[英]How can I convert integer into float in Java?
I have two integers x
and y
. 我有两个整数
x
和y
。 I need to calculate x/y
and as outcome I would like to get float. 我需要计算
x/y
,结果我想浮动。 For example as an outcome of 3/2
I would like to have 1.5. 例如,作为
3/2
的结果,我希望有1.5。 I thought that easiest (or the only) way to do it is to convert x
and y
into float type. 我认为最简单(或唯一)的方法是将
x
和y
转换为float类型。 Unfortunately, I cannot find an easy way to do it. 不幸的是,我找不到一个简单的方法来做到这一点。 Could you please help me with that?
你能帮帮我吗?
You just need to cast at least one of the operands to a float: 你只需要转换操作数为浮点的至少一个:
float z = (float) x / y;
or 要么
float z = x / (float) y;
or (unnecessary) 或(不必要的)
float z = (float) x / (float) y;
在进入进一步的计算之前,您只需要将第一个值传递给float:
float z = x * 1.0 / y;
You shouldn't use float unless you have to. 除非必须,否则不应使用浮动。 In 99% of cases, double is a better choice.
在99%的情况下,双倍是更好的选择。
int x = 1111111111;
int y = 10000;
float f = (float) x / y;
double d = (double) x / y;
System.out.println("f= "+f);
System.out.println("d= "+d);
prints 版画
f= 111111.12
d= 111111.1111
Following @Matt's comment. 关注@Matt的评论。
float has very little precision (6-7 digits) and shows significant rounding error fairly easily. float具有非常小的精度(6-7位)并且相当容易地显示出明显的舍入误差。 double has another 9 digits of accuracy.
double有另外9位精度。 The cost of using double instead of float is notional in 99% of cases however the cost of a subtle bug due to rounding error is much higher.
在99%的情况下使用double而不是float的成本是名义上的,但是由于舍入误差导致的细微错误的成本要高得多。 For this reason, many developers recommend not using floating point at all and strongly recommend BigDecimal.
出于这个原因,许多开发人员建议不要使用浮点,强烈推荐BigDecimal。
However I find that double can be used in most cases provided sensible rounding is used . 但是我发现在大多数情况下可以使用double,只要使用合理的舍入 。
In this case, int x has 32-bit precision whereas float has a 24-bit precision, even dividing by 1 could have a rounding error. 在这种情况下,int x具有32位精度,而float具有24位精度,即使除以1也可能具有舍入误差。 double on the other hand has 53-bit of precision which is more than enough to get a reasonably accurate result.
另一方面,double具有53位精度,足以获得相当准确的结果。
// The integer I want to convert //我要转换的整数
int myInt = 100;
// Casting of integer to float //将整数转换为float
float newFloat = (float) myInt
Here is how you can do it : 以下是如何做到这一点:
public static void main(String[] args) {
// TODO Auto-generated method stub
int x = 3;
int y = 2;
Float fX = new Float(x);
float res = fX.floatValue()/y;
System.out.println("res = "+res);
}
See you ! 再见 !
Sameer: 萨米尔:
float l = new Float(x/y)
will not work, as it will compute integer division of x and y first, then construct a float from it. 将无法工作,因为它将首先计算x和y的整数除法,然后从中构造一个浮点数。
float result = (float) x / (float) y;
Is semantically the best candidate. 在语义上是最好的候选人。
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