两个 int 的百分比？

发布于 2024-12-11 04:37:04 字数 489 浏览 0 评论 0原文

我想要得到两个整数，一个除以另一个得到小数或百分比。如何获得这两个整数的百分比或小数？（我不确定这是否正确......我可能还很远......）例如：

int correct = 25;
int questionNum = 100;
float percent = correct/questionNum *100;

这就是我认为我可以做到的方式，但它不起作用...我想将小数（如果有的话）转换为 100 中的百分比，例如在本例中它是 %25。有什么想法吗？

这是正确的代码（感谢 Salvatore Previti！）：（

float correct = 25;
float questionNum = 100;
float percent = (correct * 100.0f) / questionNum;

顺便说一句，我正在制作一个项目，使用它来进行测验检查程序，这就是为什么我需要百分比或小数）

原文

I want to get two ints, one divided by the other to get a decimal or percentage.
How can I get a percentage or decimal of these two ints?
(I'm not sure if it is right.. I'm probably way off...)
for example:

int correct = 25;
int questionNum = 100;
float percent = correct/questionNum *100;

This is how I thought I could do it, but it didn't work... I want to make the decimal (if there is one) into a percent out of 100 for example in this case it is %25. any ideas anyone?

Here is the correct code (thanks to Salvatore Previti!):

float correct = 25;
float questionNum = 100;
float percent = (correct * 100.0f) / questionNum;

(btw, I am making a project using this for a quiz checking program that is why I need the percentage or decimal)

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

如梦亦如幻 2024-12-18 04:37:04

如果您不添加 .0f ，它将被视为整数，并且整数除法确实与浮点除法有很大不同:)

float percent = (n * 100.0f) / v;

如果您需要其中的整数，您可以当然可以再次将 float 或 double 转换为整数。

int percent = (int)((n * 100.0f) / v);

如果您知道 n 值小于 21474836（即 (2 ^ 31 / 100)），则可以使用整数运算完成所有操作。

int percent = (n * 100) / v;

如果你得到 NaN 是因为无论你做什么，当然你都不能除以零......这是没有意义的。

If you don't add .0f it will be treated like it is an integer, and an integer division is a lot different from a floating point division indeed :)

float percent = (n * 100.0f) / v;

If you need an integer out of this you can of course cast the float or the double again in integer.

int percent = (int)((n * 100.0f) / v);

If you know your n value is less than 21474836 (that is (2 ^ 31 / 100)), you can do all using integer operations.

int percent = (n * 100) / v;

If you get NaN is because wathever you do you cannot divide for zero of course... it doesn't make sense.

回复收藏 0 原文

π浅易 2024-12-18 04:37:04

两个选项：

在乘法之后进行除法：

int n = 25;
int v = 100;
int percent = n * 100 / v;

在除法之前将 int 转换为 float

int n = 25;
int v = 100;
float percent = n * 100f / v;
//Or:
//  float percent = (float) n * 100 / v;
//  float percent = n * 100 / (float) v;

Two options:

Do the division after the multiplication:

int n = 25;
int v = 100;
int percent = n * 100 / v;

Convert an int to a float before dividing

int n = 25;
int v = 100;
float percent = n * 100f / v;
//Or:
//  float percent = (float) n * 100 / v;
//  float percent = n * 100 / (float) v;

回复收藏 0 原文

陪我终i 2024-12-18 04:37:04

其中之一必须是进入的浮点数。确保这一点的一种可能方法是：

float percent = (float) n/v * 100;

否则，您将执行整数除法，截断数字。另外，您应该使用double，除非有充分的理由使用float。

您将遇到的下一个问题是，某些百分比可能看起来像 24.9999999999999%，而不是 25%。这是由于浮点表示的精度损失造成的。你也必须决定如何处理这个问题。选项包括 DecimalFormat 来“修复”格式化或 BigDecimal 来表示精确值。

One of them has to be a float going in. One possible way of ensuring that is:

float percent = (float) n/v * 100;

Otherwise, you're doing integer division, which truncates the numbers. Also, you should be using double unless there's a good reason for the float.

The next issue you'll run into is that some of your percentages might look like 24.9999999999999% instead of 25%. This is due to precision loss in floating point representation. You'll have to decide how to deal with that, too. Options include a DecimalFormat to "fix" the formatting or BigDecimal to represent exact values.

回复收藏 0 原文