Question

我有一个名为Duration的结构，我如何更改此函数以便它返回一个Duration类型的对象？另一个问题是，如果我必须重载函数，我将如何让它接受持续时间对象？

void all(clock_t t, int &hours, int &minutes, int &seconds, int &ticks) {
    ticks = t % CLOCKS_PER_SEC;
    seconds = t / CLOCKS_PER_SEC;
    minutes = seconds / 60;
    seconds %= 60;
    hours = minutes / 60;
    minutes %= 60;
}

Answer 1

让我们说你的Duration结构是这样的：

struct Duration {
   int hours, minutes, seconds, ticks;
};

现在，您的all()方法可能如下所示：

Duration all(clock_t t) {
    Duration duration;
    duration.ticks = t % CLOCKS_PER_SEC;
    duration.seconds = t / CLOCKS_PER_SEC;
    duration.minutes = duration.seconds / 60;
    duration.seconds %= 60;
    duration.hours = duration.minutes / 60;
    duration.minutes %= 60;
    return duration;
}

这样称呼：

clock_t t = ...;
Duration duration = all(t);

要回答您的其他问题，如果您想重载all()以接受Duration输出，则它可能如下所示：

void all(clock_t t, Duration &duration) {
    duration.ticks = t % CLOCKS_PER_SEC;
    duration.seconds = t / CLOCKS_PER_SEC;
    duration.minutes = duration.seconds / 60;
    duration.seconds %= 60;
    duration.hours = duration.minutes / 60;
    duration.minutes %= 60;
}

这样称呼：

clock_t t = ...;
Duration duration;
all(t, duration);

如果重载，可以使用另一个重载来实现一个重载，以减少重复代码，如下所示：

Duration all(clock_t t) {
    Duration duration;
    duration.ticks = t % CLOCKS_PER_SEC;
    duration.seconds = t / CLOCKS_PER_SEC;
    duration.minutes = duration.seconds / 60;
    duration.seconds %= 60;
    duration.hours = duration.minutes / 60;
    duration.minutes %= 60;
    return duration;
}

void all(clock_t t, Duration &duration) {
    duration = all(t);
}

或者这个：

void all(clock_t t, Duration &duration) {
    duration.ticks = t % CLOCKS_PER_SEC;
    duration.seconds = t / CLOCKS_PER_SEC;
    duration.minutes = duration.seconds / 60;
    duration.seconds %= 60;
    duration.hours = duration.minutes / 60;
    duration.minutes %= 60;
}

Duration all(clock_t t) {
    Duration duration;
    all(t, duration);
    return duration;
}

使函数返回类类型

1 个答案: