赋值运算符的布尔和字符串重载(C ++) [英] Boolean and String Overloads of the Assignment Operator (C++)
问题描述
我定义赋值运算符的多个重载,如下所示:
I am defining multiple overloads of the assignment operator as follows:
Foo.h
class Foo
{
private:
bool my_bool;
int my_int;
std::string my_string;
public:
Foo& operator= (bool value);
Foo& operator= (int value);
Foo& operator= (const std::string& value);
};
Foo.cpp
// Assignment Operators.
Foo& Foo::operator= (bool value) {my_bool = value; return *this;}
Foo& Foo::operator= (int value) {my_int = value; return *this;}
Foo& Foo::operator= (const std::string& value) {my_string = value; return *this;}
这里是我的main.cpp > SURPRISE ):
And here's my main.cpp (see the comment marked SURPRISE
):
Foo boolFoo;
Foo intFoo;
Foo stringFoo;
// Reassign values via appropriate assignment operator.
boolFoo = true; // works...assigned as bool
intFoo = 42; // works...assigned as int
stringFoo = "i_am_a_string"; // SURPRISE...assigned as bool, not string
std::string s = "i_am_a_string";
stringFoo = s; // works...assigned as string
// works...but awkward
stringFoo = static_cast<std::string>("i_am_a_string");
问题:有人可以告诉我为什么正在评估一个uncasted字符串在布尔上下文中?
Question: Can someone tell me why an uncasted string literal is being evaluated in a boolean context?
推荐答案
C ++标准定义了第13.3节中的重载解析规则, b
$ b
The C++ standard defines overload resolution rules in chapter 13.3, there you find:
13.3.3.2排名隐含转换序列[over.ics.rank]
2 当比较隐式转换序列的基本形式(如13.3.3.1中定义)
2 When comparing the basic forms of implicit conversion sequences (as defined in 13.3.3.1)
- 标准转换序列.3.1.1)是比用户定义的转换序列或省略号转换序列更好的转换序列,
— a standard conversion sequence (13.3.3.1.1) is a better conversion sequence than a user-defined conversion sequence or an ellipsis conversion sequence, and
- 用户定义的转换序列(13.3.3.1 .2)是比省略号转换序列(13.3.3.1.3)更好的转换序列。
— a user-defined conversion sequence (13.3.3.1.2) is a better conversion sequence than an ellipsis conversion sequence (13.3.3.1.3).
这意味着编译器将更喜欢从字符串字面到 bool
或 int
(如果可用)的标准转换序列。现在,哪些标准转化相关?在你的case,这两个是相关的:
This means that the compiler will prefer a standard conversion sequence from the string literal to bool
or int
if available. Now, which standard conversions are relevant? In your case, these two are relevant:
4.2数组到指针的转换[conv.array]
1 类型NT数组或T未知边界数组的左值或右值可以转换为类型指针到T。结果是指向数组第一个元素的指针。
1 An lvalue or rvalue of type "array of N T" or "array of unknown bound of T" can be converted to a prvalue of type "pointer to T". The result is a pointer to the first element of the array.
此转换将转换字符串文字, $ c> const char [N] ,转换为 const char *
。第二个是:
This conversion turns the string literal, which is of type const char[N]
, into a const char*
. The second one is:
4.12布尔转换[conv.bool]
1 算术,无范围枚举,指针或指向成员类型的指针的prvalue可以转换为 bool
。零值,空指针值或空成员指针值将转换为 false
;任何其他值将转换为 true
。类型 std :: nullptr_t
的prval值可以转换为 bool
类型的prvalue;结果值为 false
。
1 A prvalue of arithmetic, unscoped enumeration, pointer, or pointer to member type can be converted to a prvalue of type bool
. A zero value, null pointer value, or null member pointer value is converted to false
; any other value is converted to true
. A prvalue of type std::nullptr_t
can be converted to a prvalue of type bool
; the resulting value is false
.
这就是为什么指针转换为 bool
。由于存在标准转换序列,因此不使用 std :: string
的用户定义转换。
That is the reason why the pointer is converted to bool
. Since a standard conversion sequence exists, the user-defined conversion to std::string
is not used.
解决你的问题,我建议你添加另一个重载的版本, const char *
,并将它转发到 const std :: string& / code> overload。
To solve your problem, I suggest you add another overloaded version that takes const char*
and make it forward the call to the const std::string&
overload.
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