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问题 Why is that undefined behaviour? struct s { const int id; // <-- const member s(int id): id(id) {} s& operator =(const s& m) { return *new(this) s(m); // <-- undefined behavior? } }; (Quote from the standard would be nice). This question arose from this answer. 回答1: There is nothing that makes the shown code snippet inherently UB. However, it is almost certain UB will follow immediately under any normal usage. From [basic.life]/8 (emphasis mine) If, after the lifetime of an object has ended

问题 I have the code: class A { public: A() = default; private: int i = 1; }; int main() { const A a; return 0; } It compiles fine on g++ (see ideone), but fails on clang++ with error: default initialization of an object of const type 'const A' requires a user-provided default constructor I reported this issue on LLVM bug-tracker and got it INVALID. I see it absolutly pointless to try to convince the clang developers. On the other side, I don't see the reason for such restriction. Can anyone

问题 I've seen the other similar questions and read the defect about it. But I still don't get it. Why is i = ++i + 1 well-defined in C++11 when i = i++ + 1 is not? How does the standard make this well defined? By my working out, I have the following sequenced before graph (where an arrow represents the sequenced before relationship and everything is a value computation unless otherwise specified): i = ++i + 1 ^ | assignment (side effect on i) ^ ^ | | ☆i ++i + 1 || ^ i+=1 | ^ 1 | ★assignment (side

问题 I've seen the other similar questions and read the defect about it. But I still don't get it. Why is i = ++i + 1 well-defined in C++11 when i = i++ + 1 is not? How does the standard make this well defined? By my working out, I have the following sequenced before graph (where an arrow represents the sequenced before relationship and everything is a value computation unless otherwise specified): i = ++i + 1 ^ | assignment (side effect on i) ^ ^ | | ☆i ++i + 1 || ^ i+=1 | ^ 1 | ★assignment (side

问题 Pure virtual functions are those member functions that are virtual and have the pure-specifier ( = 0; ) Clause 10.4 paragraph 2 of C++03 tells us what an abstract class is and, as a side note, the following: [Note: a function declaration cannot provide both a pure-specifier and a definition —end note] [Example: struct C { virtual void f() = 0 { }; // ill-formed }; —end example] For those who are not very familiar with the issue, please note that pure virtual functions can have definitions but

问题 In code that is supposed to be quite portable, in a many person, many environment project, I have to deal with a problem with code that negates unsigned int (and same problem in other places negating std::size_t ). The intent is to create the additive inverse of that value (modulo MAX_VAL+1 where MAX_VAL is the max value of the type) and I found other questions on this topic confirm that is the standard specified behavior. VS2012, (with compile time options I cannot easily learn) calls that

问题 Consider the two tables below: user: ID | name ---+-------- 1 | Alice 2 | Bob 3 | Charlie event: order | user ------+------------ 1 | 1 (Alice) 2 | 2 (Bob) 3 | 3 (Charlie) 4 | 3 (Charlie) 5 | 2 (Bob) 6 | 1 (Alice) If I run the following query: SELECT DISTINCT user FROM event ORDER BY "order" DESC; will it be guaranteed that I get the results in the following order? 1 (Alice) 2 (Bob) 3 (Charlie) If the three last rows of event are selected, I know this is the order I get, because it would be

问题 I am programming C++ using gcc on an obscure system called linux x86-64. I was hoping that may be there are a few folks out there who have used this same, specific system (and might also be able to help me understand what is a valid pointer on this system). I do not care to access the location pointed to by the pointer, just want to calculate it via pointer arithmetic. According to section 3.9.2 of the standard: A valid value of an object pointer type represents either the address of a byte

问题 Many times I see (and sometimes write) code similar to this example: int a=0, b=2; if( a && (b=func())!=0 ) { //... The question is: does the standard guarantee these statements? b will be not touched (and remain value 2 ) func() will not be called And vice-versa, if we write if( func()!=0 && a ) - does the standard guarantee func() will be called? I'm interested in the particular standard paragraph defining this is legitimate. UPD: my typo, changed from int a=1 to int a=0 回答1: To the exact

问题 It's a familiar fact that in C you can write "a" "b" and get "ab" . This is discussed in the C11 standard: In translation phase 6, the multibyte character sequences specified by any sequence of adjacent character and identically-prefixed string literal tokens are concatenated into a single multibyte character sequence. The phrase "character and..." would seem to suggest you can get the same results by writing 'a' "b" , but I've never come across that usage and GCC and the Microsoft compiler