将两个整数相除时如何获得浮点值？(PHP)
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How do I get a float value when dividing two integers? (PHP)
提问by user1869257
Hi I am trying to divide two integers ex: 12/13 but I always get a whole integer 1 not a decimal number.
嗨，我正在尝试将两个整数相除，例如：12/13，但我总是得到一个整数 1 而不是十进制数。
I tried type casting the values to float before hand with no success.
我尝试先将值强制转换为浮动，但没有成功。
Basically all I want is a decimal result like: 0.923...
基本上我想要的只是一个小数结果，如：0.923 ...
$x = 12;
$y = 13;
echo $value = $x / $y; //Would like to see 0.923 not 1
采纳答案by CS?
Under normal circumstances your code should return the floating value 0.923076...
在正常情况下，您的代码应该返回浮点值0.923076......
The reason you get a rounded integer might be because you have your ini settingfor "precision"set to 0, to fix this either edit your php.inior use ini_set("precision", 3);in your code before the calculation.
您获得四舍五入整数的原因可能是因为您将ini settingfor"precision"设置为0，要解决此问题，请在计算前编辑php.ini或ini_set("precision", 3);在代码中使用。
Another way to workaround this is to use BCmath:
解决此问题的另一种方法是使用BCmath：
echo $value=bcdiv($a, $b, 3);
And yet another way without using any extension is to use a little math trick by multiplying the value you want to divide by 1000to get 3 decimals.
This way you'll divide 12000by 13and the whole part will be 923, then since you multiplied by 1e3 insert a comma/dot before the last most 3 places.
另一种不使用任何扩展的方法是使用一个小数学技巧，通过乘以您想要除以的值1000来得到3 decimals.
这样，你会除以12000通过13与整体的一部分会923，然后因为你之前的最后一次最多3位乘以1E3插入逗号/点。
function divideFloat($a, $b, $precision=3) {
$a*=pow(10, $precision);
$result=(int)($a / $b);
if (strlen($result)==$precision) return '0.' . $result;
else return preg_replace('/(\d{' . $precision . '})$/', '.', $result);
}
echo divideFloat($a, $b); // 0.923
echo divideFloat($a, $b); // 0.923
回答by Raffaello
echo $value = $x / (float) $y;
if you cast the variable $yas float the interpreter use the floating point division instead of integer division.
如果将变量转换$y为浮点数，则解释器使用浮点除法而不是整数除法。
because it's default asumption to use integer division on two integer variables.
因为默认假设对两个整数变量使用整数除法。
It was different if you used $y = 13.0(a float variable as denominator):
The results is always a float number
如果您使用 $y = 13.0（作为分母的浮点变量）则不同：结果始终是浮点数
回答by felix021
All other answers are NOT RIGHT, because PHP's division will always return float values, as is stated clearly in official manual PHP: Arithmetic Operators, except for cases when two operands are both integers which can be evenly divided.
所有其他答案都不正确，因为 PHP 的除法将始终返回浮点值，正如官方手册PHP: Arithmetic Operators 中明确说明的那样，除非两个操作数都是可以被均分的整数。
In fact, the question is wrong: the code should produce 0.923..., as was expected from the questioner.
事实上，问题是错误的：代码应该产生 0.923...，正如提问者所期望的那样。
Sarcastically, the voted-down answer (came from @BulletProof47) is the only other one which is just NOT WRONG (but also meaningless). Who knows what he was thinking, but I bet everybody knows why it was voted down :D
讽刺的是，被否决的答案（来自@BulletProof47）是唯一一个没有错误（但也毫无意义）的答案。谁知道他在想什么，但我打赌每个人都知道为什么它被否决了：D
In case who is interested, the underlying function which does division in php is div_function, located in Zend/zend_operators.c, shown below:
如果有兴趣的话，php 中做除法的底层函数是div_function，位于 中Zend/zend_operators.c，如下图所示：
ZEND_API int div_function(zval *result, zval *op1, zval *op2 TSRMLS_DC) /* {{{ */
{
zval op1_copy, op2_copy;
int converted = 0;
while (1) {
switch (TYPE_PAIR(Z_TYPE_P(op1), Z_TYPE_P(op2))) {
case TYPE_PAIR(IS_LONG, IS_LONG):
if (Z_LVAL_P(op2) == 0) {
zend_error(E_WARNING, "Division by zero");
ZVAL_BOOL(result, 0);
return FAILURE; /* division by zero */
} else if (Z_LVAL_P(op2) == -1 && Z_LVAL_P(op1) == LONG_MIN) {
/* Prevent overflow error/crash */
ZVAL_DOUBLE(result, (double) LONG_MIN / -1);
return SUCCESS;
}
if (Z_LVAL_P(op1) % Z_LVAL_P(op2) == 0) { /* integer */
ZVAL_LONG(result, Z_LVAL_P(op1) / Z_LVAL_P(op2));
} else {
ZVAL_DOUBLE(result, ((double) Z_LVAL_P(op1)) / Z_LVAL_P(op2));
}
return SUCCESS;
...
回答by BulletProof47
Just use $value = (float)($x/$y); //Result will in float.
只需使用 $value = (float)($x/$y); //结果会浮动。
Cheers!
干杯!
