rethrowing exceptions: why does this happen?

[Steven T. Hatton]

I haven't thought about rethrowing an exception in a long time, and tried to
do it the wrong way. Now I'm curious about /why/ it's wrong. My
expectation was that the exception instance would simply be passed up the
call stack as the same object. It would therefore behave polymorphically .
Is the behavior shown below compiler-specific? I'm using gcc 4.0.2.

In terms of what goes on at runtime, why does the `throw e' act as if I
threw a raw std::exception?

#include <stdexcept>
#include <iostream>
void generate(){
throw std::invalid_ar gument("generat e() invalid argument exception");
}

void handle(bool broken) {
try {
generate();
} catch (std::exception & e) {
std::cout<< "handle:"<< e.what() <<std::endl;
if(broken)throw e;
throw;
}
}

void rehandle(bool broken=false) {
try {
handle(broken);
} catch (std::exception & e) {
std::cout<< "rehandle:" << e.what() <<std::endl;
if(broken)throw e;
throw;
}
}
int main() {
try {
rehandle();
} catch(std::exce ption& e) {
std::cout<< "main:"<< e.what() <<std::endl;
}

try {
rehandle(true);
} catch(std::exce ption& e) {
std::cout<< "main:"<< e.what() <<std::endl;
}
}

/*-----------output---------
handle:generate () invalid argument exception
rehandle:genera te() invalid argument exception
main:generate() invalid argument exception
handle:generate () invalid argument exception
rehandle:St9exc eption
main:St9excepti on
*/
--
NOUN:1. Money or property bequeathed to another by will. 2. Something handed
down from an ancestor or a predecessor or from the past: a legacy of
religious freedom. ETYMOLOGY: MidE legacie, office of a deputy, from OF,
from ML legatia, from L legare, to depute, bequeath. www.bartleby.com/61/

[Pete Becker]

Steven T. Hatton wrote:

>
In terms of what goes on at runtime, why does the `throw e' act as if I
threw a raw std::exception?

A throw-expression throws creates a temporary object whose type is the
static type of its argument. Since e is a reference to std::exception,
the type of the temporary object is std::exception.

Forget for the moment that e is a reference to the exception object.
Pretend that it's just some reference to an object of type
std::exception. In that case, the only legitimate copy you could make of
it would be an object of type std::exception, because you don't know
it's dynamic type.

--

-- Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com)
Author of "The Standard C++ Library Extensions: a Tutorial and
Reference." (www.petebecker.com/tr1book)

[Gianni Mariani]

Steven T. Hatton wrote:

I haven't thought about rethrowing an exception in a long time, and tried to
do it the wrong way. Now I'm curious about /why/ it's wrong. My
expectation was that the exception instance would simply be passed up the
call stack as the same object. It would therefore behave polymorphically .
Is the behavior shown below compiler-specific? I'm using gcc 4.0.2.

In terms of what goes on at runtime, why does the `throw e' act as if I
threw a raw std::exception?

Becuase "throw e" makes a copy of e and since the type is
"std::exception ", you get a slice of the "invalid_argume nt" exception
and loose any information about the exception. "throw" however throws
the same object that was caught.

[Salt_Peter]

Steven T. Hatton wrote:

I haven't thought about rethrowing an exception in a long time, and tried to
do it the wrong way. Now I'm curious about /why/ it's wrong. My
expectation was that the exception instance would simply be passed up the
call stack as the same object. It would therefore behave polymorphically .
Is the behavior shown below compiler-specific? I'm using gcc 4.0.2.

In terms of what goes on at runtime, why does the `throw e' act as if I
threw a raw std::exception?

#include <stdexcept>
#include <iostream>
void generate(){
throw std::invalid_ar gument("generat e() invalid argument exception");
}

void handle(bool broken) {
try {
generate();
} catch (std::exception & e) {
std::cout<< "handle:"<< e.what() <<std::endl;
if(broken)throw e;
throw;
}
}

void rehandle(bool broken=false) {
try {
handle(broken);
} catch (std::exception & e) {
std::cout<< "rehandle:" << e.what() <<std::endl;
if(broken)throw e;
throw;
}
}
int main() {
try {
rehandle();
} catch(std::exce ption& e) {
std::cout<< "main:"<< e.what() <<std::endl;
}

try {
rehandle(true);
} catch(std::exce ption& e) {
std::cout<< "main:"<< e.what() <<std::endl;
}
}

/*-----------output---------
handle:generate () invalid argument exception
rehandle:genera te() invalid argument exception
main:generate() invalid argument exception
handle:generate () invalid argument exception
rehandle:St9exc eption
main:St9excepti on
*/
--

You are rethrowing a copied slice of the original object.

#include <iostream>
#include <typeinfo>

struct Base
{
virtual ~Base() { }
};

struct Derived : public Base
{
};

void foo()
{
try {
throw Derived();
}
catch(const Base& r_b)
{
std::cerr << typeid(r_b).nam e();
std::cerr << std::endl;
// throw;
throw r_b; // copy a slice
}
}

int main()
{
try {
foo();
}
catch(const Base& r_b)
{
std::cerr << typeid(r_b).nam e();
std::cerr << std::endl;
}
}

/*
Derived
Base
*/

[Steven T. Hatton]

Salt_Peter wrote:

>
Steven T. Hatton wrote:

>I haven't thought about rethrowing an exception in a long time, and tried
to
do it the wrong way. Now I'm curious about /why/ it's wrong. My
expectation was that the exception instance would simply be passed up the
call stack as the same object. It would therefore behave
polymorphicall y.
Is the behavior shown below compiler-specific? I'm using gcc 4.0.2.

