Why the answer is 15 here?

问题

For the following code snippet:

$a = '5 USD';
$b = 10;
echo $a + $b;

The answer is 15. But in the variable $a if 5 is in between 'USD' or after 'USD' the output is 10. Why it is so? Can somebody explain?

回答1:

From php.net:

When a string is evaluated in a numeric context, the resulting value and type are determined as follows.

If the string does not contain any of the characters '.', 'e', or 'E' and the numeric value fits into integer type limits (as defined by PHP_INT_MAX), the string will be evaluated as an integer. In all other cases it will be evaluated as a float.

The value is given by the initial portion of the string. If the string starts with valid numeric data, this will be the value used. Otherwise, the value will be 0 (zero). Valid numeric data is an optional sign, followed by one or more digits (optionally containing a decimal point), followed by an optional exponent. The exponent is an 'e' or 'E' followed by one or more digits.

Literally:

$a + $b means numeric + numeric.
"5 USD" starts with a valid numeric data, so PHP converts it into 5.
"USD 5" or "U5SD" starts with not valid numeric data, so PHP converts it into 0.

UPDv1:

<?php
header('Content-Type: text/plain');

function plus($a, $b){
    echo $a, ' + ', $b, ' = ', $a + $b, PHP_EOL;
}

plus('5 frogs', 3);   // 5 + 3 = 8
plus('frogs: 5', 3);  // 0 + 3 = 3
plus('f5rogs', 3);    // 0 (not HEX)  + 3 = 3
plus('0xF5', 3);      // 245 (HEX)    + 3 = 248
plus('0011b', 3);     // 11 (not BIN) + 3 = 14
plus('1E5', '1.2xx'); // 100000 (FLOAT) + 1.2 (FLOAT) = 100001.2
plus('true', 2);      // 0 (not BOOL)   + 2 = 2
?>

Also, check out this: php string number concatenation messed up.

UPDv2:

There is "no way" for PHP to typecast string value to octal, regardless of zero-fill effect. Still, as mentioned before, PHP able to typecast string to hexadecimal.

<?php
header('Content-Type: text/plain');

function plus($a, $b){
    echo $a, ' + ', $b, ' = ', $a + $b, PHP_EOL;
}

plus(008, 12);     // Invalid octal, PHP assumes it is 0. Result: 12.
plus('008', 12);   // Invalid octal? No, it is decimal. Result: 20.
plus(0x0F, 1);     // Valid hexadecimal. Result: 16.
plus('0x0F', 1);   // Valid hexadecimal. Result: 16.
plus('0x0X', 1);   // Invalid hexadecimal, PHP assumes it is 0. Result: 1.
?>

It is not mentioned in "string to number conversion" docs.

回答2:

See PHP Manual Language.types.string.conversion

String conversion to numbers

When a string is evaluated in a numeric context, the resulting value and type are determined as follows.

If the string does not contain any of the characters '.', 'e', or 'E' and the numeric value fits into integer type limits (as defined by PHP_INT_MAX), the string will be evaluated as an integer. In all other cases it will be evaluated as a float.

The value is given by the initial portion of the string. If the string starts with valid numeric data, this will be the value used. Otherwise, the value will be 0 (zero). Valid numeric data is an optional sign, followed by one or more digits (optionally containing a decimal point), followed by an optional exponent. The exponent is an 'e' or 'E' followed by one or more digits.

Examples can be found in manual (Link above)

$foo = 1 + "bob-1.3e3";           // $foo is integer (1)
$foo = 1 + "bob3";                // $foo is integer (1)
$foo = 1 + "10 Small Pigs";       // $foo is integer (11)

Answering your question

$a = '5 USD'; //string starts with number so implicit cast to int it will be 5
$b = 10;
echo $a + $b; // = 15

In the other example you wrote

$a = 'USD 5'; //string does not start with number so implicit cast to int it will be 0
$b = 10;
echo $a + $b; // = 10

For more information on this conversion, see the Unix manual page for strtod(3).

回答3:

Because

php > echo (int)"a";
0

So if you feed php with

php > "USD 5" + "10"

it will cast both operands to integers

php > (int)"USD 5" + (int)"10"

and thus you will receive result of 0 + 10

来源：https://stackoverflow.com/questions/16851078/why-the-answer-is-15-here

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