[Courses] sizeof operator

[James Sutherland]

On 14 Feb 2012, at 17:03, millward wrote:

> When I define an integer array, such as
>  int array[40];
> the sizeof operator can detect the correct 
> number of bytes which make up the array
>  int number_of_bytes = sizeof( array );

It isn't "detecting" anything as such - you told it the array is "40 ints long", the compiler knows how big an 'int' is in bytes. It's the equivalent of asking a sales assistant "how long is this ten foot rope in inches?" - the assistant won't need to look at the rope itself, just know that a foot is 12 inches making the rope 120 inches long.

> How does the sizeof operator do this ?
> I can see how sizeof finds the correct byte 
> size of a  char array since it ends
> with a nul '\0' but as far as I know, the integer
> array has no special character to end it.
>    How does this sizeof operator work ?

If you try asking: "sizeof(int)" you might see what's going on. The compiler isn't measuring the size of a particular variable which holds an int - it *knows* that an int is defined to be a certain number of bytes (usually 4 or 8) on this system.

sizeof never looks at the variable itself - as "sizeof(int)" shows, there might not even be one. It's strlen which looks for a terminating null on strings.

#include <string.h>
#include <stdio.h>

int main(int argc,char **argv)
{
	char woof[10]="meow";
	printf("sizeof(woof)=%ld\n",sizeof(woof));
	printf("strlen(woof)=%ld\n",strlen(woof));
	printf("sizeof(int)=%ld\n",sizeof(int));
	return 0;
}


Output on my machine:
$ gcc -o scratch{,.c} -Wall ; ./scratch 
sizeof(woof)=10
strlen(woof)=4
sizeof(int)=4

I declared 'woof' to be ten chars long, so sizeof reports that as the size - even though it contains 'meow' and a terminating null; it's strlen which counts the characters before the null and answers 4.


James.