my c++ query for pre and post operatois

[kiritved]

hi,
#include<stdio.h>
#include<conio.h>
int main()
{
int i,j,k;
clrscr();
k=0;
k++;
i=0;
j=i+++i++;
printf("%d %d %d\n",k,i,j);

k++;
i=0;
//j=i++ +++i; this line gives an compiler error
printf("%d %d %d\n",k,i,j);
k++;
i=0;
//j=++i++i; this line gives an compiler error
printf("%d %d %d\n",k,i,j);
k++;
i=0;
j=++i+ ++i;
printf("%d %d %d\n",k,i,j);
k++;
i=0;
//j=++i++;this line gives an compiler error
printf("%d %d %d\n",k,i,j);
k++;
i=0;
j=++i;
printf("%d %d %d\n",k,i,j);
k++;
i=0;
j=+++i;
printf("%d %d %d\n",k,i,j);
k++;
i=0;
j=i+++ ++i;
printf("%d %d %d\n",k,i,j);
k++;
i=0;
j=i+++ i++;
printf("%d %d %d\n",k,i,j);
getch();
return 0;
}
given this program,i fail 2 understand how c++/c compiler handles the pre and post increment operators.
can some one explain me the logic compiler follows?
regards,
kirit ved

[scruggsy]

What do you want to know - what values will be stored in j in each case, or why the compiler rejects the commented lines?

j=++i++i

...doesn't compile because it doesn't make sense.
The pre-increment operator increments the variable's value first, then uses it.

i = 1; j = ++i; //now i==2, j==2

The post-increment operator uses the value, then increments it.

i = 1; j = i++; //now i==2, j==1

The behavior can be undefined, when you increment a variable more than once within one expression.
When in doubt, or when dealing with non-trivial expressions, you're better off doing the increment outside of the expression.

[phiefer3]

I would suggest using parenthesis to control these statements. That way it won't really matter how the compiler wants to interpret the pre and post increment operators. Just make it interpret them the way you want.

[JosAH]

I would suggest using parenthesis to control these statements. That way it won't really matter how the compiler wants to interpret the pre and post increment operators. Just make it interpret them the way you want.

That won't help you much, e.g.

int i=42;
i= (++i)+(i++);
 

The outcome is still undefined because according to the Standard, altering a
modifiable lvalue more than once before a sequence point is reached causes
undifined behaviour.

Parentheses can only alter default precedence of operators, not the order in
which operands are evaluated.

kind regards,

Jos