Should I Follow Sizeof Operator's Output Or Manually Executed Code?

Question

2.00/5 (1 vote)

See more:

Guys, basically i know the size of variables. 4/2 for int, 2 for short and 1 for char. But I just tried to find the size of a variable without sizeof().

C#

#include<stdio.h>
int main()
{
int a, b, *p1=&a, *p2=&b;
char c, d;
printf("size of a is %d\n",sizeof(a));
printf("value of p1 is:%u p2 is:%u\n",p1,p2);
printf("size of a is %d\n",(p2-p1));
return 0;
}

Using sizeof() I got 4bytes but using this which I got in many websites I got 1. What's the matter? Which one should I follow? There are many sites discussing this topic but I cannot understand!

This is my output :

CSS

size of a is 4
value of p1 is:3220066288 p2 is:3220066292
size of a is 1

Posted 31-Jul-14 1:35am

Jeevan83

Updated 31-Jul-14 1:38am

v2

Add a Solution

Comments

Philippe Mori 2-Aug-14 8:49am

p2 - p1 does not make any sense at all. It is undefined because they do not point to variables in the same array and thus the compiler is free to select any location it want.

By the way, substracting pointer give the number of elements between them. You have to learn basic things like that before using pointers arithmetic.

Jeevan83 2-Aug-14 10:06am

Okay sir :)

2 solutions

Add a Solution

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

CPallini · Accepted Answer · 2014-07-31T02:28:00

Of course you have to trust sizeof.

You code outputs '1' because of pointer arithmetic. You should write your test this way:

C

#include <stdio.h>

int main()
{
    int a[2]; // just array items are warranted to be contiguous in memory

    char * p, *q;

    printf("using sizeof operator: %d\n", sizeof(a[0]));

    p = (char *)&a[0];
    q = (char *)&a[1];

    printf("using pointers: %d\n", (q-p));

    return 0;
}

OriginalGriff · Accepted Answer · 2014-07-31T02:17:00

Solution 1

sizeof returns the size in bytes: and that is absolutely what you should follow.
When you got the result:

size of a is 1

you were subtracting pointers, which returns "the number of the item the pointer points to" between them. So if you have two pointer-to-int values, and you subtract them, you get the number of int values between the two pointers, not the number of bytes.

Think about it:

C++

p1 = p1 + 1;

Should move you to the next int value: not the second byte in the first int! :laugh:

Posted 31-Jul-14 2:17am

OriginalGriff

Comments

Philippe Mori 2-Aug-14 8:45am

And by the way, in his case p2 - p1 is not defined as they don't point to a single array. The compiler is free to select any location for a and b variables.

OriginalGriff 2-Aug-14 9:47am

Indeed - I have a problem with any pointer arithmetic:
p1 - p2
giving number of elements should be the same as:
p1 + (0 - p2)
and I have absolutely no idea what adding two pointers should give! Not to mention what the negative of a pointer is physically either... :laugh:

Should I Follow Sizeof Operator's Output Or Manually Executed Code?

2 solutions

Solution 2

Solution 1

Add your solution here

Preview 0

Should I Follow Sizeof Operator's Output Or Manually Executed Code?

2 solutions

Solution 2

Solution 1

Add your solution here

Preview 0

Existing Members

...or Join us