另一个问题

某些字母表中的" - 1963年的X3.4。Please excuse me if this has already been covered.Givenchar x[42];issizeof(x[999])any kind of error? If so, since the expression is not evaluated, howwould such an error be detected? What if the declaration wasint n = 42;char x[n];?-- Richard--"Consideration shall be given to the need for as many as 32 charactersin some alphabets" - X3.4, 1963.推荐答案 ri ***** @ cogsci.ed.ac.uk （Richard Tobin）写道：ri*****@cogsci.ed.ac.uk (Richard Tobin) writes: 如果已经涵盖，请原谅我。 给定 char x [42]; 是 sizeof（x [999]） 任何类型的错误？Please excuse me if this has already been covered.Given char x[42];is sizeof(x[999])any kind of error? [...] 我相信它完全有效，而且必须收益1. x [999]相当于*（x + 999）。添加会调用未定义的 行为，但前提是它已被评估。 我认为没有更多理由 sizeof（ x [999]） 调用UB而不是 if（0）{ x [999]; } 这样做。 - Keith Thompson（The_Other_Keith） ks *** @ mib.org < http://www.ghoti.net/~kst> 圣地亚哥超级计算机中心< * < http://users.sdsc.edu/~kst> "我们必须做点什么。这是事情。因此，我们必须这样做。 - Antony Jay和Jonathan Lynn，是部长[...]I believe it''s perfectly valid, and must yield 1.x[999] is equivalent to *(x+999). The addition would invoke undefinedbehavior, but only if it were evaluated.I see no more reason forsizeof(x[999])to invoke UB than forif (0) {x[999];}to do so.--Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>San Diego Supercomputer Center <* <http://users.sdsc.edu/~kst>"We must do something. This is something. Therefore, we must do this."-- Antony Jay and Jonathan Lynn, "Yes Minister" 9月19日，3日：09：00，Keith Thompson< ks ... @ mib.orgwrote：On Sep 19, 3:09 am, Keith Thompson <[email protected]: 我相信它完全有效，而且必须产生1个。 x [999]相当于*（x + 999）。添加会调用未定义的 行为，但前提是它已被评估。 我认为没有更多理由 sizeof（ x [999]） 调用UB而不是 if（0）{ x [999]; } 这样做。I believe it''s perfectly valid, and must yield 1.x[999] is equivalent to *(x+999). The addition would invoke undefinedbehavior, but only if it were evaluated.I see no more reason for sizeof(x[999])to invoke UB than for if (0) { x[999]; }to do so. 真的一定很好。类似的情况就是这个，可以在每个人的代码中找到数以万计的时间:: 某事* p; p = malloc（sizeof（* p））; 这是建议的分配内存的习惯用法，以及表达式 * p本身将像x [999]一样调用未定义的行为。这是好b $ b罚款，因为* p未被评估。It really must be fine. A similar situation is this one, which can befound gazillion times in everyone''s code::something* p;p = malloc (sizeof (*p));This is the recommended idiom to allocate memory, and the expression*p on its own would invoke undefined behaviour just like x[999]. It''sfine because *p is not evaluated. 2007年9月19日星期三00:16:49 +0000，Richard托宾写道：On Wed, 19 Sep 2007 00:16:49 +0000, Richard Tobin wrote: 如果已经报道，请原谅我。 鉴于 char x [42]; 是 sizeof（x [999]） 任何类型的错误？如果是这样，既然没有评估表达式，那么如何检测到这样的错误？b 如果声明是 int n = 42; char x [n];Please excuse me if this has already been covered.Given char x[42];is sizeof(x[999])any kind of error? If so, since the expression is not evaluated, howwould such an error be detected? What if the declaration was int n = 42; char x[n]; 即使在这种情况下，x'的类型是VLA，但x [999]的类型是char， 因此它不会被sizeof评估。 - Army1987（将NOSPAM替换为email） 如果你是从Windows机器发送电子邮件，请关闭微软的 傻瓜??智能行情？特征。这样你就可以避免通过你的邮件洒垃圾字符了。 - Eric S. Raymond和Rick MoenEven in this case, x''s type is a VLA, but x[999]''s type is char,so it is not evaluated by sizeof.--Army1987 (Replace "NOSPAM" with "email")If you''re sending e-mail from a Windows machine, turn off Microsoft''sstupid a??Smart Quotesa?? feature. This is so you''ll avoid sprinkling garbagecharacters through your mail. -- Eric S. Raymond and Rick Moen 这篇关于另一个问题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！