Hi all,
Just a simple question. I am wondering why the result of
echo "test " . 1 + 10 . " 123";
is “10 123”
“test 1” becomes 0… that’s right
But… Is “10 123” considered an integer? 
Thanks!
Good point
Thanks
Yes, you are right. The left most “.” is evaluated first and since it is a string operator it expects both operands to be a strings and the result is a string either. “+” interprets both operands as integers and the result is an integer too. Intval of “test 1” is 0.
But in case:
echo "0.1 test " . 1 + 10 . " 123";
The result is 10.1 123, though (int)“0.1 test” is 0, but now the “+” operator treats both operands as float values.
Well, I would say it goes like this:
step 1.
(string) "test " and (int) 1 is concatenated
result (string) “Test 1”
step 2.
(string) “Test 1” and (int) 10 will be added together.
result (int) 10.
since “test 1” doesnt start with a number, the value will be evaluated as 0
step 3
(int) 10 gets concatenated with (string) " 123", and thus converted into a string.
Result (string) “10 123”
Sure, I know!
The code I’ve posted is a sample of PHP5 Certification “PHP Basics” questions… I was just trying to understand the logic about it, not the sense 
The way I see this, would be like this:
1 string “Test "
1 calculation 1 + 10
1 string " 123”
The . concatenates strings, and the + add numbers together
Mixing the two doesnt really make any sense.
I would have expected you wanted to execute the line like this :
echo "test ". (1 + 10) ." 123";
resulting in “test 11 123”
In fact…
echo (int) "10 123";
results “10”
So, I can answer myself stating both + and . operators have same priority
Then starting from the left…
“test 1” + 10 = 0 + 10 = 10
And then…
10 . " 123" = “10 123”
Am I right?