java 为什么二次方程的根结果是 NaN ？(爪哇)
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原文地址: http://stackoverflow.com/questions/7461803/
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Why Quadratic equation's root result is NaN ? (Java)
提问by blaces
Why write out The roots are NaN and NaNin console? I've read about NaN, but I don't find the right solution how can I repair the mistakes... I've tried casting to doublethe discriminantand the rootsbut doesn't work. Can somebody help me, where and what I need to rewrite?
为什么在控制台中写出 根是 NaN 和 NaN？我读过关于 NaN 的文章，但我没有找到正确的解决方案，我该如何修复错误……我试过将判别式和根加倍，但没有用。有人可以帮助我，我需要重写的地方和内容吗？
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
Pattern newlineOrSpace = Pattern.compile(System
.getProperty("line.separator")
+ "|\s");
sc.useDelimiter(newlineOrSpace);
System.out.print("Enter a, b, c: ");
double a = sc.nextDouble();
double b = sc.nextDouble();
double c = sc.nextDouble();
// System.out.format("a = %f, b = %f, c = %f", a, b, c);
double root1;
double root2;
double discriminant;
discriminant = Math.sqrt(b * b - 4 * a * c);
if (discriminant > 0) {
System.out.println("There are no real roots ");
} else {
root1 = (-b + discriminant) / (2 * a);
root2 = (-b - discriminant) / (2 * a);
System.out.println("The roots are " + root1 + " and " + root2);
}
回答by hammar
Math.sqrt(x)returns NaNwhen xis negative, which then propagates through the rest of your code. You'll want to test for negative numbers beforetaking the square root:
Math.sqrt(x)NaN当x为负时返回，然后通过其余代码传播。在取平方根之前，您需要测试负数：
discriminant = b * b - 4 * a * c;
if (discriminant < 0) {
System.out.println("There are no real roots ");
} else {
root1 = (-b + Math.sqrt(discriminant)) / (2 * a);
root2 = (-b - Math.sqrt(discriminant)) / (2 * a);
System.out.println("The roots are " + root1 + " and " + root2);
}
回答by Jon Skeet
Firstly, let's get rid of user input as a cause for this - it's much easier if the short but complete program contains all the data we need:
首先，让我们摆脱用户输入作为造成这种情况的原因——如果简短但完整的程序包含我们需要的所有数据，那就容易多了：
public class Test {
public static void main(String args[]) {
showRoots(2.0, 10.0, 2.0);
showRoots(10.0, 1.0, 1.0);
}
private static void showRoots(double a, double b, double c) {
double discriminant = Math.sqrt(b * b - 4 * a * c);
if (discriminant > 0) {
System.out.println("There are no real roots ");
} else {
double root1 = (-b + discriminant) / (2 * a);
double root2 = (-b - discriminant) / (2 * a);
System.out.println("The roots are " + root1 + " and " + root2);
}
}
}
This shows two cases - one where there really areroots - but the program claims there aren't - and one where there really aren'treal roots, but the program prints them out as NaN. When you take the square root of a negative number, the result is NaN, which is why that's being displayed.
这显示了两种情况 - 一种情况下确实有根 - 但程序声称没有 - 一种情况下确实没有真正的根，但程序将它们打印为 NaN。当你取负数的平方根时，结果是 NaN，这就是显示它的原因。
So, the problem is how you're dealing with the discriminant. There are real roots if b2- 4acis non-negative - but you've already taken the square root at that point andreversed the nature of the condition.
所以，问题是你如何处理判别式。如果b 2- 4ac为非负数，则有实数根- 但您此时已取平方根并反转条件的性质。
So, it should be:
所以，应该是：
private static void showRoots(double a, double b, double c) {
double discriminant = b * b - 4 * a * c;
if (discriminant < 0) {
System.out.println("There are no real roots ");
} else {
double discriminantRoot = Math.sqrt(discriminant);
double root1 = (-b + discriminantRoot) / (2 * a);
double root2 = (-b - discriminantRoot) / (2 * a);
System.out.println("The roots are " + root1 + " and " + root2);
}
}
Lessons to learn:
学习经验：
- When you want to demonstrate a problem, it helps to keep it minimal; using hard-coded values is a good way of doing this
- Be careful about the order of operations - in this case, you were trying to judge something using the wrong value because you'd taken the square root too early
- Be careful with conditions and whether or not you're getting them the right way round
- 当你想演示一个问题时，它有助于保持最小；使用硬编码值是这样做的好方法
- 注意操作顺序 - 在这种情况下，您试图使用错误的值来判断某些东西，因为您过早地取平方根
- 小心处理条件以及您是否以正确的方式处理它们
EDIT: As noted in comments, there are various special cases to consider too, including when ais 0 which would otherwise lead to a division by 0 issue.
编辑：如评论中所述，还有各种特殊情况需要考虑，包括何时a为 0 否则会导致除以 0 问题。
回答by Shivan Dragon
You get that when your discriminant is negative. Like for a=1,b=2,c=3. Delta = 2*2 -4*1*3 = 4 - 12 = -8
当您的判别式为负时，您会得到它。就像 a=1,b=2,c=3。增量 = 2*2 -4*1*3 = 4 - 12 = -8
Java can't calculate square root of negative number, it doesn't know about imaginary number i.
Java 不能计算负数的平方根，它不知道虚数 i。
回答by Jiri Kriz
Do
做
double discriminant = b * b - 4 * a * c;
if (discriminant >= 0) {
discriminant = Math.sqrt(discriminant);
root1 = (-b + discriminant) / (2 * a);
root2 = (-b - discriminant) / (2 * a);
System.out.println("The roots are " + root1 + " and " + root2);
}
else {
System.out.println("There are no real roots ");
}
