为什么这里的答案是 15? [英] Why is the answer 15 here?

问题描述

对于下面的代码片段，答案是 15.

For the following code snippet the answer is 15.

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

但是在变量$a中，如果5在'USD'之间或'USD'之后> 输出为 10.为什么会这样?

But in the variable $a, if 5 is in between 'USD' or after 'USD' the output is 10. Why is it so?

推荐答案

来自 php.net:

在数字上下文中计算字符串时，结果值和类型确定如下.

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

如果字符串不包含任何字符."、e"或E"并且数值符合整数类型限制(定义为PHP_INT_MAX)，字符串将被评估为一个整数.在所有其他在这种情况下，它将被评估为浮点数.

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.

该值由字符串的初始部分给出.如果字符串以有效的数字数据开头，这将是使用的值.否则，该值将为 0(零).有效的数字数据是可选符号，后跟一个或多个数字(可选包含一个小数点)，后跟一个可选的指数.指数是一个e"或E"后跟一位或多位数字.

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.

字面意思:

1. $a + $b 表示 numeric + numeric.
2. "5 USD" 以有效的数字数据开头，因此 PHP 将其转换为 5.
3. "USD 5" 或 "U5SD" 以无效有效数字数据开头，因此 PHP 将其转换进入0.

1. $a + $b means numeric + numeric.
2. "5 USD" starts with a valid numeric data, so PHP converts it into 5.
3. "USD 5" or "U5SD" starts with not valid numeric data, so PHP converts it into 0.

<小时>

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
?>

<小时>

另外，看看这个:php字符串数字连接搞砸了.

UPDv2:

无论零填充效果如何，PHP 都无法"将字符串值类型转换为八进制.尽管如此，如前所述，PHP 能够将字符串类型转换为十六进制.

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.
?>

字符串到数字的转换"文档中未提及.

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