'invalid declarator before std::variant
I'm trying to implement an ad-hoc light weight state machine using std::variant. However, it seems that the variant fsm
isn't declared right as it fails with the following errors:
<source>: In function 'int main()':
<source>:235:40: error: invalid declarator before 'fsm'
235 | std::variant<states::A, states::B> fsm{std::in_place_index<0>};
| ^~~
<source>:238:5: error: 'fsm' was not declared in this scope
238 | fsm = std::visit([&](auto&& state){ state.on_event(eWifi::connected); }, fsm);
| ^~~
ASM generation compiler returned: 1
<source>: In function 'int main()':
<source>:235:40: error: invalid declarator before 'fsm'
235 | std::variant<states::A, states::B> fsm{std::in_place_index<0>};
| ^~~
<source>:238:5: error: 'fsm' was not declared in this scope
238 | fsm = std::visit([&](auto&& state){ state.on_event(eWifi::connected); }, fsm);
| ^~~
I can't really get behind what is wrong. Here's the code:
#include <iostream>
#include <variant>
enum class eWifi {
connected,
disconnected,
};
enum class eMQTT {
connected,
disconnected,
};
int main()
{
// int a;
struct states {
struct A {
auto on_event(eWifi evt) {
std::cout << "on_event A" << std::endl;
return B{};
}
auto on_event(eMQTT evt) {
std::cout << "on_event A" << std::endl;
return B{};
}
};
struct B {
auto on_event(eWifi evt) {
std::cout << "on_event B" << std::endl;
return A{};
}
auto on_event(eMQTT evt) {
std::cout << "on_event B" << std::endl;
return A{};
}
};
}
std::variant<states::A, states::B> fsm{std::in_place_index<0>};
// while(1) {
fsm = std::visit([&](auto&& state){ state.on_event(eWifi::connected); }, fsm);
// }
}
What is the invalid declarator before 'fsm' all about?
Solution 1:[1]
- You missed the semicolon in the declaration of
struct states
.
struct states {
struct A {
};
struct B {
};
};
- You forgot to return the result in your lambda function.
fsm = std::visit([&](auto&& state) -> std::variant<states::A, states::B> {
return state.on_event(eWifi::connected);
}, fsm);
See demo.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
---|---|
Solution 1 | Nimrod |