(is_finite($float)) is equivalent to (!is_infinite($float) && !is_nan($float)), i.e. a number can only be one of finite, infinite and NaN. You don't need to check both is_infinite() and is_nan() to see if a number is invalid or out of range.
<?php
$finite = 42;
$infinite = log(0);
$nan = acos(2);
var_dump(is_finite($finite), is_infinite($finite), is_nan($finite)); // true, false, false
var_dump(is_finite($infinite), is_infinite($infinite), is_nan($infinite)); // false, true, false
var_dump(is_finite($nan), is_infinite($nan), is_nan($nan)); // false, false, true
?>
is_finite
(PHP 4 >= 4.2.0, PHP 5, PHP 7, PHP 8)
is_finite — 判断浮点数是否是有效的有限值
返回值
如果 num 不是 NAN、INF
或 -INF，那么为 true，否则为 false。
示例
示例 #1 is_finite() 示例
<?php
$float = 1.2345;
var_dump($float, is_finite($float));
$nan = sqrt(-1);
var_dump($nan, is_finite($nan));
$inf = 1e308 * 2;
var_dump($inf, is_finite($inf));
?>以上示例会输出：
float(1.2345) bool(true) float(NAN) bool(false) float(INF) bool(false)
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User Contributed Notes 1 note
Daniel Klein ¶
7 years ago
