传递向量到函数c ++ [英] passing vector to function c++

问题描述

我有一个main.cpp test.h和test.cpp>我试图传递我的向量通过所以我可以使用它在test.cpp但我不断收到错误。

  // file：main.cpp 
 int main（）{
 vector< Item *> s; 
 //加载我的文件并赋值s [i]  - > name和s [i]  - 地址
 tester; 
} 
 
 // file：test.h 
 #ifndef TEST_H 
 #define TEST_H 
 struct Item {
 string name; 
 string address; 
}; 
 #endif 
 
 // file：test.cpp 
 int tester（Item * s []）{
 for（i = 0; i  cout<< s [i]  - > name<<<< s [i]  - > address<< endl; 
} 
 return 0; 
} 
 
 
 
 ---------------错误-------- 
文件包括从main.cpp：13：
 test.h：5：错误：âstringâ不命名类型
 test.h：6：错误：âstringâ没有命名类型
 main .cpp：在函数âptmain（）â：
 main.cpp：28：error：不能转换âstd:: vector< Item *，std :: allocator< Item *> >âtoâ¨Item**â¨for for for（）））））â>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>解决方案

A std :: vector< T> 和 T * []

更改 tester（）函数签名如下：

  // file：test.cpp 
 int tester（const std :: vector< Item>& s） std :: vector 
 //因为你不需要改变
 //在这个函数中的值
 {
 for（size_t i = 0; i  cout< s [i]  - > name<<<< s [i]  - > address<< endl; 
} 
 return 0; 
} 
  

有几种方法可以传递此 std： ：vector< T> 并且都有稍微不同的含义：

 的向量
 //，这将是本函数的范围
 void tester（std :: vector< Item *>）; 
 
 //这将使用向量的引用
 //此引用可以在
 //测试器函数中修改
 //这不涉及向量的第二个副本
 void tester（std :: vector< Item *>&）; 
 
 //这将使用向量的const引用
 //此引用不能在
 //测试器函数中修改
 //这是不涉及向量的第二个副本
 void tester（const std :: vector< Item *>&）; 
 
 //这将使用向量的指针
 //这不涉及向量的第二个副本
 // caveat：使用裸指针可能是危险的， 
 //如果可能的话，应该避免非平凡的情况
 void tester（std :: vector< Item *> *）; 
  

I have a main.cpp test.h and test.cpp> I am trying to pass my vector through so i can use it in test.cpp but i keep getting errors.

   //file: main.cpp
    int main(){
        vector <Item *> s;
         //loading my file and assign s[i]->name and s[i]-address
         tester(s);
    }

    //file: test.h
    #ifndef TEST_H
    #define TEST_H
    struct Item{
        string name;
        string address;
    };
    #endif

    //file: test.cpp
    int tester(Item *s[]){
        for (i=0; i<s.sizeof();i++){
            cout<< s[i]->name<<"  "<< s[i]->address<<endl;
        }
        return 0;
    }



    ---------------errors--------
    In file included from main.cpp:13:
    test.h:5: error: âstringâ does not name a type
    test.h:6: error: âstringâ does not name a type
    main.cpp: In function âint main()â:
    main.cpp:28: error: cannot convert âstd::vector<Item*, std::allocator<Item*> >â to âItem**â for argument â1â to âint tester(Item**)â

解决方案

A std::vector<T> and T* [] are not compatible types.

Change your tester() function signature as follows:

//file: test.cpp
int tester(const std::vector<Item>& s)   // take a const-reference to the std::vector
                                         // since you don't need to change the values 
                                         // in this function
{
    for (size_t i = 0; i < s.size(); ++i){
        cout<< s[i]->name<<"  "<< s[i]->address<<endl;
    }
    return 0;
}

There are several ways you could pass this std::vector<T> and all have slightly different meanings:

// This would create a COPY of the vector
// that would be local to this function's scope
void tester(std::vector<Item*>); 

// This would use a reference to the vector
// this reference could be modified in the
// tester function
// This does NOT involve a second copy of the vector
void tester(std::vector<Item*>&);

// This would use a const-reference to the vector
// this reference could NOT be modified in the
// tester function
// This does NOT involve a second copy of the vector
void tester(const std::vector<Item*>&);

// This would use a pointer to the vector
// This does NOT involve a second copy of the vector
// caveat:  use of raw pointers can be dangerous and 
// should be avoided for non-trivial cases if possible
void tester(std::vector<Item*>*);

这篇关于传递向量到函数c ++的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！