问题描述:

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
```

您可能感兴趣的文章：

- drupal adding onchange event to node form
- android - current time to time picker
- reporting services - SSRS & asp.net - passing parameters from .net to ssrs in report viewer
- uiactivityindicatorview - Using UIActivityIndicator in objective-c
- sql - UNION the results of multiple stored procedures
- php - Reading a specific line from a text file
- clr - Any implementation of an Unrolled Linked List in C#?
- Finding Hudson Log Files
- Forward to a payment-gateway together with POST data using cURL (or any other PHP server side solution)
- WCF in Winforms app - is it always single-threaded?

随机阅读：

**推荐内容**-

**热点内容**