language-lawyer

问题 I have the following example #include <stdlib.h> #include <stdio.h> #include <stddef.h> typedef struct test{ int a; long b; int c; } test; int main() { test *t = (test*) malloc(offsetof(test, c)); t -> b = 100; } It works fine, but Im not sure about it. I think I have UB here. We have a pointer to an object of a structure type. But the object of the structure type is not really valid. I went through the standard and could not find any definition of this behavior. The only section I could find

问题 I have three classes: B , D and G . D is a B and G is a D . Both B and D are abstract. B is from a third party. B has a non-pure, virtual method that G needs to implement (to be a D ). Can I and is it good practice to redefine/override a virtual function to be pure virtual? Example: class B // from a third party { public: virtual void foo(); }; class D : public B { public: void foo() override = 0; // allowed by gcc 4.8.2 virtual void bar() = 0; }; class G : public D { public: // forgot to

问题 Am I allowed to use the NULL pointer as replacement for the value of 0 ? Or is there anything wrong about that doing? Like, for example: int i = NULL; as replacement for: int i = 0; As experiment I compiled the following code: #include <stdio.h> int main(void) { int i = NULL; printf("%d",i); return 0; } Output: 0 Indeed it gives me this warning, which is completely correct on its own: warning: initialization makes integer from pointer without a cast [-Wint-conversion] but the result is still

问题 Am I allowed to use the NULL pointer as replacement for the value of 0 ? Or is there anything wrong about that doing? Like, for example: int i = NULL; as replacement for: int i = 0; As experiment I compiled the following code: #include <stdio.h> int main(void) { int i = NULL; printf("%d",i); return 0; } Output: 0 Indeed it gives me this warning, which is completely correct on its own: warning: initialization makes integer from pointer without a cast [-Wint-conversion] but the result is still

问题 Am I allowed to use the NULL pointer as replacement for the value of 0 ? Or is there anything wrong about that doing? Like, for example: int i = NULL; as replacement for: int i = 0; As experiment I compiled the following code: #include <stdio.h> int main(void) { int i = NULL; printf("%d",i); return 0; } Output: 0 Indeed it gives me this warning, which is completely correct on its own: warning: initialization makes integer from pointer without a cast [-Wint-conversion] but the result is still

问题 In C++17, this code is illegal: constexpr int foo(int i) { return std::integral_constant<int, i>::value; } That's because even if foo can be evaluated at compile-time, the compiler still needs to produce the instructions to execute it at runtime, thus making the template instantiation impossible. In C++20 we will have consteval functions, which are required to be evaluated at compile-time, so the runtime constraint should be removed. Does it mean this code will be legal? consteval int foo(int

问题 In C++17, this code is illegal: constexpr int foo(int i) { return std::integral_constant<int, i>::value; } That's because even if foo can be evaluated at compile-time, the compiler still needs to produce the instructions to execute it at runtime, thus making the template instantiation impossible. In C++20 we will have consteval functions, which are required to be evaluated at compile-time, so the runtime constraint should be removed. Does it mean this code will be legal? consteval int foo(int

问题 When looking for evidence of unsigned long being enough to hold size_t for the purpose of being argument to printf I ran into two fact(oid)s. First there's this answer stating that long is indeed not guaranteed to be large enough for size_t . On the other hand I saw this answer suggesting to use printf("%lu", (unsigned long)x) in pre C99, x being of size_t . So the question is could you assume or were you guaranteed that long were enough to hold size_t in pre C99 . The other question is

问题 When looking for evidence of unsigned long being enough to hold size_t for the purpose of being argument to printf I ran into two fact(oid)s. First there's this answer stating that long is indeed not guaranteed to be large enough for size_t . On the other hand I saw this answer suggesting to use printf("%lu", (unsigned long)x) in pre C99, x being of size_t . So the question is could you assume or were you guaranteed that long were enough to hold size_t in pre C99 . The other question is

问题 When looking for evidence of unsigned long being enough to hold size_t for the purpose of being argument to printf I ran into two fact(oid)s. First there's this answer stating that long is indeed not guaranteed to be large enough for size_t . On the other hand I saw this answer suggesting to use printf("%lu", (unsigned long)x) in pre C99, x being of size_t . So the question is could you assume or were you guaranteed that long were enough to hold size_t in pre C99 . The other question is