Explain About Type Juggling In Php

  • 5 Feb, 2012

    PHP does not require (or support) explicit type definition in variable declaration; a variable's type is determined by the context in which that variable is used. That is to say, if you assign a string value to variable $var, $var becomes a string. If you then assign an integer value to $var, it becomes an integer.An example of PHP's automatic type conversion is the addition operator '+'. If any of the operands is a float, then all operands are evaluated as floats, and the result will be a float. Otherwise, the operands will be interpreted as integers, and the result will also be an integer. Note that this does NOT change the types of the operands themselves; the only change is in how the operands are evaluated.$foo += 2; // $foo is now an integer (2)$foo = $foo + 1.3; // $foo is now a float (3.3)$foo = 5 + "10 Little Piggies"; // $foo is integer (15)$foo = 5 + "10 Small Pigs"; // $foo is integer (15)If the last two examples above seem odd, see String conversion to numbers.If you wish to change the type of a variable, see settype().If you would like to test any of the examples in this section, you can use the var_dump() function.Note: The behavior of an automatic conversion to array is currently undefined.Since PHP (for historical reasons) supports indexing into strings via offsets using the same syntax as array indexing, the example above leads to a problem: should $a become an array with its first element being "f", or should "f" become the first character of

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