问题
While developing an application, I had the following problem. I wanted to return an empty std::list<string>
when a given function pointer was null, or the result of that function otherwise. This is a simplified version of my code:
typedef std::list<std::string> (*ParamGenerator)();
std::list<std::string> foo() {
/* ... */
ParamGenerator generator = ...;
if(generator)
return generator();
else
return {};
}
However, I usually like to use the ternary (?:
) operator in these cases, so I tried using it this way (as usual):
return generator ? generator() : {};
But got this error:
somefile.cpp:143:46: error: expected primary-expression before ‘{’ token
somefile.cpp:143:46: error: expected ‘;’ before ‘{’ token
Does this mean I can't use the ternary operator to return objects created using their constructor from an initializer_list
? Is there any particular reason for that?
回答1:
Standard writes in 8.5.4.1: List-initialization
Note: List-initialization can be used
- as the initializer in a variable definition (8.5)
- as the initializer in a new expression (5.3.4)
- in a return statement (6.6.3)
- as a function argument (5.2.2)
- as a subscript (5.2.1)
- as an argument to a constructor invocation (8.5, 5.2.3)
- as an initializer for a non-static data member (9.2)
- in a mem-initializer (12.6.2)
- on the right-hand side of an assignment (5.17)
Nothing of them is a ternary operator. The more minimalistic return 1?{}:{};
is invalid too, what you want is impossible.
Of course you can explicitly call the constructor std::list<std::string>{}
, but I would recommend to write out the if
-else
-block as you already did.
回答2:
When you do {}
the compiler has no knowledge of the type you are expecting, so it's just a meaningless expression that the compiler doesn't know what to do with. Both sides of the :
are evaluated separately, and only then will the compiler complain if the types don't match. I would just do this:
return generator ? generator() : std::list<std::string>();
回答3:
If you really like the ternary operator, you can try something like this:
return generator ? generator() : decltype(generator()) { "default value", "generator was empry" };
it will work even if you change the return types later.
回答4:
Another possibility is to define a wrapper function for the conditional operator:
template<class T> T& conditional(bool b, T&x, T&y) { return b ? x : y; }
template<class T> const T& conditional(bool b, const T&x, const T&y) { return b ? x : y; }
That allows you to call:
return conditional(generator, generator(), {});
来源:https://stackoverflow.com/questions/9982681/ternary-operator-c11-constructor-from-initializer-list