c program

why does this program output 3,2,515?

Code:

#include<stdio.h>
int main()
{
    union a
    {
        short int i;
        char ch[2];
    };
    union a u;
    u.ch[0]=3;
    u.ch[1]=2;
    printf("%d %d %d\n",u.ch[0],u.ch[1],u.i);
    return 0;
}

because you posted in the C++ forum when this is clearly C code. did you really not see the C forum directly below this one? you should learn the difference between C and C++. it is fairly significant.

that being said, a short int is (usually) the same size as a char[2], but they represent totally different things in memory. what you really have is a short int sharing memory space with a char[2], and since your array contains 3 and 2, you get (2 * 256) + 3 = 515.

Moved thread

Try hex output to help with illustration

Code:

   printf("%x %x %x %d \n", u.ch[0], u.ch[1], u.i, u.i );

/*output:
3 2 203 515

Your answer lies in how data is stored in bits.
Here's an illustration of your two bytes with the 3 an 2 in them in binary.
See how the bit values are combined to determine tha two byte (short) value:

        2            3
0000 0010    0000 0011    <- bits with 1 are "set" or "on"

       52    1631 8421  <- each set bit place is worth this much. (read vertically)
       15    2426
       26    8

512  <- add up the bit values.
  2
  1
---
515

*/

this is clearly C code

which part of this program is incompatible with C++? seems to compile fine on VC++ and G++ with all warnings enabled.

Originally Posted by dmh2000

which part of this program is incompatible with C++? seems to compile fine on VC++ and G++ with all warnings enabled.

This part is NOT proper C++. Real C++ programmers use <cstdio> instead; unless using an ancient C++ Compiler.

Code:

#include<stdio.h>

Tim S.
Note: I am not a real C++ Programmer and I know the difference between C and C++ Headers.

You could always try
printf("Answer=%lu\n", (unsigned long)sizeof('a') );

Within a C program and a C++ program.