In the PHP manual, operator precedence section, there is this example:

```
// mixing ++ and + produces undefined behavior
$a = 1;
echo ++$a + $a++; // may print 4 or 5
```

I understand the behavior is undefined because of the following reason:

Since `x + y = y + x`

the interpreter is free to evaluate `x`

and `y`

for addition in any order in order to optimize speed and/or memory. I concluded this after looking at the C code example in this article.

My question is that the output of the above mentioned PHP code should be `4`

no matter which way the expression and sub-expressions are evaluated:

- op1 = ++$a => $a = 2, op1 = 2; op2 = $a++ => op2 = 2, $a = 3; 2 + 2 = 4
- op1 = $a++ => op1 = 1, $a = 2; op2 = ++$a => op2 = 3, $a = 3; 1 + 3 = 4

Where does the 5 come from? Or should I learn more about how the operators work?

Edit:

I have been staring at Incrementing/Decrementing Operators section but still could not figure out why 5.

++$a: Pre-increment — Increments $a by one,

returns $a.then

$a++: Post-increment — Returns $a,increments $a by one.then

```
a = 1;
++ (preincrement) gives a = 2 (higher precedence than +, and LR higher precedence than postincrement)
++ (postincrement) gives a = 3 (higher precedence than +)
+ (add) gives 2 + 3 = 5
```

$a is initially set to 1. The ++$a then preincrements $a before using it in the formula, setting it to 2, and pushing that value onto the lexer stack. The $++ is then executed, because incrementor has a higher precedence than +, and that value is also pushed that result onto the lexer stack; and the addition that then takes place adds the lexer stack’s 2 result to the lexer stack’s 3 result giving a result of 5, which is then echoed. The value of $a once the line has executed is 3.

**OR**

```
a = 1;
++ (preincrement) gives a = 2 (higher precedence than +, and LR higher precedence than postincrement)
+ (add) gives 2 + 2 = 4 (the value that is echoed)
++ (postincrement) gives a = 3 (incremented __after__ the variable is echoed)
```

$a is initially set to 1. When the formula is parses, the ++$a preincrements $a, setting it to 2 before using it in the formula (pushing the result to the lexer stack). The result from the lexer stack and the current value of $a are then added together giving 4; and this value is echoed. Finally, $a is postincremented, leaving a value of 3 in $a.

### Answer：

Yes it will give you 5 because the **right side operator works first by its priority/precendence** and after that the sum(+) operator will work.

So first increment makes it to 2 and second makes it to 3 and after that both will sum and outputs you the result as 5

```
$result = ++$a + $a++;
```

`++$a`

outputs as 2

`$a++`

outputs as 2 3 only but internally it wll be incremented.

finally sum will happens as 2+3 = 5

### Answer：

Mark, I believe you are wrong!

Post-increment: Returns $a, then increments $a by one. (from documentation)

So there is no way to get $a value of 3 in sum operation.

Tags: phpphp