Math.round(msgBody.length() / 160 + 0.5) 计算不正确？
因此，我尝试动态计算给定消息长度将发送的短信数量。 SMS 消息被分成 160 字节（字符）块。我使用 MOD 160 <= 1 因为我需要考虑添加和减去文本。出于性能原因，我只想在边界附近进行 numMsgs 计算：0, 1, 160, 161, 320, 321, ...
问题是在 msgBody 长度为 160 时，舍入操作的计算结果为 2 (160/160 = 1 + 0.5 四舍五入）。在 160 + 1 的倍数处，它应该计算为下一个最大整数，因为任何 160 字节 + 1 字节都等于整个附加消息。
我通过使用 OR 运算符 and == 1 OR == 159 完成了这项工作。它正确地递增和递减，但它仅以 160 -1 的倍数递减，这是不正确的。
另外，我可以将 IF 逻辑放入外部逻辑中，以便在 MOD 计算结果为 0（零）时简单地减去 1，但这看起来很笨拙，我宁愿学习我可能从我的箭袋中丢失的模式:)
if (msgBody.length() % 160 <= 1) {
numMsgs.setText(Math.round(msgBody.length() / 160 + 0.5));
}
So I'm trying to dynamically calculate the number of SMS messages that will be sent for a given message length. SMS messages are chunked into 160 byte (character) chunks. I use MOD 160 <= 1 because I need to account for adding and subtracting text. For performance reasons I only want to do the numMsgs calculations near the boundaries: 0, 1, 160, 161, 320, 321, ...
The problem is at a msgBody length of 160 the rounding operation evaluates to 2 (160/160 = 1 + 0.5 rounded up). At multiples of 160 + 1 it should evaluate to the next highest integer because any 160 bytes + 1 byte equals a whole additional message.
I have made this work by using an OR operator and == 1 OR == 159. It increments and decrements correctly, but it only decrements at multiples of 160 -1 which is not correct.
Also, I can put IF logic inside the outer logic to simply subtract 1 if the MOD evaluates to 0 (zero), but that seemed kludgey and I'd rather learn the pattern I may be missing from my quiver :)
if (msgBody.length() % 160 <= 1) {
numMsgs.setText(Math.round(msgBody.length() / 160 + 0.5));
}
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（3）
msgBody.length() / 160是一个除法表达式，两边都是整数 - 因此它将执行整数除法。我怀疑这就是你做错的事情，但说实话，这并不完全清楚。 （您声称它没有正确评估，但您没有提供示例输入/输出/预期输出供我们查看。）说实话，最简单的方法是像这样进行舍入：
仍然执行整数除法，但是+159 意味着如果输入不完全是 160 的倍数，它将向上舍入。
我建议您可以执行该计算无论边界如何 - 但请确保您如果与现有值相比没有变化，则实际上不会更新 UI。它会非常便宜——一个整数加法和一个整数除法。
msgBody.length() / 160is a division expressions with both sides being integers - so it will perform integer division. I suspect that's what you're doing wrong, but it's not entirely clear to be honest. (You claim it's not evaluating correctly, but you haven't given a sample input / output / expected output for us to look at.)To be honest, the simplest approach is to round up like this:
That's still performing integer division, but the +159 means it will round up if the input isn't exactly a multiple of 160.
I suggest you could perform that calculation regardless of the boundaries - but make sure you don't actually update the UI if there's no change compared with the existing value. It's going to be very cheap - one integer add and one integer divide.
如果有消息，则至少会发送 1 条短信。
你想要
If there is a message, then at least 1 SMS will be sent.
You want
只要使用正确的公式，使用
int进行加法和除法就足够了。取模可能比除法还要慢。但是，我不清楚您期望什么结果...(msgBody.length() + 159) / 160还不够吗？Just adding and dividing using
intis enough with the right formula. Modulo is probably even slower than dividing. However, it's not clear to me what result you expect...(msgBody.length() + 159) / 160isn't enough?