Question

我正在重新编写一些C ++而且我老实说生锈了。我觉得如果我知道如何正确地说出来，我会快速回答我的问题，但我仍然感谢你的帮助。

sanitycheck.cpp：

#include <string>    
using namespace std;

typedef struct STR_1 {    
  int val_a, val_b;

  STR_1 (int a, int b)
  { val_a = a; val_b = b; }    
} STR_1;

typedef struct STR_2{    
  string name;
  STR_1 myStr1;

  STR_2 (string n, STR_1 s)
  { name=n; myStr1 = s; }
} STR_2;

int main(){

  return 0;
} // end main

当我尝试使用g++ -o sanitycheck ./test/sanitycheck.cpp进行编译时，我得到以下内容，

./test/sanitytest.cpp: In constructor ‘STR_2::STR_2(std::string, STR_1)’:
./test/sanitytest.cpp:25:3: error: no matching function for call to ‘STR_1::STR_1()’
   { name=name; myStr1 = &s; }
   ^
./test/sanitytest.cpp:25:3: note: candidates are:
./test/sanitytest.cpp:11:3: note: STR_1::STR_1(int*, int*)
   STR_1 (int *a, int *b)
   ^
./test/sanitytest.cpp:11:3: note:   candidate expects 2 arguments, 0 provided
./test/sanitytest.cpp:7:16: note: STR_1::STR_1(const STR_1&)
 typedef struct STR_1 {
                ^
./test/sanitytest.cpp:7:16: note:   candidate expects 1 argument, 0 provided
./test/sanitytest.cpp:25:23: error: no match for ‘operator=’ (operand types are ‘STR_1’ and ‘STR_1*’)
   { name=name; myStr1 = &s; }
                       ^
./test/sanitytest.cpp:25:23: note: candidate is:
./test/sanitytest.cpp:7:16: note: STR_1& STR_1::operator=(const STR_1&)
 typedef struct STR_1 {
                ^
./test/sanitytest.cpp:7:16: note:   no known conversion for argument 1 from ‘STR_1*’ to ‘const STR_1&’

我不清楚的一件事是STR_1 myStr1; STR_2为什么需要首先调用STR_1构造函数？我无法使用

初始化这两种类型

int main()
{
    STR_1 bob = STR_1(5,6);
    STR_2 tom = STR_2('Tom',bob);
    return 0;
}

谢谢！

Answer 1

除非从对OP的评论中的链接中已经不清楚：这里，

typedef struct STR_2{    
  string name;
  STR_1 myStr1;

  STR_2 (string n, STR_1 s)  // here the myStr1 default constructor is called
  { name=name; myStr1 = s; }
} STR_2;

要求STR_1是默认的可构造性。为了解决这个问题，你必须在构造函数的初始化列表中构造成员STR_1 myStr1;：

  STR_2 (string n, STR_1 s) : name(n), myStr1(s) {}

DEMO

这将调用编译器生成的STR_1复制构造函数，而不是默认构造函数（通过提供自定义构造函数来抑制其自动生成）。

另一种选择是使用指向STR_1的指针：

typedef struct STR_2{    
  string name;
  std::unique_ptr<STR_1> myStr1;

  STR_2 (string n, STR_1 s)
  { name=name; myStr1 = std::make_unique<STR_1>(s); }  //just for the sake of explanation
                                                       //again, this would be better
                                                       //done in the initializer list
} STR_2;

然而，出于某个原因，我会偏离第一种选择。

从构造函数初始化结构

1 个答案: