unary increment operator execution

[agnesteena]

Can Somebody please explain how the output for the program below comes to 45 and 46 respectively.

#include<stdio.h>
#include<conio.h>
void main()
{
int x=10,y=10;
clrscr();
//printf("%d\n",x+(++x)+x+(x++)+x);
//printf("%d",(x++)+(++x)+x);


x=(++x)+(x++)+x+x;
printf("%d\n%d\n",x,(++y)+(y++)+y+y);
//printf("value of x is %d",x+(++x)+x+(++x));
getch();
}

the output obtained for the above program is 45 and 46.

Thanks in Advance.

[laserlight]

The behaviour is undefined: C FAQ on expressions

In other words, instead of asking how the output came to be that, just don't do it. If you must figure it out, have the compiler output assembly instead and check the assembly output. Alternatively, just come up with some sequence of increments that happens to correspond to the output that you see.

[Hodor]

Well, I dunno. Seems quite readable to me

Edit: Apart from undefined behaviour, can you think of a single reason (just one) why anybody would write code like that? Is this a competition to see who can write the most absurd, incorrect and stupid C code possible?

[Salem]

The code is broken, for the reasons already stated.

I get different answers.

$ cat foo.c
#include<stdio.h>
int main()
{
  int x=10,y=10;
  x=(++x)+(x++)+x+x;
  printf("%d\n%d\n",x,(++y)+(y++)+y+y);
  return 0;
}
$ gcc -Wall foo.c
foo.c: In function ‘main’:
foo.c:5:4: warning: operation on ‘x’ may be undefined [-Wsequence-point]
foo.c:5:4: warning: operation on ‘x’ may be undefined [-Wsequence-point]
foo.c:6:24: warning: operation on ‘y’ may be undefined [-Wsequence-point]
foo.c:6:24: warning: operation on ‘y’ may be undefined [-Wsequence-point]
foo.c:6:24: warning: operation on ‘y’ may be undefined [-Wsequence-point]
$ ./a.out 
45
44

For a more in-depth analysis (more compilers, simpler program), read this thread.
Post and Pre Increment

[agnesteena]

Thank you for the timely reply .....it was informative....