按其返回类型,强制转换规则重载函数 [英] Overloaded functions by its return type, cast rules

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本文介绍了按其返回类型,强制转换规则重载函数的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述



1. abybody可以解释为什么C ++函数不能被它的

返回类型重载?返回类型的优先级高于强制转换规则。

例如:


char input(); //可以编译为nameinput $ char $ void

int input(); //可以编译成名称输入$ int $ void

....


int i = 3 +''0''+输入(); //可以编译为:

int(3)+ int(''0'')+ input $ int $ void()

char ch = 3+' '0''+输入(); //可编译为:

char(3)+ char(''0'')+ input $ char $ void()


是否存在可以理解的障碍?这样的

重载有一些利润,例如,我们可能不会写:

char input_char();

int input_int() ;


2.仅按回报类型进行投射。我对吗?例如:


A a;

struct B {静态B& operator +(const A&,const B&); };

B b;

struct B {静态C& operator +(const B&,const C&); };

C c;

struct D {

静态D& operator +(const D&,const D&);

D(const C&);

};


D d = A + b + C; //给我


d.operator =(D :: operator +(D :: operator +(D(a)+ D(b)),D(c)));

或声明的class_name :: operator +;改变演员

d.operator =(D(C :: operator +(B :: operator +(a,b),c)));


或这是常见问题吗?


1. Can abybody explain me why C++ function can not be overloaded by its
return type? Return type can has priority higher than casting rules.
For example:

char input(); //can be compiled to name "input$char$void"
int input(); //can be compiled to name "input$int$void"
....

int i= 3+''0''+input(); //can be compiled to:
int(3)+int(''0'')+input$int$void()
char ch= 3+''0''+input(); //can be compiled to:
char(3)+char(''0'')+input$char$void()

Are any understandable obstacles exist? There is some profit for such
overloading, for example, we would may not write:
char input_char();
int input_int();

2. Cast is only by return type. Am i right? For example:

A a;
struct B{ static B& operator+ (const A&, const B&); };
B b;
struct B{ static C& operator+ (const B&, const C&); };
C c;
struct D{
static D& operator+ (const D&, const D&);
D(const C&);
};

D d=a+b+c; //give me

d.operator=( D::operator+( D::operator+( D(a)+D(b) ), D(c) ) );
or declared "class_name::operator+" change cast
d.operator=( D( C::operator+ ( B::operator+(a,b), c ) ) );

Or is it FAQ question?

推荐答案

char


void"

int input(); //可以编译为名称input
void"
int input(); //can be compiled to name "input


int


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