Why am I getting error while initializing a struct with curly braces?

Question

I am using the below code and getting error. I don't understand why I am getting this error.

prog.cpp: In function ‘int main()’:
prog.cpp:15:44: error: could not convert ‘{"foo", true}’ from 
                       ‘<brace-enclosed initializer list>’ to ‘option’
                       option x[] = {{"foo", true},{"bar", false}};
                                            ^
prog.cpp:15:44: error: could not convert ‘{"bar", false}’ from 
                       ‘<brace-enclosed initializer list>’ o ‘option’

The code

#include <iostream>
#include <string>
 
struct option
{
    option();
    ~option();
 
    std::string s;
    bool b;
};
 
option::option() = default;
option::~option() = default;

int main()
{
    option x[] = {{"foo", true},{"bar", false}};
}

Answer 1

When you provide^† the default constructor and destructor, you are making the struct be a non-aggregate type, hence aggregate initialization is not possible.

However, you can check if a type is an aggregate using the standard std::is_aggregate_v trait. (Since c++17).

See here for your case. It is not an aggregate, as you provided^† those constructors.

You have the following three ways to make this work:

• Remove the constructors and you are good to go.

  struct option
  {
     std::string s;
     bool b;
  };
  

• Default the constructors inside the struct (i.e. declaring^†).

  struct option 
  {    
     std::string s;
     bool b;
  
     option() = default;
     ~option() = default;
  };
  

• Otherwise, you need to provide a suitable constructor in your struct.

  struct option 
  {
     std::string mStr;
     bool mBool;
  
     option(std::string str, bool b)
        : mStr{ std::move(str) }
        , mBool{ b }
     {}
  
     // other constructors...
  };
  

---

^† The following post explains when the constructor will be defaulted, when it is considered as user-declared and user-provided clearly: (Credits @NathanOliver)

C++ zero initialization - Why is `b` in this program uninitialized, but `a` is initialized?