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;
}
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.
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.
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.
sizeof returns size_t not int 2.) The preferred pattern to calculate the size of an array by the array itself is: char a[42]; size_t s = sizeof arr / sizeof *arr; For completeness: sizeofis an operator and not a function, no need for parenthesis, it does not get "called".size_t len turns out to be 0 In that case if I do int i = len - 1;. What will i contain?T a[N]; this expression sizeof a / sizeof *a; will never by <1 by definition, as the definttion T a[0]; is not allowed by the C Standard. Recent versions of GCC allow it as an extension, but make it to have the size of T.i <= 0
the code can never enter the loop since the initial value of i is greater than zero.
sizeof returns the size in bytes, so i will start at (probably) 12 or 24.
for loop is like the condition of a while loop: As long as it is true, the loop repeats.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;
};
sizeof arr / sizeof arr[0] would work fine.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;
}
arr[3] won't return 3. arr[3] is not a valid element of the array (index 3 is out of bounds).sizeof(arr) does not return 0. What are you talking about?
sizeof(arr) will be 12 or 24 on common C implementations.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.
sizeof(anArray) (if its known as an array like here and not decayed to a pointer), returns the size of the whole array in bytes.
for(int i = sizeof(arr); i <= 0; i--)-->for(int i = sizeof(arr) - 1; i >= 0; i--)sizeof(arr)? What do you thinksizeof(arr)is in this case?