk where subtracting pointers in C actually gives the result in a large integer type called ptrdiff_t - beginners need not worry about that one, so just cast.

Answer 3

First of all, you will read from out of array range, since your array last element is 5, and you read up to 6, which can lead in segmentation faults. #Ludin is right saying that you should change
for (int i = 0; i <= 6; i++) // reads from 0 to 6 range! It is roughly equal to for (int i = 0; i == 6; i++)
to:
for (int i = 0; i < 6; i++) // reads from 0 to 5
The last occurrence of this number I want to return as address.
You are printing only value of 3, not address. To do so, you need to use & operator.
If said number cannot be found I want to use a NULL-pointer
I don't understand, where do you want to return nullpointer? Main function can't return nullpointer, it is contradictory to its definition. To do so, you need to place it in separate function, and then return NULL.
If you want to return last occurence, then I would iterate from the end of this array:
for (int i = 5; i > -1; i--) {
if (ary[i] == 3) {
printf("place in array: %u\n", i); // to print iterator
printf("place in memory: %p\n", &(ary[i])); // to print pointer
break; // if you want to print only last occurence in array and don't read ruther
}
else if (i == 0) {
printf("None occurences found");
}
}

Answer 4

If you want to return an address you need yo use a function instead of writing code in main
As you want to return the address of the last occurence, you should iterate the array from last element towards the first element instead of iterating from first towards last elements.
Below are 2 different implementations of such a function.
#include <stdio.h>
#include <assert.h>
int* f(int n, size_t sz, int a[])
{
assert(sz > 0 && a != NULL);
// Iterate the array from last element towards first element
int* p = a + sz;
do
{
--p;
if (*p == n) return p;
} while(p != a);
return NULL;
}
int* g(int n, size_t sz, int a[])
{
assert(sz > 0 && a != NULL);
// Iterate the array from last element towards first element
size_t i = sz;
do
{
--i;
if (a[i] == n) return &a[i];
} while (i > 0);
return NULL;
}
int main(void)
{
int n = 3;
int ary[] = { 1,3,7,8,3,9 };
size_t elements = sizeof ary / sizeof ary[0];
int* p;
p = g(n, elements, ary); // or p = f(n, elements, ary);
if (p != NULL)
{
printf("Found at address %p - value %d\n", (void*)p, *p);
}
else
{
printf("Not found. The function returned %p\n", (void*)p);
}
return 0;
}

Answer 5

Working on the specified requirements in your question (i.e. a function that searches for the number and returns the address of its last occurrence, or NULL), the code below gives one way of fulfilling those. The comments included are intended to be self-explanatory.
#include <stdio.h>
// Note that an array, passed as an argument, is converted to a pointer (to the
// first element). We can change this in our function, because that pointer is
// passed BY VALUE (i.e. it's a copy), so it won't change the original
int* FindLast(int* arr, size_t length, int find)
{
int* answer = NULL; // The result pointer: set to NULL to start off with
for (size_t i = 0; i < length; ++i) { // Note the use of < rather than <=
if (*arr == find) {
answer = arr; // Found, so set our pointer to the ADDRESS of this element
// Note that, if multiple occurrences exist, the LAST one will be the answer
}
++arr; // Move on to the next element's address
}
return answer;
}
int main(void)
{
int num = 3; // Number to find
int ary[6] = { 1,3,7,8,3,9 }; // array to search
size_t arrlen = sizeof(ary) / sizeof(ary[0]); // Classic way to get length of an array
int* result = FindLast(ary, arrlen, num); // Call the function!
if (result == NULL) { // No match was found ...
printf("No match was found in the array!\n");
}
else {
printf("The address of the last match found is %p.\n", (void*)result); // Show the address
printf("The element at that address is: %d\n", *result); // Just for a verification/check!
}
return 0;
}

Answer 6

Lots of answers so far. All very good answers, too, so I won't repeat the same commentary about array bounds, etc.
I will, however, take a different approach and state, "I want to use a NULL-pointer" is a silly prerequisite for this task serving only to muddle and complicate a very simple problem. "I want to use ..." is chopping off your nose to spite your face.
The KISS principle is to "Keep It Simple, St....!!" Those who will read/modify your code will appreciate your efforts far more than admiring you for making wrong decisions that makes their day worse.
Arrays are easy to conceive of in terms of indexing to reach each element. If you want to train in the use of pointers and NULL pointers, I suggest you explore "linked lists" and/or "binary trees". Those data structures are founded on the utility of pointers.
int main( void ) {
const int n = 3, ary[] = { 1, 3, 7, 8, 3, 9 };
size_t sz = sizeof ary/sizeof ary[0];
// search for the LAST match by starting at the end, not the beginning.
while( sz-- )
if( ary[ sz ] == n ) {
printf( "ary[ %sz ] = %d\n", sz, n );
return 0;
}
puts( "not found" );
return 1; // failed to find it.
}
Consider that the array to be searched is many megabytes. To find the LAST match, it makes sense to start at the tail, not the head of the array.
Simple...