PHP Increment Decrement operator understanding and considerations

Increment decrement operator

Before increment ++++ $a $ A increments by 1 and then returns $ A

After increment + + + $a first return $ A, then $ A since 1

Before decrement----$a $ A self minus 1 and then return $ A

After decrement--$a--returns $ A, then $ A minus 1

The first thing to note: The increment/decrement operator does not affect the Boolean value. Decreasing the null value also has no effect, but the result of incrementing null is 1.

In other words: in an increment/decrement operation, the operand is not converted to an integer before it is calculated. If the operand is a Boolean value, the result is returned directly.

Increment/Decrement Boolean value:

$a = true;var_dump (+ + $a); BOOL (true) $a = true;var_dump (--$a); BOOL (true) $b = False;var_dump (+ + $b); BOOL (false) $b = False;var_dump (--$b); BOOL (FALSE)

Increment/decrement null:

$a = null;var_dump (+ + $a); Int (1) $a = null;var_dump (--$a); Null

When dealing with arithmetic operations of character variables, PHP inherits the Perl habit, not the C.

For example, in Perl

$a = ' Z '; $a + +;

Will change $ A to ' AA ', and in C,

A = ' Z '; a++;

Will turn a into ' [' (the ASCII value of ' Z ' is 90, ' [' ASCII value is 91).

Note that character variables can only be incremented, cannot be decremented, and only plain letters (A-Z and A-Z) are supported.

For example:

$a = "9d9"; Var_dump (+ + $a);  String (3) "9E0"

But here's another trap:

$a = "9E0"; echo + + $a;  10

Install the above rules, you should output 9E1, but this is the output of 10. Why?

If we write like this, the big people will know why.

$a = "9E0"; Var_dump (+ + $a);  Float (10)

The type of $a is floating point, that is, 9E0 is the scientific notation for floating-point numbers, that is, 9 * 10^0 = 9, and 9 self-increment, the result is of course 10.

Reference: string conversion to numeric value

And now the question is:

$l = "Z99"; $l + +;

How much is the result? The result is "AA00" according to the Perl language rules.

There is one more important note:

Increment/Decrement Other character variables are not valid and the original string does not change.

That's not going to explain.

One final note:

$a = ' 012 '; $a ++;var_dump ($a);

The result is ' 013 '? 13? 11?

The result of this paragraph is int (13), and the string ' 012 ' is not treated as an octal.

$a = 012;   Octal, decimal is 10$b = "012"; Convert to Integer to decimal 12

What if it starts with 0x?

$a = ' 0x1A '; $a ++;var_dump ($a);   Int (27)

0 is not considered to be octal, but the beginning of 0x is considered hexadecimal.

In the official PHP document, there is another octal trap for integer integers:

Var_dump (01090); Octal 010 = Decimal 8

This is explained in the manual as:

Warning if an illegal number (that is, 8 or 9) is passed to the octal number, the remaining digits are ignored.