Thread: Explanation on a simple program

hey everyone,
Can anyone explain to me why x=4 in the following code?

Code:

#include <stdio.h>


int main()
{
    int x,y,z;
    
    x=2;
    y=2;
    
    x=2*(y++);
    z=2*(++y);
    
    printf("x=%d\ny=%d\nz=%d\n", x,y,z);
}

Yes - but this should be a simple enough of an exercise to be able to solve yourself.

Try figuring it out and tell us what you think the value of 'x' should be.

Originally Posted by Matticus

Yes - but this should be a simple enough of an exercise to be able to solve yourself.

Try figuring it out and tell us what you think the value of 'x' should be.

from what i see:
x=2;
x= 2*3=6; because y++=3;

Note that there is a distinct difference between "++y" and "y++". The first is a "pre-increment" and the second is a "post-increment".

"Pre-increment" increments the value before it is used in an expression.
"Post-increment" increments the value after it is used in an expression.

Does this help clear up the confusion, or do you need further explanation?

(3.12a)

Originally Posted by Matticus

Note that there is a distinct difference between "++y" and "y++". The first is a "pre-increment" and the second is a "post-increment".

"Pre-increment" increments the value before it is used in an expression.
"Post-increment" increments the value after it is used in an expression.

Does this help clear up the confusion, or do you need further explanation?

(3.12a)

okay so why is the result 4 when it should be 8 then.
since ++y is already 3 before it's used then y++ so y=4
2*4=8
correct me if i'm wrong please

since ++y is already 3 before it's used then y++ so y=4

But in the program you posted, "y++" comes first.

In that case, the current value of 'y' is used before it is incremented.

So on line 11, x = 2 * 2 = 4. Now that the expression is complete, 'y' is incremented to three.

ohh okay so if i understand

x=2*(y++)= 2*2
y=3;

is that the case?

Yes, it seems you have it!
Now analyze line 12 and see if you understand how that result is determined.

okay well since now y=3 and we have ++y which is initialized before it's use y=4 therefor answer is 8

Correct! Well done!

(Note: The terminology in your post should be "incremented" and not "initialized")

Originally Posted by Matticus

Correct! Well done!

(Note: The terminology in your post should be "incremented" and not "initialized")

thanks for the help

You're welcome.