Understanding and precautions for PHP increment and decrement operators

Increment and decrement operator

increment before ++++$a$a increments by 1, then return $a

increment afterward++$a++ returns first $a, then $a increases by 1

before decrementing----$a$a decreases by 1, then returns to $a

and then decrements--$a--first returns $a , then $a decrements by 1

The first note: Increment/decrement operators do not affect Boolean values. Decrementing a NULL value has no effect, but increasing NULL results in 1.

In other words: During increment/decrement operations, the operands will not be converted into integers before operation. 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 on character variables, PHP follows the habits of Perl instead of C of.

For example, in Perl

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

will turn $a into 'AA', while in C,

a = 'Z';
a++;

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

Note that character variables can only be incremented, not decremented, and only support pure letters (a-z and A-Z).

For example:

$a="9D9"; 
var_dump(++$a);  // string(3) "9E0"

However, there is another trap here:

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

After installing the above rules, 9E1 should be output, but 10 is output here. Why?

If we write it this way, most people will know why. The type of

$a = "9E0"; 
var_dump(++$a);  // float(10)

$a is Floating point, that is to say, 9E0 is the scientific notation of floating point numbers, that is, 9 * 10^0 = 9, incrementing by 9, Of course the result is 10.

Reference:StringConvert to numerical value

Now the question comes again:

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

What is the result? The result is "AA00" according to the rules of the perl language.

There is another note:

Increasing/decrementing other character variables will be invalid, and the original string will not change.

I won’t explain this.

Last note:

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

Is this result '013'? 13? 11?

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

$a = 012;   // 八进制，十进制为 10
$b = "012"; // 转换为整数为十进制 12

What if it starts with 0x?

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

Those starting with 0 are not considered octal, but those starting with 0x are considered hexadecimal.

在 PHP 官方文档中 Integer 整型 还有另一个八进制陷阱：

var_dump(01090); // Octal 010 = Decimal 8

The manual explains this as:

Warning
如果向八进制数传递了一个非法数字（即 8 或 9），则后面其余数字会被忽略。

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