如果枚举不能适应无符号整数类型，会发生什么情况？

本文介绍了如果枚举不能适应无符号整数类型，会发生什么情况？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 29岁程序员，3月因学历无情被辞！ 根据 Bathsheba 的要求，以及作为后续问题，如果枚举不能适合整数类型会发生什么？： 由于枚举定义如下： b 否，根据标准，它是完全有效的，只有未固定的底层类型有一个限制，无法表示所有的枚举值：它没有说明一个不能表示所有枚举值的固定的底层类型。 问题 oops 和未定义？ > undefined ：标准没有说明这种情况。 可能的值foo :: undefined ： 最高可能值（ true ）： undefined 和 oops 应为底层类型的最大值。 最低可能值（ false ）：底层类型的最小值。注意：在有符号整数中，它不会匹配当前的整数溢出行为（未定义的行为）。 随机值（？）：编译器将选择一个值。 所有这些值的问题是它可能导致两个具有相同值的字段（例如 foo :: True == （例如 undefined = 2） 之间的区别（） ）和implicit初始值设定项（例如 True = true，undefined ） 到标准：换句话说： 枚举栏：bool {undefined = 2}; 相当于 枚举栏：bool {undefined = static_cast< bool>（2）}; 然后 bar :: undefined true 。在隐式初始化器中，不是这种情况：这个标准段落仅仅关于 initializer ，而不是关于隐式初始化器。 摘要 这样，枚举 b $ b 根据问题和意见，这在GCC和clang中无效，但对MSVS-2015有效（带有警告）。 As requested by Bathsheba and as a follow up question to "What happens if an enum cannot fit into an integral type?":Asuming an enum is defined as follows :enum foo : unsigned int{ bar = UINT_MAX, oops};Is the value of oops defined or isn't it?MSVS2015 compilation:warning C4340: 'oops': value wrapped from positive to negative valuewarning C4309: 'initializing': truncation of constant valuewarning C4369: 'oops': enumerator value '4294967296' cannot be represented as 'unsigned int', value is '0'MSVS2015 output:bar = 4294967295oops= 0gcc 4.9.2 compilation:9 : note: in expansion of macro 'UINT_MAX'bar = UINT_MAX,^10 : error: enumerator value 4294967296l is outside the range of underlying type 'unsigned int'oops^Compilation failedgcc 4.9.2 output//compilation failed 解决方案 This is a very interesting question. The simple answer is that this is literally undefined: the standard doesn't say anything about this case.To have a better example, consider this enum: enum foo : bool { True=true, undefined };According to the standard:Therefore, the value of foo::undefined in our example is 2 (true+1). Which can not be represented as a bool.Is it ill-formed?No, according to the standard, it is perfectly valid, only not-fixed underlying type have a restriction about not being able to represent all of the enumerator values:It says nothing about a fixed underlying type that can not represent all the enumerator values.What is the value of original question's oops and undefined?It is undefined: the standard doesn't say anything about this case.Possible values for foo::undefined:Highest possible value (true): undefined and oops should be underlying type's maximum value.Lowest possible value (false): the underlying type's minimum value. Note: In signed integers, it would not match the current behavior for Integer overflow (undefined behavior).Random value (?): the compiler will choose an value.The problem with all of these values is that it may result two fields with the same value (e.g. foo::True == foo::undefined).The difference between initializer (e.g. undefined=2) and "implicit" initializer (e.g. True=true, undefined)According to the standard:In other words: enum bar : bool { undefined=2 };is equivalent to enum bar : bool { undefined=static_cast<bool>(2) };And then bar::undefined will be true. In an "implicit" initializer, it would not be the case: this standard paragraph say it about only initializer, and not about "implicit" initializer.SummaryWith this way, it is possible for an enum with a fixed-underlying-type to have unrepresentable values.Their value is undefined by the standard.According to the question and comments, this is not valid in GCC and clang but valid for MSVS-2015 (with a warning). 这篇关于如果枚举不能适应无符号整数类型，会发生什么情况？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！ 上岸，阿里云！