datatype deduction

Without the prior knowledge that integers store 2 byte in C,or char store 4 byte,float reserves 'so and so bytes',...etc
Is it possible to deduce directly from a C program which datatype store what amount of free memory?

I tried this code but it was without any results.
(As array of any datatype stores consecutive memory locations I declared )

Code:

int a[2];
char b[2];
float c[2];

(Then I substracted the adress and printed )

Code:

printf("For integer data type : Bytes stored = %d",(&a[1])-(&a[0]));

The result should come 4 but it was in vain,it was showing 1 as output which is wrong.....

In case of deduction of other datatype also ,I went on with the same logic ,but the result in each case was erronious

Please help,
Refer if some shorter methods available...

Use sizeof to see the size of a data type

Code:

int i;
float f;
printf("%d %d", sizeof(i), sizeof(f));

Code:

printf("Size of integer in bytes = %d\n",sizeof(int));

Originally Posted by progmateur

Without the prior knowledge that integers store 2 byte in C,or char store 4 byte,float reserves 'so and so bytes',...etc
Is it possible to deduce directly from a C program which datatype store what amount of free memory?

Use the sizeof operator.

Originally Posted by progmateur

The result should come 4 but it was in vain,it was showing 1 as output which is wrong.....

You're subtracting pointers, so of course the difference is 1 since the elements are adjacent.

But in memory loc ,it skips 4 bytes for the adjascent location.
I was differentiating the location adress !!
It should show 4!

When I was printing like this:
printf("%x %x",&a[1],&a[0]);

The output was "FFFE FFFA"

without using any inbuilt functions .....please!

Originally Posted by progmateur

But in memory loc ,it skips 4 bytes for the adjascent location.
I was differentiating the location adress !!
It should show 4!

That is how subtraction of addresses work, but not how subtraction of pointers work. That said, there is a way to get your idea to work: you need to subtract pointers of a different type.

Originally Posted by progmateur

without using any inbuilt functions .....please!

Use the sizeof operator.

If I store the pointer adress to another variable,yet there was an error.
Please provide me the soln without sizeof function!

Originally Posted by progmateur

If I store the pointer adress to another variable,yet there was an error.

What did you try?

Originally Posted by progmateur

Please provide me the soln without sizeof function!

But sizeof isn't a function!

By the way, why do you want to do this?

Please Hold on Ill be posting the code in a minute!......

Smells like a (badly written) homework assignment to me....

I also distinctly remember this exact problem on this board about a month or two ago.

Originally Posted by laserlight

What did you try?


But sizeof isn't a function!

By the way, why do you want to do this?

Code:

#include <stdio.h>

main()
{
int *a,d,e;
char *b;
float *c;
d=&a;
a++;
e=&a;
printf("%d",e-d);
getch();
return 0;
}

There was a nonportable pointer conversion warning!
Please debug it!
It was on in one of our assignments,
Thats why I wanted to do it
There It wasnt mentioned to deduce the datatype programatically
but Im trying just the same

Don't do this: your pointers do not point to anything, so you end up with undefined behaviour. Note that d and e are ints, not pointers to int.

If you insist on the idea of using pointer subtraction, then I suggest that you focus on:

Code:

printf("For integer data type : Bytes stored = %d",(&a[1])-(&a[0]));

Think: what type of pointers must you subtract in order for the subtraction to compute the difference as number of bytes rather than number of ints between the two pointers?

frankly speaking,I have no clue about that
Thats why I posted it in the forum

Sigh. Cast the pointers to char*. It is guaranteed that sizeof(char) == 1, so when you cast the pointers to char*, the subtraction will give the number of bytes.