Reputation: 157
In question about scope of exception it is stated by Aj. that throw and catch clauses will create copies of the exception (unless reference is used I guess)
I tried myself a small toy code and I don't understand the result. here :
//g++ 7.4.0
#include <iostream>
using namespace std;
struct Some_error {
Some_error(float code):err_code(code){ cout << "Some_error(" << err_code << ")\n"; }
~Some_error() { cout << "~Some_error(" << err_code << ")\n"; }
Some_error(const Some_error& o):err_code(o.err_code+0.1) { cout << "Some_error(copy::" << err_code << ")\n"; }
Some_error(Some_error&& o):err_code(std::move(o.err_code)+.01){ cout << "Some_error(move::" << err_code << ")\n"; }
int get_code() const { return err_code; }
private : float err_code;
};
int do_task() {
if ( false ) return 42; else throw Some_error {1};
cout << "end do_task\n" ;
}
void taskmaster(){
try { auto result = do_task(); cout << "the answer is " << result << "\n" ; }
catch (Some_error e) { cout << "catch Some_error : " << e.get_code() << "\n" ; }
cout << "end taskmaster\n" ;
}
int main() { taskmaster(); }
the trace I get is as follows :
Some_error(1)
Some_error(copy::1.1)
catch Some_error : 1
~Some_error(1.1)
~Some_error(1)
end taskmaster
Now first, as I used no reference here, according to Aj., I would expect 2 copies to happen.
And second, there was a copy, that set err_code to 1.1, but the display is still 1.
Remark: just to be complete, I changed the catch to : catch(Some_error& e), and then the trace looks fine to me :
Some_error(1)
catch Some_error : 1
~Some_error(1)
end taskmaster
Upvotes: 0
Views: 46
Reputation: 545508
I would expect 2 copies to happen.
Why? Only one copy is made by the catch
block. Where would the second copy happen?
set err_code to 1.1, but the display is still 1.
Because get_code
returns an int
, so the floating point value gets truncated.
Upvotes: 1