如何在不检查大小或超出范围的情况下获取字符串的前n个字符？ [英] How do I get the first n characters of a string without checking the size or going out of bounds?

问题描述

如何在不首先进行大小检查（内联可接受）或冒险 n 字符> IndexOutOfBoundsException ？

How do I get up to the first n characters of a string in Java without doing a size check first (inline is acceptable) or risking an IndexOutOfBoundsException?

推荐答案

这是一个简洁的解决方案：

Here's a neat solution:

String upToNCharacters = s.substring(0, Math.min(s.length(), n));

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^{意见：虽然此解决方案是整洁的，我认为它实际上不太可读比使用的解决方案 / else 以明显的方式。如果读者没有看到这个技巧，他/她必须更难思考来理解代码。 IMO，代码的含义在中更明显，如果 / else 版本。有关更清晰/更易读的解决方案，请参阅@ paxdiablo的回答。}

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^{Opinion: while this solution is "neat", I think it is actually less readable than a solution that uses if / else in the obvious way. If the reader hasn't seen this trick, he/she has to think harder to understand the code. IMO, the code's meaning is more obvious in the if / else version. For a cleaner / more readable solution, see @paxdiablo's answer.}

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