Question

    #include<iostream>
    #include <conio.h>
    using namespace std;

   struct book
    {  int bookid;
       char title[20];
       float price;
    }b2;

 int main()
   {  
     b2={100,"c++ by saurabh",105.2}; //values initialised during variable declaration

     cout<<"\n"<<b2.bookid;
     cout<<b2.title<<"  "<<b2.price;
     return 0;
     getch();
   }

以上代码显示输出错误，如下所示：

C：\ Users \ shandude \ Documents \ codeblock \ cprog \ struct2.cpp | 13 | error：不匹配＆＃39; operator =＆＃39; （操作数类型是＆＃39; book＆＃39;和＆＃39;＆＃39;）|

C：\ Users \ shandude \ Documents \ codeblock \ cprog \ struct2.cpp | 5 |注意：没有来自＆＃39;＆＃39;的参数1的已知转换到＆＃39; const book＆amp;＆＃39; |

Answer 1

您可以使用：

title

<强> PS

我建议将成员变量std::string更改为char。使用struct book { int bookid; std::string title; float price; };数组来表示用户代码中的字符串在2017年是不合时宜的。

rfxtrx.service

Answer 2

您正在做的不是初始化，而是赋值，因为b2之前已经声明过。您需要在声明变量的位置进行初始化：

   struct book
   {   int bookid;
       char title[20];
       float price;
   } b2 = {100,"c++ by saurabh",105.2}; //values initialised during variable declaration

 int main()
 {  
     cout<<"\n"<<b2.bookid;
     cout<<b2.title<<"  "<<b2.price;
     return 0;
     getch();
 }

Answer 3

What complier do you use?

After removing #include <conio.h> and replacing float with double, Clang and VC++ both accept this code while GCC complains. I think it is a bug for GCC.

Although this is not an initialization, it is equivalent to call the assignment operator with the initializer-list as an argument. The parameter of the assignment operator is const book&, and use this initializer-list to initialize this reference is well-defined. The program is also well-defined.

Answer 4

您正尝试按ol初始化b2。您可以阅读reference以了解如何初始化它有很多方法。一个简单的方法是：

list initialization

不匹配＆＃39;运营商=＆＃39; （操作数类型是＆＃39; book＆＃39;和＆＃39; <brace-enclosed initializer =“”list =“”>＆＃39;）

4 个答案: