如何知道浮点值是否大于integer.max_value

Question

// The given input
String input = "99999999.99";
// We need only 2 decimals (in case of more than 2 decimals is in input)
Float value = Float.valueOf(input);
DecimalFormat df = new DecimalFormat("#0.00");
input = df.format(value);
value = new Float(input);
// Now we have a clear 2 decimal float value
// Check for overflow
value *= 100; // Multiply by 100, because we're working with cents
if (value >= Integer.MAX_VALUE) {
   System.out.println("Invalid value");
}
else {
///
}

该语句不起作用，条件失败。

比较具有整数值的浮子的正确方法？

Answer 1

您的代码没有错。999999999.99乘以100等于999999999。比max int == 2147483647（2^31 -1）更大。如果您想能够存储较大的整数使用long。

如果不是您的问题，请进一步详细说明。

Answer 2

尝试以下：

public static void main(final String[] args) {
    // The given input
    String input = "99999999.99";
    // We need only 2 decimals (in case of more than 2 decimals is in input)
    Float value = Float.parseFloat(input);
    DecimalFormat df = new DecimalFormat("#0.00");
    DecimalFormatSymbols custom = new DecimalFormatSymbols();
    custom.setDecimalSeparator('.');
    df.setDecimalFormatSymbols(custom);
    input = df.format(value);
    value = new Float(input);
    // Now we have a clear 2 decimal float value
    // Check for overflow
    value *= 100; // Multiply by 100, because we're working with cents
    if (value >= Integer.MAX_VALUE) {
        System.out.println("Invalid value");
    } else {
        ///
    }
}

祝您有美好的一天