如何传递std :: vector< void *> *作为函数的std :: vector< mystruct *> * [英] How to pass std::vector<void*> * as std::vector<mystruct*>* to a function

问题描述

如何传递std :: vector< void *> *作为std :: vector< MyStruct *> *到一个函数？



函数声明是void SetDataReferences（std :: vector< MyStruct *> * pVector ）;



如果我传递std :: vector< void *> * as std :: vector< MyStruct *> *我收到以下错误



错误C2664：'SetDataReferences'：无法从'std :: vector<转换参数2 ; _Ty> *'到'std :: vector< _Ty> *'



如何解决这个问题？



我的尝试：



我不知道它的替代方案。

解决方案

如果你确定矢量你传递的实际上包含指向有效 MyStruct 实例的指针， reinterpret_cast [ ^ ]会做的伎俩。例如，尝试：

  #include   <   iostream  >  
   #include   <   vector  >  
  使用  namespace  std; 
 
  struct  Foo 
 {
 string s; 
  int  i; 
}; 
 
 
  void  reportFooVector（vector< Foo *> * pvf）
 {
  for （ auto  p：* pvf）
 cout<< p-> s<<  ，<< p-> i<< ENDL; 
} 
 
 
  int  main（）
 {
 Foo foo {  foo， 5 }; 
 Foo goo {  goo， 10 }; 
 
 vector< void *> vf {& foo，&咕}; 
 
 reportFooVector（reinterpret_cast< vector< Foo *> *>（& vf））; 
} 

替代方案非常明确：以正确的类型创建和传递参数。



你应该非常严肃地对待这个警告，因为它尖叫编码缺陷。

  //   TODO：更改输入的创建代码参数 

Hi,

How to pass std::vector<void*> * as std::vector<MyStruct*>* to a function?

The function declaration is void SetDataReferences(std::vector<MyStruct*>* pVector);

if I pass std::vector<void*> * as std::vector<MyStruct*>* I am getting below error

error C2664: 'SetDataReferences' : cannot convert parameter 2 from 'std::vector<_Ty> *' to 'std::vector<_Ty> *'

How to resolve this?

What I have tried:

I dont know what is the alternative for it.

解决方案

Provided you are sure the vector you are passing actually contains pointer to valid MyStruct instances, a reinterpret_cast[^] would do the trick. Try, for instance:

#include <iostream>
#include <vector>
using namespace std;

struct Foo
{
  string s;
  int i;
};


void reportFooVector( vector <Foo*> * pvf )
{
  for (auto p : *pvf)
    cout << p->s << ", " << p->i << endl;
}


int main()
{
  Foo foo{ "foo",  5};
  Foo goo{ "goo", 10};

  vector <void *> vf { &foo, & goo};

  reportFooVector( reinterpret_cast< vector < Foo * > * >(&vf));
}

The alternative is very clear: creating and passing the parameter in the right type.

You should take this warning very serious because it screams for a coding flaw.

//TODO: Change the creation code of input parameter

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