catch float type

[Caruso]

I'm trying to understand try, throw and catch. And I ran into an issue with catching a float.

With this code I got a runtime error. Is it the compiler (MS visual studio 2015) or I'm I actually violating c++ rules?

int main(){
try {
throw 20.0;
}
catch (float i) {
std::cout << i;
}
return 0;
}

It tells me that my of type float has a value of -107374176.
placing a f behind 20.0. Or change catch(float i) into catch (double i).
It works fine.

int main(){
try {
throw 20.0f;
}
catch (float i) {
std::cout << i;
}
return 0;
}

[Cubbi]

20.0 is a double, not a float, and so catch(float i) can't catch it. An uncaught exception is a runtime error.

20.0f is a float, so catch(float i) can catch it.

Normal C++ code throws library exception types: throw std::runtime_error("stuff happened"); and catches them by their common base type catch(const std::exception& e) { std::cout << e.what() << '\n'; }

[Caruso]

Thank you very much. it is very helpful.

I did run a check to deepen my understanding.

std::cout << sizeof(20.0)
//becomes 8, therefore double
std::cout << sizeof(20.0f)
//becomes 4 therefore float

[mbozzi]

A better way to check that in general might have been

# include <type_traits>
... 
static_assert(std::is_same<decltype(20.0), double>::value, "20.0 is not a double");

Because there is no guarantee that different types don't have the same size.

[gunnerfunner]

max: sorry to sound pedantic but I'd probably re-write the static_assert message as something like

first template parameter is not a double

[mbozzi]

That's not pedantic at all, and you're totally right. I'm afraid I have been making that mistake consistently....

FWIW, I was very glad that in C++17 you can omit the error message as in a normal assert().

Fixed ;)