Lambda capture causes incompatible operand types error?

Question

Consider the following code:

main()
{
    bool t;
    ...
    std::function f = t ? [](bool b) { return b; } : [](bool b) { return !b; };          


        

Answer 1

The type of a lambda is "a unique, non-union class type" called the closure type. Each lambda is implemented as a different type, local to the scope of declaration, which has an overloaded operator () to call the function body.

Example: if you write this:

auto a=[t](bool b){return t==b;};
auto b=[t](bool b){return t!=b;};

Then the compiler compiles this (more or less):

class unique_lambda_name_1 
{
 bool t; 
public:
 unique_lambda_name_1(bool t_) t(_t) {}
 bool operator () (bool b) const { return t==b; }
} a(t); 
class unique_lambda_name_2
{
 bool t;
public: 
 unique_lambda_name_2(bool t_) t(_t) {}
 bool operator () (bool b) const { return t!=b; }
} b(t); 

a and b have different types and can't be used in the ?: operator.

However, §5.1.2(6) says, that the closure type of a lambda with no capture has a non-explicit, public conversion operator, which converts the lambda to a function pointer - non-closures can be implemented as simple functions. Any lambda with the same argument and return types can be converted to the same type of pointer and so the ternary ?: operator can be applied to them.

Example: the non-capture lambda:

auto c=[](bool b){return b;};

is implemented like this:

class unique_lambda_name_3
{
 static bool body(bool b) { return b; }
 public:
 bool operator () (bool b) const { return body(b); }
 operator decltype(&body) () const { return &body; }
} c; 

which means that this line:

auto x = t?[](bool b){return b;}:[](bool b){return !b;};

means actually this:

// a typedef to make this more readable
typedef bool (*pfun_t)(bool); 
pfun_t x = t?((pfun_t)[](bool b){return b;}):(pfun_t)([](bool b){return !b;});