PHP一些优先级的问题

直接看代码

<?php
echo '1'.print(2)+3,"
";

 不错，就是这么简单，但是很少有人能正确回答

我们执行一下

[root@localhost test]# php -dvld.active=1  test7.php    
Finding entry points
Branch analysis from position: 0
Jump found. Position 1 = -2
filename:       /data/www/test/test7.php
function name:  (null)
number of ops:  6
compiled vars:  none
line     #* E I O op                           fetch          ext  return  operands
-------------------------------------------------------------------------------------
   2     0  E >   ADD                                              ~0      2, 3
         1        PRINT                                            ~1      ~0
         2        CONCAT                                           ~2      '1', ~1
         3        ECHO                                                     ~2
         4        ECHO                                                     '%0A'
   3     5      > RETURN                                                   1

branch: #  0; line:     2-    3; sop:     0; eop:     5; out1:  -2
path #1: 0, 
511

 让人很诧异吧

换一种写法

<?php
echo '1',print(2)+3,"
";

 执行结果

[root@localhost test]# php -dvld.active=1  test7.php 
Finding entry points
Branch analysis from position: 0
Jump found. Position 1 = -2
filename:       /data/www/test/test7.php
function name:  (null)
number of ops:  6
compiled vars:  none
line     #* E I O op                           fetch          ext  return  operands
-------------------------------------------------------------------------------------
   2     0  E >   ECHO                                                     '1'
         1        ADD                                              ~0      2, 3
         2        PRINT                                            ~1      ~0
         3        ECHO                                                     ~1
         4        ECHO                                                     '%0A'
   3     5      > RETURN                                                   1

branch: #  0; line:     2-    3; sop:     0; eop:     5; out1:  -2
path #1: 0, 
151

 再换一种

<?php
echo print(2)+3,"
";

 执行结果

[root@localhost test]# php -dvld.active=1  test7.php 
Finding entry points
Branch analysis from position: 0
Jump found. Position 1 = -2
filename:       /data/www/test/test7.php
function name:  (null)
number of ops:  5
compiled vars:  none
line     #* E I O op                           fetch          ext  return  operands
-------------------------------------------------------------------------------------
   2     0  E >   ADD                                              ~0      2, 3
         1        PRINT                                            ~1      ~0
         2        ECHO                                                     ~1
         3        ECHO                                                     '%0A'
   3     4      > RETURN                                                   1

branch: #  0; line:     2-    3; sop:     0; eop:     4; out1:  -2
path #1: 0, 
51

代码

<?php

$a=3;
$b=4;

if($a=3 || $b=4)
{
        $a++;
        $b++;
}

var_dump($a);
var_dump($b);

 结果

[root@dev test]# php -dvld.active=1 get.php 
Finding entry points
Branch analysis from position: 0
Jump found. Position 1 = 3, Position 2 = 5
Branch analysis from position: 3
Jump found. Position 1 = 7, Position 2 = 12
Branch analysis from position: 7
Jump found. Position 1 = 12
Branch analysis from position: 12
Jump found. Position 1 = -2
Branch analysis from position: 12
Branch analysis from position: 5
filename:       /data/www/www.erongtu.com/test/get.php
function name:  (null)
number of ops:  17
compiled vars:  !0 = $a, !1 = $b
line     #* E I O op                           fetch          ext  return  operands
-------------------------------------------------------------------------------------
   3     0  E >   ASSIGN                                                   !0, 3
   4     1        ASSIGN                                                   !1, 4
   6     2      > JMPNZ_EX                                         ~2      3, ->5
         3    >   ASSIGN                                           $3      !1, 4
         4        BOOL                                             ~2      $3
         5    >   ASSIGN                                           $4      !0, ~2
   7     6      > JMPZ                                                     $4, ->12
   8     7    >   POST_INC                                         ~5      !0
         8        FREE                                                     ~5
   9     9        POST_INC                                         ~6      !1
        10        FREE                                                     ~6
  10    11      > JMP                                                      ->12
  12    12    >   SEND_VAR                                                 !0
        13        DO_FCALL                                      1          'var_dump'
  13    14        SEND_VAR                                                 !1
        15        DO_FCALL                                      1          'var_dump'
  14    16      > RETURN                                                   1

branch: #  0; line:     3-    6; sop:     0; eop:     2; out1:   3; out2:   5
branch: #  3; line:     6-    6; sop:     3; eop:     4; out1:   5
branch: #  5; line:     6-    7; sop:     5; eop:     6; out1:   7; out2:  12
branch: #  7; line:     8-   10; sop:     7; eop:    11; out1:  12
branch: # 12; line:    12-   14; sop:    12; eop:    16; out1:  -2
path #1: 0, 3, 5, 7, 12, 
path #2: 0, 3, 5, 12, 
path #3: 0, 5, 7, 12, 
path #4: 0, 5, 12, 
bool(true)
int(5)

 未完待续……