Is this code well defined by using sizeof operator? - c

Question

#include <stdio.h>
void test(int arr[]);
int main ()
{
int *arr[3];
int ar1[2] = { 1, 2 };
int ar2[3] = { 3, 4, 5 };
int vla[ar2[1]];
arr[0] = ar1;
arr[1] = ar2;
arr[2] = vla;
for (int i = 0; i < 3; i++)
{
test(arr[i]);
}
}
void test(int arr[])
{
printf("%zu\r\n", sizeof(arr));
}
As the title say: is this valid? and if so, is there a way to make test() output different values by using sizeof on function argument handed in like on array?

Answer 1

Yes it is valid and will give the size of an int pointer (that's what the type of the arr parameter is, there are no array parameters in C) in bytes. 1
To get the size of an array in bytes with sizeof, the argument has to be an actual array, not a pointer.
Consequently, it's not possible to get the size of an array passed as a parameter inside a function without passing an extra parameter specifying the size.
So,
is there a way to make test() output different values by using sizeof on function argument handed in like on array?
no. (Without editing the function body.)
1 If you're wondering how it's possible to pass an array as a pointer parameter, it's because of the implicit array-to-pointer conversion, known as array decaying.

Answer 2

Is this:
void test(int arr[]) {
printf("%d\r\n", sizeof(arr));
}
valid?
It is NOT a valid1 code, however it is equivalent to: sizeof(int*), not size of the array, arr, because as you pass the array as an argument it decays to a pointer2 i.e. it looses the information about the size of the whole array. The only way to convey this information into a function is through an additional, second, parameter.
Is there a way to make test() output different values by using sizeof on function argument handed in like on array?
If you want to print the size of the array you are passing, you could use something like: sizeof(array) as an second argument of the function, where the function signature would be:
void test(int arr[], size_t size);
you could insert the above as second argument and get the size of the first argument.
1. There is a type mismatch between size_t(the returning type of sizeof()) and int (the expected type matching the specifier "%d" in the printf(),, which leads to undefined behaviour.
2. arr[] outside the function represents the whole array as an object and that is why if you apply sizeof() you will get the size of the whole array, however, when passed as argument, arr decays to a pointer of the array's first element and if you apply sizeof() you'll get the size of the type of the pointer.

Answer 3

Is this code well defined by using sizeof operator?
Pedantically: No
sizeof() returns type size_t. "%d" expects type int. Using mis-match type results in undefined behavior (UB). Step 1: Use matching specifier
// printf("%d\r\n", sizeof(arr));
printf("%zu\r\n", sizeof(arr));
// or
printf("%d\r\n", (int) sizeof(arr));
The following prints the size of a pointer arr. The code in main() is not relevant.
void test(int arr[]) {
printf("%zu\r\n", sizeof(arr));
}
// or simply: remove (), \r of dubious value
printf("%zu\n", sizeof arr);