I am trying to solve this out:
x = 10; y = 0; z = 5;
y = z * x++
y = z * ++x
The instruction maunal (500 pages long) that I am reading all it tells me is that the answer to the x++ problem is:
y = 50 and x = 11.
I know how to get the x++ all you have to do is add an increment of 1.
x = x + 1.
Can someone please help me, because I cant figure out what I am doing wrong! I am not understanding where the book got 50 from!
The problem you are having here deal with post-incrementation and pre-incrementation.
When doing x++, you are post-incrementing... This means that the incrementation will only occur after the statement has been evaluated.
So, given the following code:
x = 10; y = 0; z = 5;
y = z * x++;
JavaScript does this:
x = 10; y = 0; z = 5;
y = z * x++;
// Ignore Post-Increment, Evalutate
y = z * x;
y = 5 * 10;
y = 50;
// Now Increment x - POST-INCREMENT
x = x + 1;
x = 10 + 1;
x = 11;
When doing ++x, you are pre-incrementing... This means that the incrementation will occur before the statement is evaluated:
x = 10; y = 0; z = 5;
y = z * ++x;
// Do Pre-Increment
x = x + 1;
x = 10 + 1;
x = 11;
// Evaluate
y = z * x;
y = 5 * 11;
y = 55;
The result of a post-increment expression is the value before the variable has been incremented.
So this:
y = z * x++;
is roughly equivalent to:
y = z * x; // y = 50
x = x + 1; // x = 11 (the increment is done afterwards)
On the other hand, a pre-increment operation evaluates to the value after the increment. So this:
y = z * ++x;
is roughly equivalent to this:
x = x + 1; // x = 11 (the increment is done first)
y = z * x; // y = 55
x++ increase x by after value is delivered:
y = z * x++ // 5 * 10 // and 10 become 11 after operation
++x increase x by before value is delivered:
y = z * ++x // 5 * 11
fancy names post and pre incrementations
You should understand the difference between post-increment and pre-increment operators here.
In the y = z * x++ case, x if first used in the expression and later incremented. Thus the value of the expression becomes y = 5 * 10 after which the value of x is incremented to 11.
In the other case, the pre-increment operator is used. So it's first incremented and used later. Thus the value becomes 55 in that case.
It is post incrememt operator. x++ use value of x, then increment it ++x ( pre increment) increment x, then use it
++ after the variable is a post-increment operation, that is, x is incremented after the assignment is made.
In other words, y is calculated to be 50 and then x is incremented.
Example:
a * (b++)
It's the same as:
a * b;
b = b + 1; //POST increment
but
a * (++b);
is the same like this:
b = b + 1; //PRE increment
a * b;
First you have
x = 10;
y = 0;
z = 5;
then you do:
y = z * x++ = 5 * 10 = 50
after that you get:
y = 50;
x = 11; (because x = x+1)
then:
y = z * ++x = 5 * 11 = 55
z * x; x = x + 1;, so you only have to write the style that better suits you.