CODEWITHSUNDEEP
c++methodsthrow
Vladislav Krejcirik
Vladislav Krejcirik

Reputation: 623

C++ method declaration with throw(...)

I have a question regarding declaration class method in C++. I usually using declaration method without providing throw (will throw anything). But I saw somewhere declaration like this:

void method(int param) throw (...);

Does it have any sense? What is difference?

Upvotes: 1

Views: 757

Answers (2)

Puppy
Puppy

Reputation: 146910

It's a Microsoft extension, which basically means "This function may throw something", which is equivalent to having no specification at all. The value of adding it is questionable.

Upvotes: 2

Lightness Races in Orbit
Lightness Races in Orbit

Reputation: 385104

Well it's not valid C++ so, no, it doesn't "have any sense":

g++ -std=c++11 -O2 -Wall -pedantic -pthread main.cpp && ./a.out
main.cpp:1:31: error: expected type-specifier before '...' token
 void method(int param) throw (...);

(nor in C++03)

The only place you can write ... in an exception specifier is after the type-id in a dynamic-exception-specification in order to form a pack expansion ([C++11: 15.4/16]), like so:

template <typename ...T>
void method(int param) throw (T...) {}

int main()
{
    method<int, bool>(42);
    // ^ somewhat like invoking a `void method(int) throw(int, bool)`
}

Upvotes: 3

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