Dudu it works fine for variables.....not for data types.shows
compiletime error...check this code.
#include<iostream>
//#define size(T)(*((char*)&T+1-*(char*)&T));
#define SIZEOF(var) (size_t)(&var+1) - (size_t)(&var)
using namespace std;
int main()
{
char a;
cout<<sizeof(char); //original sizeof .... it outputs 1
cout<<SIZEOF(char);
cin.sync();
cin.get();
return 0;
}
by the way what that sync() do?
On 5/31/10, sharad kumar <[email protected]> wrote:
> #include<iostream>
> //#define size(T)(*((char*)&T+1-*(char*)&T));
> #define SIZEOF(var) (size_t)(&var+1) - (size_t)(&var)
>
> using namespace std;
> int main()
> {
> char a;
> cout<<SIZEOF(a);
> cin.sync();
> cin.get();
> return 0;
> }
>
>
>
> On Mon, May 31, 2010 at 1:14 PM, debajyotisarma
> <[email protected]>wrote:
>
>> This is not about algorithms,but related to C programming.
>>
>> How to implement sizeof operator?
>>
>> macro for this
>> #define my_sizeof(a) (char*)(&a+1)-(char*)&a
>>
>> this works fine of variables
>> int a;
>> printf("%d",my_sizeof(a)); //or even for user defined
>> structures
>>
>> but it will not work for data types
>>
>> like
>>
>> printf("%d",my_sizeof(int));
>>
>> so please get another solution.
>> function will be preferable.not macro
>>
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>
>
> --
> yezhu malai vaasa venkataramana Govinda Govinda
>
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