Thread: Help me in simple programs

I have 2 problem--

Code:

#include<stdio.h>
void main()
{
int x=33,y=10;
printf("%d");
}

why printf always prints the last initialized variable.

Code:

#include<stdio.h>
void main()
{
int x=2,i;
i=x++*2/++x*2;
printf("%d",i);
}

how above expression is evaluated to 4..
please give brief explanation.

1. Variables are passed to functions on the stack. Local variables are declared on the stack. The last initialised variable was the last thing pushed to the stack before the call to printf. (Probably, it could just happen that the same value was on the stack, you'd have to look at the assembler output to know 100%).

2. The result of that expression is undefined. It could yield absolutely any result from 4 to opening up your disk drive or turning off your computer. It just happens to be 4 this particular time that you have run it.

thanks!! But in second program it always prints 4 no matter what the value of x has. i want to know -how??

1. Compiler should have given an error - you are not passing an expected argument. Output, if it compiles, is indeterminate.
2. You may change the same variable twice because order of execution is not defined. Therefore output is indeterminate.

I hope these are not examples given to teach C. They are clearly just plain wrong.