在C ++ 11中左移（<<）一个负整数未定义的行为？

本文介绍了在C ++ 11中左移（&lt;&lt;）一个负整数未定义的行为？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 是否向左移动一个否定 int 在C ++ 11中未定义的行为？ 这里是从5.8：令我困惑的部分是： $否则，如果E1具有有符号类型和非负值，并且E1×2E2 在结果类型中是可表示的，那么这是结果值; 否则行为是未定义的。 这应该解释为意味着左移负数是UB？或者它只是意味着如果您的结果为负，并且结果不适合结果类型， 它是UB？ 此外，前面的子句说：这使得明确的是，对于一个的操作数使用负数是UB。如果是UB对其他操作数使用负数，我希望在这里也清楚。 所以，底线是： -1<< 1 未定义的行为？ @Angew 提供了对标准版的解释，简洁地表达了一个可能的（可能的）有效解释。其他人质疑这个问题是否真的关于语言行为是未定义的的适用性与我们的（StackOverflow的）使用短语未定义的行为。 @ Angew对Standardese的解释是： if（typeof（E1）== unsigned integral） value = E1 * 2 ^ E2％blah blah; else if（typeof（E1）== signed integral&& E1> = 0&& else value = undefined; 这个问题真正归结为这是正确的解释： value = E1 left-shift-by（E2） switch（typeof（E1）） { case unsigned积分： value = E1 * 2 ^ E2％blah blah; break; 表示签名整数： if（E1> = 0） { if（representable（E1 * 2 ^ E2）） { value = E1 * 2 ^ E2; } else { value = undefined; } } break; } ？ Sidenote，从psudocode的角度来看这一点在我看来，@ Agnew的解释是正确的。解决方案是的，我会说它是undefined。如果我们将standardese转换为伪代码： if（typeof（E1）== unsigned integral） value = E1 * 2 ^ E2％blah blah; else if（typeof（E1）== signed integral&& E1> = 0&& else value = undefined; 我会说明他们为什么要明确右手操作数，而不是左边的一个是你引用的paragrpah（右边的操作数的情况）适用于左右移位。 对于左边的操作数，裁决不同。左移负值是未定义的，右移是实现定义的。 Is left-shifting a negative int Undefined Behavior in C++11?The relevant Standard passages here are from 5.8:The part that confuses me is:Should this be interpreted to mean that left-shifting any negative number is UB? Or does it only mean if you LS a negative and the result doesn't fit in the result type, then it's UB?Moreover, the preceding clause says:This makes it explicit that using a negative number for one of the operands is UB. If it were UB to use a negative for the other operand, I would expect that to be made clear here as well.So, bottom line, is:-1 << 1Undefined Behavior?@Angew provided a psudocode interpretation of the Standardese which succinctly expresses one possible (likely) valid interpretation. Others have questioned whether this question is really about the applicability of the language "behavior is undefined" versus our (StackOverflow's) use of the phrase "Undefined Behavior." This edit is to provide some more clarification on what I'm trying to ask.@Angew's interpretation of the Standardese is:if (typeof(E1) == unsigned integral) value = E1 * 2^E2 % blah blah;else if (typeof(E1) == signed integral && E1 >= 0 && representable(E1 * 2^E2)) value = E1 * 2^E2;else value = undefined;What this question really boils down to is this -- is the correct interpretation actually:value = E1 left-shift-by (E2)switch (typeof(E1)){case unsigned integral : value = E1 * 2^E2 % blah blah; break;case signed integral : if (E1 >= 0) { if (representable(E1 * 2^E2)) { value = E1 * 2^E2; } else { value = undefined; } } break;}?Sidenote, in looking at this in terms of psudocode makes it fairly clear in my mind that @Agnew's interpretation is the correct one. 解决方案 Yes, I would say it's undefined. If we translate the standardese to pseudo-code:if (typeof(E1) == unsigned integral) value = E1 * 2^E2 % blah blah;else if (typeof(E1) == signed integral && E1 >= 0 && representable(E1 * 2^E2)) value = E1 * 2^E2;else value = undefined;I'd say the reason why they're explicit about the right-hand operand and not about the left-hand one is that the paragrpah you quote (the one with the right-hand operand case) applies to both left and right shifts.For the left-hand operand, the ruling differs. Left-shifting a negative is undefined, right-shifting it is implementation-defined. 这篇关于在C ++ 11中左移（&lt;&lt;）一个负整数未定义的行为？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！