constexpr int i = 42; // type of i is const int // i and j must be defined outside any function <==i和j必须定义在任何函数体之外 constexpr const int *p = &i; // p is a constant pointer to the const int i constexpr int *p1
/* We can see a FIELD_DECL in a pointer-to-member expression. */ case FIELD_DECL: case PARM_DECL: case USING_DECL: return true; case AGGR_INIT_EXPR: case CALL_EXPR: /* -- an invocation of a function other than a constexpr function or a constexpr constructor. */ { tree fun = ...
/* We can see a FIELD_DECL in a pointer-to-member expression. */ case FIELD_DECL: case PARM_DECL: case USING_DECL: return true; case AGGR_INIT_EXPR: case CALL_EXPR: /* -- an invocation of a function other than a constexpr function or a constexpr constructor. */ { tree fun = ...
constexpr int *q = 0; // const pointer, non-const data 与其它常量指针类似,const指针既可以指向常量也可以指向一个非常量: int j = 0; constexpr int i = 2; constexpr const int *p = &i; // const pointer, const data constexpr int *p1 = &j; // const pointer, non-const data 1. ...
constint(*pf)(); // pf is a non-const pointer and points to a function which has no arguments and returns a const int ... const指针的解读规则差不多就是这些了…… 指针自身为const表示不可对该指针进行赋值,而指向物为const则表示不可对其指向进行赋值。因此可以将引用看成是一个自身为const的...
If a lambda is implicitly or explicitlyconstexpr, and you convert it to a function pointer, the resulting function is alsoconstexpr: C++ autoIncrement = [](intn) {returnn +1; };constexprint(*inc)(int)= Increment; See also C++ Language Reference ...
const的语义 C++中的const的⽬的是通过编译器来保证对象的常量性，强制编译器将所有可能违背const对象的常量性的操作都视为error。对象的常量性可以分为两种：物理常量性（即每个bit都不可改变）和逻辑常量性（即对象的表现保持不变）。C++中采⽤的是物理常量性，例如下⾯的例⼦：struct A { int *ptr;};...
const int (*pf)(); // pf is a non-const pointer and points to a function which has no arguments and returns a const int ... const指针的解读规则差不多就是这些了…… 指针自身为const表示不可对该指针进行赋值,而指向物为const则表示不可对其指向进行赋值。因此可以将引用看成是一个自身为const...
const int (*pf)(); // pf is a non-const pointer and points to a function which has no arguments and returns a const int ... const指针的解读规则差不多就是这些了…… 指针自身为const表示不可对该指针进行赋值,而指向物为const则表示不可对其指向进行赋值。因此可以将引用看成是一个自身为const...
std::addressof(dereference of a const pointer rvalue reference) erroneously returns 0 inside constexpr function Closed - FixedView solution16 0Votes HWHan Wang - Reported Sep 07, 2020 8:32 PM Simple code as below. This bug might be dangerous in generic code. ...