Strange echo, print behaviour in PHP?

Question

The following code outputs 43211, why?

  echo print(\'3\').\'2\'.print(\'4\');

Answer 1

Much of the confusion is due to the placement of parentheses around the arguments to print. As you know, parentheses are optional with language constructs; what you probably didn't know is that they're removed during parsing.

Evaluation order

Let's remove the parentheses first:

echo print '3' . '2' . print '4';

And illustrate the actual order of evaluation:

echo (print ('3' . '2' . (print '4')))
^     ^      ^                     ^
3     2      1--------->>----------1

In the heart of this you will find a concatenation of strings or string representations; this is evaluated first:

'3' . '2' . (print '4')

The first two elements are concatenated:

'32' . (print '4')

Then, the value of (print '4') is evaluated; after printing its argument '4', the return value of print itself is always int(1); this is cast into a string '1' and concatenated with the other elements:

'321'

This concludes the first step. The second step passes the temporary results to another print statement:

print '321'

As before, '321' is printed and now int(1) is returned for the last step:

echo 1

Proof

You can confirm this behaviour when you look at the opcodes that are generated (output column is added for clarity):

line     # *  op          return  operands        output
------------------------------------------------+-------
   1     0  >   CONCAT      ~0      '3', '2'    |
         1      PRINT       ~1      '4'         | 4
         2      CONCAT      ~2      ~0, ~1      | 4
         3      PRINT       ~3      ~2          | 4321
         4      ECHO        ~3                  | 43211

Explanation

• "3" and "2" are concatenated - "32" - and stored into ~0.
• "4" is printed and the return value int(1) is stored into ~1.
• ~0 and ~1 are concatenated - "321" - and stored into ~2.
• "321" is printed and the return value is stored into ~3.
• int(1) is printed as "1" due to string casting.