arrays again :-)

[al3x4nder]

look this code:

#include <stdio.h>
#define length(arr) sizeof(arr)/sizeof(arr[0])

void func(int arr[]){
printf("func(): length(arr) = %d\n", length(arr));
}

void main(void){
int i, arr[] = {1,2,3};
printf("main(): length(arr) = %d\n", length(arr));
func(arr);
}

and, please, explain me, why length() macro
works with differences in different context?
stupid question? oh...

 

[Chris Dollin]

look this code:

#include <stdio.h>
#define length(arr) sizeof(arr)/sizeof(arr[0])

void func(int arr[]){

As is well-known, declaring a function argument as a simple array is
treated as declaring it as a /pointer/ argument, so that this is the
same as writing `void func( int *arr )`.

printf("func(): length(arr) = %d\n", length(arr));

Hence this will /not/ print the length of the array.

}
void main(void){

Declaring `main` as returning `void` is unnecesarily non-portable;
don't do it.

 

[Antonio Contreras]

look this code:

#include <stdio.h>
#define length(arr) sizeof(arr)/sizeof(arr[0])

void func(int arr[]){
printf("func(): length(arr) = %d\n", length(arr));
}

void main(void){
int i, arr[] = {1,2,3};
printf("main(): length(arr) = %d\n", length(arr));
func(arr);
}

and, please, explain me, why length() macro
works with differences in different context?
stupid question? oh...

When you pass an array as a parameter to a function you're really
passing a pointer to its first element. In fact when an array is used
in an expression except as the operand of the sizeof or the & operator,
the array identifier allways decays into a pointer to the first element
of the array, including when the arrays is one parameter in a function
call.

The declaration of func you gave is totally equivalent to:

void func (int *arr)
[...]

Inside the function arr is of type int*, not of type int[3]. So inside
the function, the macro evaluates to:

length(arr) = sizeof(arr) / sizeof(arr[0]) = sizeof (int*) / sizeof
(int) = some value that will vary from implementation to
implementation. 1 & 2 are usual values, but you can expect about
anything.

In main the macro expands to:

length(arr) = sizeof(arr) / sizeof(arr[0]) = sizeof (int[3]) / sizeof
(int) = 3 irregardless of implementation

HTH

 

[Jens Theisen]

look this code:

#include <stdio.h>
#define length(arr) sizeof(arr)/sizeof(arr[0])

void func(int arr[]){
printf("func(): length(arr) = %d\n", length(arr));
}

void main(void){
int i, arr[] = {1,2,3};
printf("main(): length(arr) = %d\n", length(arr));
func(arr);
}

Within func, arr is of type int* and it's size is that of that pointer,
not of that of the array. You will have to pass the length of the array
along with it I'm afraid. That's one of the reasons c strings are null
terminated: That's the way to determine the length of that array.

The output of func is platform dependet. I guess it will usually be 1 or
2.

Jens

 

[August Karlstrom]

look this code:

#include <stdio.h>
#define length(arr) sizeof(arr)/sizeof(arr[0])

The defensive approach is

#define length(arr) (sizeof (arr) / sizeof ((arr)[0]))

void func(int arr[]){
printf("func(): length(arr) = %d\n", length(arr));
}

Only a pointer to the first element of arr is passed to func; the length
needs to be passed separately. That's why you see function signatures
such as

int sum(int a[], int len);


August

 

[al3x4nder]

August Karlstrom напиÑав:

look this code:

#include <stdio.h>
#define length(arr) sizeof(arr)/sizeof(arr[0])

The defensive approach is

#define length(arr) (sizeof (arr) / sizeof ((arr)[0]))

void func(int arr[]){
printf("func(): length(arr) = %d\n", length(arr));
}

Only a pointer to the first element of arr is passed to func; the length
needs to be passed separately. That's why you see function signatures
such as

int sum(int a[], int len);


August

thanks