Operator Precedence Problem

Consider the following:

Code #1

#include <stdio.h>

int main() {
int a=1, b=2, c=3, n;
n = (( ++ a ) * b ) - ( c -- );
printf("a = %d, b = %d and c = %d\n", a, b, c);
printf("with n = %d", n);
return 0;
}

The output for Code #1 is:

a = 2, b = 2, c = 2
with n = 1

I am trying to figure out exactly how this code is read and performed by the computer.

Now, consider an earlier example that I attempted.

Code #2

#include <stdio.h>

int main() {
int i = 3, j = 8, k;
k = j +++ i;
k /= 2;

printf(%d %d %d", j, i, k);
return 0;
}

the output for Code #2 is:

9 3 5

From this I concluded that the j++ operation takes effect after the line statement k = j +++ i;

This is because if j became 9 during the statement, then the final value of k would be 6.

However this does not seem to match the output of Code #1, as if ++a and c-- did not take effect on the number until after the line, then the value of n would be:

(( 1 ) * 2 ) - ( 3 ) = -1

Furthermore, of both took effect during the line statement, then the value of n would be:

(( 2) * 2 ) - ( 2) = 2

I must be interpreting this wrong. Could someone please help me?

Code:

int a=1, b=2, c=3, n;
n = (( ++ a ) * b ) - ( c -- );

This reads as ((2) * 2) - (3). a is incremented and evaluates to the incremented value, 2. b obviously stays the same. c is decremented, but evaluates to the value before it was decremented since you're doing c-- instead of --c.

I just figured it out because of what you said!

This was what I didn't know:

i++ or i-- is incremented/decremented after an expression is evaluated
++i or --i is incremented/decremented before an expression is evaluated

It all makes sense now

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