sizeof static arrays
Counting elements in a static array with sizeof is convenient, but arrays are second class citizens in C/C++ and they silently decay to pointers at the first chance. So, code like this won’t work as intended:
#include <stdio.h>
void func(int * foo)
{
printf("func : %d\n", sizeof(foo));
}
int main()
{
int foo[20];
printf("main : %d\n", sizeof(foo));
func(foo);
}
Because inside the function we get the size of a pointer, not the size of the array it points to, the result will be:
main : 80
func : 4
Okay, but is there a way to go around this? Let’s try to change the function. Unfortunately, all of the following attempts are equivalent and will still display the size of a pointer.
void func(int * foo)
void func(int foo[])
void func(int foo[20])
void func(int foo[40]))
Which is not only inconvenient but downright dangerous in the last case, because the compiler doesn’t care about the complete array definition and won’t issue a warning if you are calling the function with an array of the wrong size, and reading the code can be misleading.
So what? Here comes the trick…
void func(int (&foo)[20])
Meet the static array reference, this improbable syntax will carry the original array type around, and sizeof will return the desired value of 20 times the size of an integer. And it won’t compile if the array has the wrong size, sweet.
Sources:
- Reddit topic (link)
- Kevin Heifner blog (link)
