为什么$ x = 5; $ X +++ $ X ++; 在PHP中等于11？

Question

According to the opcodes it should be 12. Am I getting it wrong?

number of ops:  8
compiled vars:  !0 = $x
line    #* E I O op                  fetch      ext   return  operands
-------------------------------------------------------------------------
  3     0  E >   EXT_STMT                                                 
        1        ASSIGN                                         !0, 5
  5     2        EXT_STMT                                                 
        3        POST_INC                               ~2      !0
        4        POST_INC                               ~3      !0
        5        ADD                                    ~4      ~2, ~3
        6        ECHO                                           ~4
  7     7      > RETURN                                         1

branch: #  0; line:     3-    7; sop:     0; eop:     7; out1:  -2
path #1: 0,

Edit

Also ($x++)+($x++); returns the same result (11). Actually this was the main reason for the question and opcode investigation.

Answer 1

It took me a few reads, but $x=5; $x++ + $x++; $x=5; $x++ + $x++; works like this:

In the case of a $x++, it first 'gets used', then increased:

• Set $x to 5
• Place $x onto stack (which is 5)
• Increment( ++ ) ($x is now 6, stack=[ 5 ])
• Add $x onto stack (stack=[5, 6 ], so 5+6 -> $x=11)
• Adding is done, that outcome is 11
• Increment $x( ++ ) (which is isn't used further, but $x is now 7)

Actually, in this specific example, if you would echo $x; it would output 7. You never reassign the value back to $x, so $x=7 (you incremented it twice);

Answer 2

$x = 5;
$a = $x++ + $x++;

the expression line will be executed like this: 1st occurrence of $x++ in the statement will increment $x value by 1 so it will become 6 and

in 2nd occurrence, $x will be having value 6;

So $a = 5 + 6;

So final result $a will be 11.

Answer 3

++ has higher precedence than + operator

(x++) will return the value of x first then increment it by 1

$x = 2
$x++ // return 2, then increment it to 3

x+++x++ is evaluated like the following

1. Get x value first which is 5
2. Then it will be incremented to 6
3. But first x value will be 5 because (x++) statement will return 5 first then increment the value
4. Then + operator is encountered
5. Next x will have 6 as value not 7 for the same reason (x++) will return the x value first and then increment it
6. So 5+6 is 11
7..At the end, x value will be 7

Same goes for ($x++)+($x++)

grouping operator () has left to right associatevity. First ($x++) executes first.

$x = 5
($x++) returns 5 and then increment $x by 1. Same as before. 

then last ($x++) executes. It returns 6 and then increment $x to 7

so same 5+6 // 11 is returned back

Answer 4

The post increment operator increment the variable, but returns its old value.

So $x++ is equivalent to:

($temp = $x, $x = $x + 1, $temp)

When you do it twice in an expression, it's like:

echo ($temp1 = $x, $x = $x + 1, $temp1) + ($temp2 = $x, $x = $x + 1, $temp2);

The first part sets $temp1 = 5 and increments $x to 6 .

The second part sets $temp2 = 6 and increments $x to 7.

Then it does $temp1 + $temp2 and echoes the result, which is 5 + 6 = 11 .

Answer 5

You are using the post-increment operator ($x++). If you would like to use the incremented value for the addition you should use the pre-increment operator (++$x).

Therefore if $x = 5

$x++ + $x++ equals 5+6 = 11
++$x + $x++ equals 6+6 = 12
$x++ + ++$x equals 5+7 = 12
$++x + ++$x equals 6+7 = 13

Yet in all cases, x is equal to 7 afterward.

Answer 6

$x = 5;
echo $x++ + $x++;

prints 11 as first $x++ returns 5 and then after that it increments, the second $x++ returns 6 and the only value in incremented. so actual addition is 5+6 which gives 11.

Answer 7

++$x + $x++ will be 12

And ++$x + ++$x will be 13

when you use $x++ $x get +1 as soon it's value is use, but the value that will gonna be use is the one it has before the increment, so when yo do:

$x=5; $x+++$x++;

$x+++$x++ is 11, but $x will be 7