I have the following code where a variable is being initialized with the result of a function call. This function throws so I set up a try-catch to catch the exception. For some reason the exception is still showing up on the screen even after the catch clause runs.
#include <iostream>
#include <stdexcept>
int f() { throw std::invalid_argument("threw"); return 50; }
struct S
{
S()
try : r(f())
{
std::cout << "works";
}
catch(const std::invalid_argument&)
{
std::cout << "fails";
}
int r;
};
int main()
{
S s;
}
This code prints "fails" after showing the exception:
terminate called after throwing an instance of 'std::invalid_argument' what(): threw
Why is the exception still thrown? I have the same code set up in main and it works without fail:
int main()
{
try { throw std::invalid_argument("blah"); }
catch(const std::invalid_argument&) { }
}
So why does it fail when being used in an initializer list?
Because if an exception is thrown, Code in the finally clause will execute as the exception propagates outward, even if the exception aborts the rest of the method execution; Code after the try/catch block will not get executed unless the exception is caught by a catch block and not rethrown.
A thrown object may match several catch block but only the first catch block that matches the object will be executed. A catch-block will catch a thrown exception if and only if: the thrown exception object is the same as the exception object specified by the catch-block.
Bookmark this question. Show activity on this post. Java does not allow us to catch the exception twice, so we got compile-rime error at //1 .
Without using throws When an exception is cached in a catch block, you can re-throw it using the throw keyword (which is used to throw the exception objects). If you re-throw the exception, just like in the case of throws clause this exception now, will be generated at in the method that calls the current one.
Constructors with function try blocks (like what you have for S
) automatically rethrow any exceptions caught by the catch
block. Consequently, after the catch
catches the exception, it rethrows it. This behavior is different from normal catch
handlers, which don't do this. I think the rationale is that if construction of a data member or base class fails, the object has failed to construct. The purpose of the catch
handler is just to do any extra cleanup before the exception propagates outward.
Hope this helps!
From the C++11 Standard, 15.3/15:
The currently handled exception is rethrown if control reaches the end of a handler of the function-try-block of a constructor or destructor.
The reasons are well explained in the GOTW Jerry links under your question, but summarily: imagine if it hadn't rethrown, then the next lines after S s;
will presumably attempt to use s
despite it never having completed construction, and when s
leaves scope the constructor will have arranged a call to the destructor for s
- potentially freeing never-initialised pointers, releasing never-taken locks etc..
By way of contrast, if you let the data member be default initialised then assign to it from a try/catch block in the constructor body, the state of the object including bases and data members can potentially be kept in some coherent state: it's up to you as the programmer to decide whether that state's ok - i.e. whether you'll use the try/catch block inside the constructor body, and have later member functions handle a potentially default-constructed data member.
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