Printing last element in an array using if statement - c

Question

I am currently trying to learn about using pointers and functions together in C, which I don't think is easy.
I am trying to print the last element in an array, it actually does the opposite and prints the first element.
I know people normally use for loops, but I can't figure out how to do that with exactly this kind of problem and therefore I thought that I would try it out with an if statement instead.
Edit:
Why is if statement not working in this case? It seems logic that it should work...
My main.c file:
#include <stdio.h>
#include <stdlib.h>
#include "functions.h"
#define Size 7
int main(void)
{
int array1[] = { 11, 88, 5, 9, 447, 8, 68, 4 };
maxValue(array1, Size);
return 0;
}
My functions.h file:
#pragma once
#ifndef FUNCTIONS_H
#define FUNCTIONS_H
int maxValue(const int *, int);
#endif
My functions.c file:
#include "functions.h"
#include <stdio.h>
#include <stdlib.h>
int maxValue(const int *array1, int Size)
{
int max = array1[0];
if (max < array1[Size]) {
Size++;
max = array1[Size];
}
printf("Max value: %d \n", max);
}

Answer 1

Why is if statement not working in this case? It seems logic that it should work...? Because here
if (max < array1[Size]) { }
Size is defined as 7 and you are comparing array1[0] with array1[7] i.e 11 < 4 -> false, hence it doesn't enter into if block, so the last printf executes and that prints max. But its not a correct logic if if blocks becomes true then further Size++ will cause accessing out of bound array elements which cause undefined behavior.
int maxValue(const int *array1, int Size)
{
int max = array1[0];
if (max < array1[Size]) { /* 11 < 4 false, skips if block */
//Size++; /* this is wrong as Size++ here and next accessing array1[Size] cause UB due to accessing out of bound array element */
max = array1[Size];
}
printf("Max value: %d \n", max); /* max is stills array1[0] i.e 11 */
}

Answer 2

Let's simulate what the CPU does when it enters the maxValue function with those arguments. 1. The variable max is assigned the value of array1[0], which is 11.
2. If max (11) is less than array1[7] (4). It is not, so the if block is not executed.
3. Print max: print 11.
Another thing: Your program causes undefined behaviour. Let's take an example where array1[0] is 3, instead of 11. The if block will be executed (3 < 4), so:
Size is incremented to 8.
max is assigned array1[8]. Since the last index in array1 is 7 (that is how you declared the array), you are accessing a memory adress which you are not supposed to access. This is undefined behaviour.

Answer 3

The names maxValue() and max are misleading and confusing what you are trying to do. lastValue() and last would make much more sense.
However what you are trying to do makes no sense in C because arrays are of known length, so you can access the last element directly:
int array1[] = { 11, 88, 5, 9, 447, 8, 68, 4 };
int array_length = sizeof(array1) / sizeof(*array1) ;
printf("Lastvalue: %d \n", array1[array_length - 1] ) ;
However you cannot do this in a function because arrays are not first class data types in C and when passed to a function will "degrade" to a simple pointer without any information regarding the size of the array pointed to. The calling function having the size information must pass that too (as you have done, but then appeared to get very confused):
void printLast( int* array, int length )
{
printf( "Lastvalue: %d \n", array1[length - 1] ) ;
}
It is difficult to see why you thought you might need any other code or what your maxValue() function is intended to achieve. The "logic" which you say "should work" is thus:
If value of the first array element is less than the value of the last array element, then print the undefined value one past the end of the array; otherwise print the first element of the array.
If you wanted to print the last element, then you simply print it, the value of the first element has nothing to do do with it. Either way you should not index past the end of the array - that value is undefined.