更新时间:2022-06-27 00:20:47
每当创建一个新对象时,都会调用一个构造函数。 static_cast
总是会立即生成一个新的临时对象(但是请参阅James McNellis的注释)或
,或通过调用用户定义的转换。 (但是在
命令中有一个所需类型的对象返回,用户定义的
转换操作符必须调用一个构造函数。)
Any time a new object is created, a constructor is called. A static_cast
always results in a new, temporary object (but see comment by James McNellis) either
immediately, or through a call to a user defined conversion. (But in
order to have an object of the desired type to return, the user defined
conversion operator will have to call a constructor.)
当目标是类类型时,根据定义,C风格的转换和函数式
转换与单个参数是一样的 static_cast
。如果函数式转换具有零个或多个
参数,那么它将立即调用构造函数;在这种情况下不考虑用户定义的
转换运算符。 (并且可以
对选择调用这种类型转换提出质疑。)
When the target is a class type, C style casts and functional style
casts with a single argument are, by definition, the same as a
static_cast
. If the functional style cast has zero or more than one
argument, then it will call the constructor immediately; user defined
conversion operators are not considered in this case. (And one could
question the choice of calling this a "type conversion".)
对于记录,用户定义的转换操作符可能是
call:
For the record, a case where a user defined conversion operator might be called:
class A
{
int m_value;
public
A( int initialValue ) : m_value( initialValue ) {}
};
class B
{
int m_value;
public:
B( int initialValue ) : m_value( initialValue ) {}
operator A() const { return A( m_value ); }
};
void f( A const& arg );
B someB;
f( static_cast<A>( arg ) );
在这种特殊情况下,转换是不必要的,转换
会隐式在其缺席。但在所有情况下:隐式
转换, static_cast
,C风格转换((A)someB
) function
style cast( A(someB)
), B :: operator A()
将被调用。)
In this particular case, the cast is unnecessary, and the conversion
will be made implicitly in its absence. But in all cases: implicit
conversion, static_cast
, C style cast ((A) someB
) or functional
style cast (A( someB )
),
B::operator A()
will be called.)