Question:
I have the following array :
$array = array(10,20,30);
print_r($array);
Exit:
Array
(
[0] => 10
[1] => 20
[2] => 30
)
If I print with echo before print_r :
echo 'primeiro: ';
print_r($array);
Exit:
primeiro: Array
(
[0] => 10
[1] => 20
[2] => 30
)
If I print concatenating, the print_r is printed before:
echo 'primeiro: ' . print_r($array);
Exit:
Array
(
[0] => 10
[1] => 20
[2] => 30
)
primeiro: 1
Also, print that 1 in front of the primeiro .
Why does it happen ? What is this 1 ?
Answer:
Because print_r returns before:
print_r returns a value. Because of this, echo waits for all operations within its parameters before executing in order to print the value that print_r returned.
This would be analogous to doing a mathematical operation inside echo : first you solve the operation inside, and then demonstrate the value.
Because it returns 1 :
According to the print_r documentation:
When the return parameter is TRUE, this function will return a string. Otherwise, the returned value will be TRUE.
Since the return parameter has not been defined, and it is FALSE by default, the print_r function is returning TRUE. This is converted to "1" within echo .