preincrement and post increment evaluation problem

[siree sha]

void main()
{
 int a=1;
printf("%d",a++ + a++);
}

the output for this is 3

but when same post increment is written as

void main()
{
 int a=1;
 int j;
clrscr();
 j=a++ + a++;
printf("%d",j);
}

output is shown as 2.
can anybody clarrify?

[weaknessforcats]

There is no rule the compiler has to follow as to when it changes the value of a variable in a statement. Therefore, if you change the variable more then once in the same statement, different compilers will give different results. It's what is called indeterminate results.

[Banfa]

Not just different compilers and its not just changing the value more than once. Accessing a variable more than once between sequence points where 1 or more of the accesses is to change the variables value results in undefined behaviour. So a + a++is as undefined as a++ + a++

Since the behaviour is undefined in theory (or though rarely in practice) the result could be different every-time you run the program.