undefined-behavior

问题 In this example, is the c-style cast to int& followed by an assignment to kind of hack the interface of class A undefined behavior? class A { public: A() : x(0) { } ~A() { std::cout << x << std::endl; } const int& getX() { return x; } private: int x; }; int main() { A a; int& x = (int&)a.getX(); x = 17; std::cout << x << std::endl; } Output: 17 17 If so, what part of the standard can i refer to? Also, is there any reason why this compiles without warnings? (i tested with c++14 on cpp.sh with

问题 What if I have something like this: int a = 20; int min = INT_MIN; if(-a - min) //do something Assume that INT_MIN if positive is more than INT_MAX. Would min ever be converted by the compiler to something like -min as in -INT_MIN, which could be undefined? 回答1: You are right that unary minus applied to INT_MIN can be undefined, but this does not happen in your example. -a - min is parsed as (-a) - min . Variable min is only involved in binary subtraction, and the first operand only needs to

问题 This question already has answers here : Why are these constructs using pre and post-increment undefined behavior? (14 answers) Closed 5 years ago . I have this code to take a string of the form bla_2 and separate it: void separate(char* str, char* word, int* n) { int i = 0; while(str[i] != '_') { word[i] = str[i++]; } *n = str[++i] - '0'; } I got: warning: operation on ‘i’ may be undefined [-Wsequence-point] But I am only changing i via ++ operator, I am not assigning anything to. So, why is

问题 Derived from this question ... Given the declaration in line 1 of main , are the second and third printf statements considered undefined behavior because they point to locations not owned by this process? struct my_structure { int i; }; void main() { struct my_structure variable = {20}; struct my_structure *p = &variable; printf("NUMBER: %d\n", p++->i); printf("NUMBER: %d\n", p++->i); printf("NUMBER: %d\n", p++->i); } 回答1: In this case, first printf() is OK per C11 6.5.6.8 printf("NUMBER: %d

问题 In an Stack Overflow answer here, Kerrek posts the following code. Foo && g() { Foo y; // return y; // error: cannot bind ‘Foo’ lvalue to ‘Foo&&’ return std::move(y); // OK type-wise (but undefined behaviour, thanks @GMNG) } GManNickG points out that this results in undefined behavior. Kerrek adds True; you don't really return && from anything but forward and move. It's a contrived example. What confuses me is that the C++11 standard uses a function call that returns an rvalue reference as an

问题 It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. Closed 7 years ago . I was asking myself if these line of code could produce undefined behaviour in C and in C++. I tried to answer to each point reading what the standard says about array subscripting (C 6.5.6 - 8). I didn't posted

问题 I was self-studying CSAPP and got a strange result when I ran into a strange issue during the run of a assertion test. I'm not sure what to start this question with, so let me get the code first (file name visible in comments): // File: 2.30.c // Author: iBug int tadd_ok(int x, int y) { if ((x ^ y) >> 31) return 1; // A positive number and a negative integer always add without problem if (x < 0) return (x + y) < y; if (x > 0) return (x + y) > y; // x == 0 return 1; } // File: 2.30-test.c //

问题 This question already has answers here : issue with assignment operator inside printf() (3 answers) How to understand the behaviour of printf statement in C? (5 answers) Why are these constructs using pre and post-increment undefined behavior? (14 answers) Closed 7 months ago . I've been studying C for about a year now, and I came across this above when I was just playing around. I first thought maybe it's a case of assignment precedence (i.e. x=10 happens first), but then I tried printf("%d

问题 This question already has answers here : Sequence Point - Xor Swap on Array get wrong result (1 answer) How does XOR variable swapping work? (9 answers) Closed 5 years ago . What is the difference between these two macros? #define swap(a, b) (((a) ^ (b)) && ((a) ^= (b) ^= (a) ^= (b))) Or #define swap(a, b) (((a) ^ (b)) && ((b) ^= (a) ^= (b), (a) ^= (b))) I saw the second macro here but couldn't understand why it wasn't written like the first one? Is there a special reason that I missed? 回答1:

问题 Apparently, the behaviour of the right shift operation: a >> b is undefined in C and C++ when b >= sizeof(a)*CHAR_BIT (whereas in the normal case, the "new bits" introduced from the left due to the right shift are equal to zero). Why is this undefined behaviour better than setting the result to zero when b >= sizeof(a)*CHAR_BIT ? 回答1: We can get an idea of why languages choose undefined behavior from Why Language Designers Tolerate Undefined Behavior and it says: This answer came from two