NumberFormat / DecimalFormat将某些浮点值视为长整型，而不是双精度 [英] NumberFormat/DecimalFormat treats certain floating-point values as longs instead of doubles

问题描述

NumberFormat / DecimalFormat似乎没有以＃。0格式（其中＃是任何数字）作为双精度解析字符串。
以下代码说明了这一点：

NumberFormat/DecimalFormat doesn't seem to parse strings with the "#.0" format (where # is any number) as a double. The following code illustrates this:

#include <cstdio>
#include <iostream>
#include <unicode/decimfmt.h>
#include <unicode/numfmt.h>
#include <unicode/unistr.h>
#include <unicode/ustream.h>

int main() {
    UErrorCode status = U_ZERO_ERROR;
    // DecimalFormat doesn't work either
    NumberFormat* f = NumberFormat::createInstance(status);
    f->setGroupingUsed(false);
    f->setParseIntegerOnly(false);

    UnicodeString str("2.0"); // Change to "2.#" for it to work, where # is any non-zero number
    Formattable formattable;
    f->parse(str, formattable, status);
    if (U_FAILURE(status)) {
        printf("ERROR: %s\n", u_errorName(status));
        delete f;
        return 1;
    } else {
        if (formattable.getType() == Formattable::kDouble) {
            printf("kDouble: %f\n", formattable.getDouble());
        } else if ((formattable.getType() == Formattable::kLong)) {
            printf("kLong: %d\n", formattable.getLong());
        } else {
            printf("ERROR: unexpected type: %d\n", formattable.getType());
        }
    }

    str.remove(); // Clear the string
    f->format(2.0f, str);
    std::cout << "formatted: \"" << str << '\"' << std::endl; // Outputs "2"
    delete f;

    return 0;
}

解析2.0，Formattable的类型是2（ Formattable :: Type :: kLong ）。当解析2.1时，Formattable的类型为1（ Formattable：Type :: kDouble ） - 两个字符串。

When parsing "2.0", the Formattable's type is 2 (Formattable::Type::kLong). When parsing "2.1" the Formattable's type is 1 (Formattable:Type::kDouble) - as it should be for both strings.

当你尝试将浮点格式化为UnicodeString时也会出现问题（例如float 2.0 格式化为2）。

Issues also arise when you try to format a float as a UnicodeString (e.g. a float 2.0 gets formatted to "2").

因此：如何解析/作为ICU中的整数？

So: How can I parse/format any double number without it being interpreted as an integer in ICU?

推荐答案

可以调用 formattable.getDouble（status） - getType返回long的唯一原因是特定的值将适合长。

You can call formattable.getDouble(status) - the only reason getType returns long is that the particular value will fit in a long.

至于格式，如果在格式化之前调用 f-> setMinimumFractionDigits（1）; 2.0与您的代码，设置最小数字为2给出2.00等。

As to format, if you call f->setMinimumFractionDigits(1); before format, you will get "2.0" with your code, setting the min digits to 2 gives "2.00" etc.

hth

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