本文整理汇总了C++中Period::withYears方法的典型用法代码示例。如果您正苦于以下问题：C++ Period::withYears方法的具体用法？C++ Period::withYears怎么用？C++ Period::withYears使用的例子？那么, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类Period的用法示例。
在下文中一共展示了Period::withYears方法的1个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C++代码示例。
示例1: err
Period *Period::normalizedStandard(const PeriodType *type) {
type = DateTimeUtils::getPeriodType(type);
int64_t millis = getMillis(); // no overflow can happen, even with Integer.MAX_VALUEs
millis += (((int64_t) getSeconds()) * ((int64_t) DateTimeConstants::MILLIS_PER_SECOND));
millis += (((int64_t) getMinutes()) * ((int64_t) DateTimeConstants::MILLIS_PER_MINUTE));
millis += (((int64_t) getHours()) * ((int64_t) DateTimeConstants::MILLIS_PER_HOUR));
millis += (((int64_t) getDays()) * ((int64_t) DateTimeConstants::MILLIS_PER_DAY));
millis += (((int64_t) getWeeks()) * ((int64_t) DateTimeConstants::MILLIS_PER_WEEK));
Period *result = new Period(millis, type, ISOChronology::getInstanceUTC());
int years = getYears();
int months = getMonths();
if (years != 0 || months != 0) {
int64_t totalMonths = years * 12L + months;
if (type->isSupported(DurationFieldType::YEARS_TYPE)) {
int normalizedYears = FieldUtils::safeToInt(totalMonths / 12);
result = result->withYears(normalizedYears);
totalMonths = totalMonths - (normalizedYears * 12);
}
if (type->isSupported(DurationFieldType::MONTHS_TYPE)) {
int normalizedMonths = FieldUtils::safeToInt(totalMonths);
result = result->withMonths(normalizedMonths);
totalMonths = totalMonths - normalizedMonths;
}
if (totalMonths != 0) {
string err("Unable to normalize as PeriodType is missing either years or months but period has a month/year amount: ");
err.append(toString());
throw UnsupportedOperationException(err);
}
}
return result;
}