var_dump((int) 010); //Output 8
var_dump((int) "010"); //output 10
I am unable to understand the PHP data type conversion method from string to integer.
First one is octal notation so the output is as I expect. But what about the when converting "010" to integer? I expect that it should also output 8.
How can I get 8 as the returned value from var_dump((int) "010");?
Casting to an integer using (int) will always cast to the default base, which is 10.
Casting a string to a number this way does not take into account the many ways of formatting an integer value in PHP (leading zero for base 8, leading "0x" for base 16, leading "0b" for base 2). It will simply look at the first characters in a string and convert them to a base 10 integer. Leading zeroes will be stripped off because they have no meaning in numerical values, so you will end up with the decimal value 10 for (int)"010".
Converting an integer value between bases using (int)010 will take into account the various ways of formatting an integer. A leading zero like in 010 means the number is in octal notation, using (int)010 will convert it to the decimal value 8 in base 10.
This is similar to how you use 0x10 to write in hexadecimal (base 16) notation. Using (int)0x10 will convert that to the base 10 decimal value 16, whereas using (int)"0x10" will end up with the decimal value 0: since the "x" is not a numerical value, anything after that will be ignored.
If you want to interpret the string "010" as an octal value, you need to instruct PHP to do so. intval("010", 8) will interpret the number in base 8 instead of the default base 10, and you will end up with the decimal value 8. You could also use octdec("010") to convert the octal string to the decimal value 8. Another option is to use base_convert("010", 8, 10) to explicitly convert the number "010" from base 8 to base 10, however this function will return the string "8" instead of the integer 8.