asked on

sizeof() operator for a structure

there r 2 ques:
1.what will be the size of an empty structure?
structure test
{
};
main()
{
struct test *p;
p = (struct test *)malloc(sizeof(struct test));
printf("%d",sizeof(struct test));
}

It prints 1.But wat does that mean.

2.and if i print like printf("%d"); gives me -1? why is this so.

Where can i get answers for questions like this.

thanks

SOLUTION

Sjef Bosman

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ASKER CERTIFIED SOLUTION

migoEX

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migoEX

sorry, sjef_bosman
I didn't see your answer

Sjef Bosman

Happens in EE all the time, I'm used to it :) Sometimes good, somewhat reassuring to the asker.

Sjef

Sys_Prog

struct Data {
} ;

int main () {
printf ( "%d", sizeof ( struct Data ) ) l
}

would print 1....because of the following

Consider before printf, u declare an array of the struct Data

struct Data arr[10] ;

As u know, array elemnets are stored in contiguous memrory locations...so each elemnt should have a distinct address
Thus in our case, if the size would have been 0, then each elemnet would not have had a unique addr

Hence a non zero size

Amit

avizit

#include <stdio.h>
struct Data {
} ;

int main () {
printf ( "%d", sizeof ( struct Data ) ) ;
}

=========
the above prints 0 on my machine. ( linux /gcc)

Sjef Bosman

I even modified the code to
struct Data {
} data;

int main () {
printf ( "%d\n", sizeof ( struct Data ) );
printf ( "%d\n", sizeof ( data ) );
}

and Linux prints two lines with zeroes... The outcome is not defined in the C language, and hence compiler-dependent.

SOLUTION

avizit

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Sys_Prog

Oh.....Sorry Guys........I didn't notice the TA...I thought I was answering C++ question

Amit