sizeof integer array

[rahul8143]

hello,
why following 2 codes does not give same result that is 10?
1)
#include <stdio.h>
#define DIM(array) sizeof(array)/sizeof(int)
void main()
{
int arr[10];
printf("The dimension of the array is %d", DIM(arr));
}
2)
#include <stdio.h>
int DIM(int array[])
{
return (sizeof(array)/sizeof(int));
}
void main()
{
int arr[10];
printf("The dimension of the array is %d", DIM(arr));
}

[Suman]

ra*******@gmail .com wrote:

hello,
why following 2 codes does not give same result that is 10?
1)
#include <stdio.h>
#define DIM(array) sizeof(array)/sizeof(int) #define DIM(array) sizeof array / sizeof array[0]
is probably better. void main() use int main() {
int arr[10];
printf("The dimension of the array is %d", DIM(arr));
}
2)
#include <stdio.h>
int DIM(int array[]) should ideally return size_t. {
return (sizeof(array)/sizeof(int));
} array is no longer the object you have in main, just a copy. void main()
{
int arr[10];
printf("The dimension of the array is %d", DIM(arr));
}

[Suman]

ra*******@gmail .com wrote:

hello,
why following 2 codes does not give same result that is 10?
1)
#include <stdio.h>
#define DIM(array) sizeof(array)/sizeof(int)
void main()
{
int arr[10];
printf("The dimension of the array is %d", DIM(arr));
}
2)
#include <stdio.h>
int DIM(int array[])
{
return (sizeof(array)/sizeof(int)); From the draft of the C99 standard:
When applied to a parameter declared to have array [...] type,
the sizeof operator yields the size of the adjusted (pointer) type.
}
void main()
{
int arr[10];
printf("The dimension of the array is %d", DIM(arr));
}

[Antonio Contreras]

ra*******@gmail .com wrote:

hello,
why following 2 codes does not give same result that is 10?
1)
#include <stdio.h>
#define DIM(array) sizeof(array)/sizeof(int)
void main()
{
int arr[10];
printf("The dimension of the array is %d", DIM(arr));
}
2)
#include <stdio.h>
int DIM(int array[])
{
return (sizeof(array)/sizeof(int));
}
void main()
{
int arr[10];
printf("The dimension of the array is %d", DIM(arr));
}

When you pass an array as a parameter to a function it decays into a
pointer to the first element of the array. In the second case, inside
the DIM function, sizeof(array) is equivalent to siezof(int *), which
is the reason why the second method does not give the correct answer.

BTW, it would've been nice of your part to post the results of each
program.

[Krishanu Debnath]

Suman wrote:

ra*******@gmail .com wrote:

<snip>

2)
#include <stdio.h>
int DIM(int array[])

should ideally return size_t.

{
return (sizeof(array)/sizeof(int));
}

array is no longer the object you have in main, just a copy.

Nonsense. array decay to pointer to first element of actual 'array'
in function parameter context.

Krishnau

[Suman]

Krishanu Debnath wrote:

Suman wrote:

ra*******@gmail .com wrote:

<snip>

2)
#include <stdio.h>
int DIM(int array[])

should ideally return size_t.

{
return (sizeof(array)/sizeof(int));
}

array is no longer the object you have in main, just a copy.

Nonsense. array decay to pointer to first element of actual 'array'
in function parameter context.

Poor wording on my part, which is why I posted a second time.

[Martin Ambuhl]

ra*******@gmail .com wrote:

hello,
why following 2 codes does not give same result that is 10?
1)
#include <stdio.h>
#define DIM(array) sizeof(array)/sizeof(int)
void main()

^^^^
You're dead already.

[Emmanuel Delahaye]

ra*******@gmail .com wrote on 05/08/05 :

why following 2 codes does not give same result that is 10?

Because an array is not a pointer (and reverse).

--
Emmanuel
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