将指针传递到临时对象 [英] Passing a pointer to temporary object
问题描述
我们知道我们可以通过const引用将临时对象传递给函数,如下所示:
class A
{
public:
A(int _b = 0)
{
b = _b;
}
int b;
};
void foo(A& a){printf(%d,ab);}
void cfoo(const A& a){printf(%d,ab); }
int main(void)
{
// foo(A(4));不编译
cfoo(A(5));
}
但是通过指针传递呢?
为什么要编译?
void pfoo(A * pa){pa-> b = 19;}
int main(void)
{
pfoo(& A(5));
}
但是如何传递匿名变量指针?为什么会编译?
你可能使用的编译器不符合C ++标准。
可以不使用r值(临时)对象的地址。这不应该编译。
但是, operator&
重载,以便它可以在临时对象上调用,例如:
struct A
{
A * operator&(){return this; }
};
在C ++ 11中,临时对象可以是绑定到r值引用。之后,r值引用的行为类似于l值,因此临时对象的地址可以取:
struct A {};
void foo(A *);
void foo(A&& a){foo(& a); }
int main(){
foo(A {});
}
We know that we can pass temporary objects to functions by const reference, like this:
class A { public: A(int _b = 0) { b = _b; } int b; }; void foo(A& a) {printf("%d", a.b);} void cfoo(const A& a) {printf("%d", a.b);} int main(void) { //foo(A(4)); doesn't compile cfoo(A(5)); }
but what about passing by pointer? why does this compile?
void pfoo(A* pa) {pa->b = 19;} int main(void) { pfoo(&A(5)); }
解决方案but what about passing anonymous variables pointer? why does this compile?
You are probably using a compiler that does not honour C++ standard.
No address of an r-value (temporary) object can be taken. That should not compile.
However,
operator&
can be overloaded, so that it can be invoked on a temporary object, e.g.:struct A { A* operator&() { return this; } };
In C++11 a temporary object can be bound to an r-value reference. After that r-value reference behaves like an l-value and hence the address of a temporary object can be taken:
struct A {}; void foo(A*); void foo(A&& a) { foo(&a); } int main() { foo(A{}); }
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