PHP parser: braces around variables

Question

I was wondering, how is the semantics of braces exactly defined inside PHP? For instance, suppose we have defined:

$a = \"foo\";

then what

Answer 1

The brackets delimit where the variable name ends; this example should speak for itself.

$a = "hi!"

echo "$afoo";  //$afoo is undefined

echo "${a}foo";  //hi!foo

echo "{$a}foo";  //hi!foo

Also, this should spit out a warning; you should use

${'a'}

Otherwise it will attempt to assume a is a constant.

Answer 2

Also you can use braces to get Char in the position $i of string $text:

$i=2;
$text="safi";

echo $text{$i}; // f

Answer 3

There are a lot of possibilities for braces (such as omitting them), and things get even more complicated when dealing with objects or arrays.

I prefer interpolation to concatenation, and I prefer to omit braces when not necessary. Sometimes, they are.

You cannot use object operators with ${} syntax. You must use {$...} when calling methods, or chaining operators (if you have only one operator such as to get a member, the braces may be omitted).

The ${} syntax can be used for variable variables:

$y = 'x';
$x = 'hello';
echo "${$y}"; //hello

The $$ syntax does not interpolate in a string, making ${} necessary for interpolation. You can also use strings (${'y'}) and even concatenate within a ${} block. However, variable variables can probably be considered a bad thing.

For arrays, either will work ${foo['bar']} vs. {$foo['bar']}. I prefer just $foo[bar] (for interpolation only -- outside of a string bar will be treated as a constant in that context).