Using printf( ) in a for loop - c

Question

When attempting this (code below), the console doesn't even request an input value, then spits out a random number (likely from a number previously stored at the location).
Why does this not work and how can i fix it?
int main( ) {
int arr[3];
for(int i = sizeof(arr); i <= 0; i--) {
scanf("%d", &arr[i]);
}
printf("%d", arr[2]);
return 0;
}

Answer 1

There are two things that are wrong.
Assuming you want the number of elements in the array, it is done, using:
size_t len = sizeof(arr) / sizeof(*arr)
sizeof gives the actual size (number of bytes allocated for arr.
You should start with len - 1 and not len.
NOTE: Array indexing is 0 based not 1 based, so the array elements are indexed from 0 to 2 But you would have tried to access arr[3], which can result in undefined behaviour.
You wrote i <= 0. So, i starts from let's say 2, is 2 <= 0 ? NO!
Hence it will never go inside the loop. The correct condition is i >= 0
int len = sizeof(arr) / sizeof(*arr);
for(int i = len - 1; i >= 0; i--)
Well, I don't know why you are taking reverse order input, but a general convention is to take input using:
size_t len = sizeof(arr)/sizeof(*arr);
for (i = 0; i < len; i++)
{
// take input
}
EDIT:
From other comments it seems that you don't understand the for loop.
Have a look in this answer
Please comment for any further clarification.

Answer 2

i <= 0
the code can never enter the loop since the initial value of i is greater than zero.

Answer 3

It is important to note that in C, other than languages like Java/Python, you must explicitly know the length of the array, sizeof will NOT give you the amount of items in the array.
int main() {
int arr[3];
int itemsInArray = 3;
for(int i = itemsInArray-1; i >= 0; i--) {
scanf("%d", &arr[i]);
}
printf("%d", arr[2]);
return 0;
};

Answer 4

Since i-- will decrease the value of i , and the condition for loop is i <=0 to start the loop the i must be 0 or negative.Since arr[3] will return 12(3 elements and each has 4 bytes(int has 4 bytes)), the value will be posivite,greater than 0 so we need to change the loop condition to check if i is positive
#include <stdio.h>
int main( ) {
int arr[3]={0};
int i = sizeof(arr)/sizeof(arr[0]) -1;
for(; i >= 0; i--) {
scanf("%d", &arr[i]);
}
printf("%d", arr[2]);
return 0;
}

Answer 5

There are a couple of issues with your code: first of all, sizeof(arr) won't return "3" as you probably thought; "arr" is a pointer to arr[0], so you are requesting the size of an int pointer.
Secondly, i <= 0 prevent the loop to even be executed.
Finally, please set array to zero while declarating, as best practice, ie:
int arr[3] = {0};
EDIT: i wrong-spelled my thoughts: you are requesting the size of the whole memory area allocated for the array.
Comments below are corrected though.