在 for 语句的终止表达式中使用数字文字有什么区别?

What's the difference when using numeric literal in termination expression of a for statement?

为什么这段代码:

String value = JOptionPane.showInputDialog("Enter x"); //Input = 100
int x = Integer.parseInt(value);
double result = 1;

for (int i = 1; i <= x; i++) //used variable "x" here
{
    result += (x * 1.0) / fact(i);
    x *= x;
}

public static int fact(int n) {
    int fact = 1;
    for (int i = 1; i <= n; i++) {
        fact *= i;
    }
    return fact;
}

与这个工作方式不同?

String value = JOptionPane.showInputDialog("Enter x"); //Input = 100
int x = Integer.parseInt(value);   
double result = 1;

for (int i = 1; i <= 100; i++) //and here I used the value "100"
{
    result += (x * 1.0) / fact(i);
    x *= x;
}

public static int fact(int n) {
    int fact = 1;
    for (int i = 1; i <= n; i++) {
        fact *= i;
    }
    return fact;
}

我所做的唯一更改是在我的终止表达式中使用值 100 而不是使用变量 x

当我 运行 第一个代码时,我得到:

9.479341033333334E7

然而,对于第二个我总是得到

NaN

为什么?

两个片段的区别是:

for (int i = 1; i <= x; i++) {

对比

for (int i = 1; i <= 100; i++) {

在第一种情况下,x 每次都会变大!最终,它会在 x overflows and becomes 0, which will be much sooner than in the second case. For an explanation as to why this results in 0 instead of some other random number, see:

时停止

在第二种情况下,当i = 34时,fact(n)将return0,所以双除法是(0 * 1.0) /0,结果是NaN。将任何双精度添加到 NaN 时,都会变成 NaN,这就是为什么第二个片段会导致 NaN。参见:In Java, what does NaN mean?