引用定义对象的替代名称.引用类型引用"另一种类型.我们通过编写 &d
形式的声明符来定义引用类型,其中 d
是要声明的名称.
A reference defines an alternative name for an object. A reference type "refers to"
another type. We define a reference type by writing a declarator of the form &d
,
where d
is the name being declared.
接下来是引用不是对象.相反,引用只是已存在对象的另一个名称.所以我们将使用这些引用来通过引用传递参数,以便它直接影响实际参数.
The next thing is a reference is not an object. Instead, a reference is just another name for an already existing object. So we'll use these references to pass the parameter by reference so that it directly effect the actual parameters.
问题:当我们在函数名称之前使用引用 (&
) 会发生什么?
Question:
What happens when we use a reference (&
) before a function name?
我有点困惑,因为在我看来它会返回返回的别名(变量名).还是我错了?.
I'm a little bit confused, as in my opinion it will return the alias of return (variable name). Or am I wrong?.
std::vector<std::string> &split(const std::string &s, char delim,
std::vector<std::string> &elems) {
std::stringstream ss(s);
std::string item;
while (std::getline(ss, item, delim)) {
elems.push_back(item);
}
return elems;
}
在 C++ 中,当在函数名之前使用引用符号 (&
) 时函数的声明,它与函数的返回值相关联,意味着该函数将通过引用返回.
In C++, when the ref-sign (&
) is used before the function name in the declaration of a function it is associated with the return value of the function and means that the function will return by reference.
int& foo(); // Function will return an int by reference.
当不在声明上下文中使用时,将引用符号放在函数名称之前会导致调用 address-of 运算符返回函数的地址.这可以用于例如创建一个指向函数的指针.
When not used within a declaration context, putting the ref-sign before a function name results in calling the address-of operator returning the address of the function. This can be used to e.g. create a pointer to a function.
// Some function.
int sum(int a, int b) {
return a + b;
}
int main() {
// Declare type func_ptr_t as pointer to function of type int(int, int).
using func_ptr_t = std::add_pointer<int(int, int)>::type;
func_ptr_t func = ∑ // Declare func as pointer to sum using address-of.
int sum = func(1, 2); // Initialize variable sum with return value of func.
}
在 C 中,&
的唯一用途是用于地址运算符.C 语言中不存在引用.
In C, the only use of &
is for the address-of operator. References does not exist in the C language.
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