In terms of what goes on at runtime, why does the `throw e' act as if I
threw a raw std::exception?

#include <stdexcept>
#include <iostream>
void generate(){
throw std::invalid_ar gument("generat e() invalid argument exception");
}

void handle(bool broken) {
try {
generate();
} catch (std::exception & e) {
std::cout<< "handle:"<< e.what() <<std::endl;
if(broken)throw e;
throw;
}
}

void rehandle(bool broken=false) {
try {
handle(broken);
} catch (std::exception & e) {
std::cout<< "rehandle:" << e.what() <<std::endl;
if(broken)throw e;
throw;
}
}
int main() {
try {
rehandle();
} catch(std::exce ption& e) {
std::cout<< "main:"<< e.what() <<std::endl;
}

try {
rehandle(true);
} catch(std::exce ption& e) {
std::cout<< "main:"<< e.what() <<std::endl;
}
}

/*-----------output---------
handle:generat e() invalid argument exception
rehandle:gener ate() invalid argument exception
main:generate( ) invalid argument exception
handle:generat e() invalid argument exception
rehandle:St9ex ception
main:St9except ion
*/
--

You are rethrowing a copied slice of the original object.

#include <iostream>
#include <typeinfo>

struct Base
{
virtual ~Base() { }
};

struct Derived : public Base
{
};

void foo()
{
try {
throw Derived();
}
catch(const Base& r_b)
{
std::cerr << typeid(r_b).nam e();
std::cerr << std::endl;
// throw;
throw r_b; // copy a slice
}
}

int main()
{
try {
foo();
}
catch(const Base& r_b)
{
std::cerr << typeid(r_b).nam e();
std::cerr << std::endl;
}
}

/*
Derived
Base
*/

I believe Pete Becker had it right. Even if I pass the same time as the
original event, it comes out empty. What happens is that the compiler
borrows the constructor from the class used as the catch parameter, and
constructs a new instance.
--
NOUN:1. Money or property bequeathed to another by will. 2. Something handed
down from an ancestor or a predecessor or from the past: a legacy of
religious freedom. ETYMOLOGY: MidE legacie, office of a deputy, from OF,
from ML legatia, from L legare, to depute, bequeath. www.bartleby.com/61/

[Salt_Peter]

Steven T. Hatton wrote:

Salt_Peter wrote:

Steven T. Hatton wrote:

I haven't thought about rethrowing an exception in a long time, and tried
to
do it the wrong way. Now I'm curious about /why/ it's wrong. My
expectation was that the exception instance would simply be passed up the
call stack as the same object. It would therefore behave
polymorphically .
Is the behavior shown below compiler-specific? I'm using gcc 4.0.2.

re>

In terms of what goes on at runtime, why does the `throw e' act as if I
threw a raw std::exception?

#include <stdexcept>
#include <iostream>
void generate(){
throw std::invalid_ar gument("generat e() invalid argument exception");
}

void handle(bool broken) {
try {
generate();
} catch (std::exception & e) {
std::cout<< "handle:"<< e.what() <<std::endl;
if(broken)throw e;
throw;
}
}

void rehandle(bool broken=false) {
try {
handle(broken);
} catch (std::exception & e) {
std::cout<< "rehandle:" << e.what() <<std::endl;
if(broken)throw e;
throw;
}
}
int main() {
try {
rehandle();
} catch(std::exce ption& e) {
std::cout<< "main:"<< e.what() <<std::endl;
}

try {
rehandle(true);
} catch(std::exce ption& e) {
std::cout<< "main:"<< e.what() <<std::endl;
}
}

/*-----------output---------
handle:generate () invalid argument exception
rehandle:genera te() invalid argument exception
main:generate() invalid argument exception
handle:generate () invalid argument exception
rehandle:St9exc eption
main:St9excepti on
*/
--

You are rethrowing a copied slice of the original object.

#include <iostream>
#include <typeinfo>

struct Base
{
virtual ~Base() { }
};

struct Derived : public Base
{
};

void foo()
{
try {
throw Derived();
}
catch(const Base& r_b)
{
std::cerr << typeid(r_b).nam e();
std::cerr << std::endl;
// throw;
throw r_b; // copy a slice
}
}

int main()
{
try {
foo();
}
catch(const Base& r_b)
{
std::cerr << typeid(r_b).nam e();
std::cerr << std::endl;
}
}

/*
Derived
Base
*/

I believe Pete Becker had it right. Even if I pass the same time as the
original event, it comes out empty. What happens is that the compiler
borrows the constructor from the class used as the catch parameter, and
constructs a new instance.

In the case where you throw a base or attempt reconstruction of the
derived object, yes.
In the case you *rethrow* the derived exception, however, the catch
block in main captures the original derived object via a pointer to its
base.
output:
/*
Derived
Derived
*/
To "pass" or rethrow the original "event":
throw; // rethrow the exception

--
NOUN:1. Money or property bequeathed to another by will. 2. Something handed
down from an ancestor or a predecessor or from the past: a legacy of
religious freedom. ETYMOLOGY: MidE legacie, office of a deputy, from OF,
from ML legatia, from L legare, to depute, bequeath. www.bartleby.com/61/