## Monday, July 9, 2007

### How would you find the size of structure without using sizeof()?

Try using pointers

struct MyStruct
{
int i;
int j;
};

int main()
{
struct MyStruct *p=0;
int size = ((char*)(p+1))-((char*)p);
printf("\nSIZE : [%d]\nSIZE : [%d]\n", size);
return 0;
}

Krazy for Amazon said...
This comment has been removed by the author.
Anonymous said...

What if struct size if more than the MAX int value, i.e you can't keep that value in an integer.

Anonymous said...

i just want to know y have u used (char *) typecasting in ur pgm?

Anonymous said...

how will this work ?

subburajr said...

This may give you appx value...still depends on the padding

struct NODE_T p;
char s;
int size = (char*)&p-&s;

Anonymous said...

This doesn't work
for struct test {
int i;
char j;
}

the solution shows the size = 8 but it is 5

Pavel Nikitin said...
This comment has been removed by the author.
Pavel Nikitin said...

struct MyStruct *p=NULL;
p++;
printf("size = %d",p);

vikky said...

typedef struct node
{
int data;
flaot a;
char b;
}node;

void main()
{
node a[2]={0};
printf("size = %d",a[1]-a[0]);
}

Correct me if I am wrong.

jshutzman said...

struct el {
int y;
char s[30];
};

el arr[3];
el *p1=arr;
el *p2=&arr[1];
long a1, a2;
cout << "The address of the first structure element: " << p1 << "\n";
cout << "The address of the second structure element: " << p2 << "\n";
a1=(long)p1; a2=(long)p2;
printf("Size is: %x in Hexadecimal\n", a2-a1);

//If you simply print p2-p1 you'll get 1 (pointer //arithmetics). I used long to match pointer size of 4 //bytes (different machines could differ, I use //Macbook pro, dual core (2009) 64 bit OS.
//This gives 24 in Hex which is 36 in Dec. so there is //little overhead for alignment on the structure //elements

PunkFloyd said...

Short and sweet.

struct moop {...};

size_t my_size = (size_t)&((moop*)NULL)[1];

sandeep kumar said...

Thanks for the info. It is very helpful.

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#include
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last = &a[1].x;
start = &a[0].x;
printf("%d",last-start);
}

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