Hello,
i learning to code without any prior experience to it whatsoever. Here is my confusion regarding the function sizeof().
So, when declaring an array, the name of array is just a pointer to its first element. Then, why is the output in the following code is 12 and not 4, which is the actual size of the variable "movie1"?
Code:int main() { char movie1[] = "Hello World"; printf("%d", sizeof(movie1)); return 0; }
A couple of things.
Despite the function-like syntax, sizeof is an operator (like + or -), not a function.
The parentheses are only required when referring to types.
Eg.
> So, when declaring an array, the name of array is just a pointer to its first elementCode:#include<stdio.h> int main() { char movie1[] = "Hello World"; printf("%zd\n", sizeof movie1); return 0; }
Not quite.
The name of the array is the name of the array.
The "array becomes a pointer to the first element" happens in most circumstances, but not all.
One example as you have found is when using an array name in a sizeof expression.
Arrays and Pointers
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
Thank You, Salem.
