Terminate called after throwing an instance of an exception, core dumped
I am going over C++ exceptions and am running into an error that I am unsure of why it is giving me issues:
#include <iostream>
#include <exception>
class err : public std::exception
{
public:
const char* what() const noexcept { return "error"; }
};
void f() throw()
{
throw err();
}
int main()
{
try
{
f();
}
catch (const err& e)
{
std::cout << e.what() << std::endl;
}
}
When I run it, I get the following runtime error:
terminate called after throwing an instance of 'err'
what(): error
Aborted (core dumped)
If I move the try/catch logic completely to f(), i.e.
void f()
{
try
{
throw err();
}
catch (const err& e)
{
std::cout << e.what() << std::endl;
}
}
And just call it from main (without the try/catch block in main), then there is no error. Am I not understanding something, as it relates to throwing exceptions from functions?
c++ c++11 exception
edited Jan 2 at 18:48
Ron
10.5k21834
asked Jan 2 at 18:40
HectorJ
675
add a comment |
I am going over C++ exceptions and am running into an error that I am unsure of why it is giving me issues:
#include <iostream>
#include <exception>
class err : public std::exception
{
public:
const char* what() const noexcept { return "error"; }
};
void f() throw()
{
throw err();
}
int main()
{
try
{
f();
}
catch (const err& e)
{
std::cout << e.what() << std::endl;
}
}
When I run it, I get the following runtime error:
terminate called after throwing an instance of 'err'
what(): error
Aborted (core dumped)
If I move the try/catch logic completely to f(), i.e.
void f()
{
try
{
throw err();
}
catch (const err& e)
{
std::cout << e.what() << std::endl;
}
}
And just call it from main (without the try/catch block in main), then there is no error. Am I not understanding something, as it relates to throwing exceptions from functions?
c++ c++11 exception
edited Jan 2 at 18:48
Ron
10.5k21834
asked Jan 2 at 18:40
HectorJ
675
9
void f() throw()says that the function doesn't throw exceptions. And then you do.
– Neil Butterworth
Jan 2 at 18:42
2
Remove thethrow(). You are saying "this function never throws" but then you violate that promise and throw anyway. That leads tostd::terminatesince you broke the rules of the language - at that point you cannot expect anything.
– Jesper Juhl
Jan 2 at 18:57
add a comment |
I am going over C++ exceptions and am running into an error that I am unsure of why it is giving me issues:
#include <iostream>
#include <exception>
class err : public std::exception
{
public:
const char* what() const noexcept { return "error"; }
};
void f() throw()
{
throw err();
}
int main()
{
try
{
f();
}
catch (const err& e)
{
std::cout << e.what() << std::endl;
}
}
When I run it, I get the following runtime error:
terminate called after throwing an instance of 'err'
what(): error
Aborted (core dumped)
If I move the try/catch logic completely to f(), i.e.
void f()
{
try
{
throw err();
}
catch (const err& e)
{
std::cout << e.what() << std::endl;
}
}
And just call it from main (without the try/catch block in main), then there is no error. Am I not understanding something, as it relates to throwing exceptions from functions?
c++ c++11 exception
edited Jan 2 at 18:48
Ron
10.5k21834
asked Jan 2 at 18:40
HectorJ
675
I am going over C++ exceptions and am running into an error that I am unsure of why it is giving me issues:
#include <iostream>
#include <exception>
class err : public std::exception
{
public:
const char* what() const noexcept { return "error"; }
};
void f() throw()
{
throw err();
}
int main()
{
try
{
f();
}
catch (const err& e)
{
std::cout << e.what() << std::endl;
}
}
When I run it, I get the following runtime error:
terminate called after throwing an instance of 'err'
what(): error
Aborted (core dumped)
If I move the try/catch logic completely to f(), i.e.
void f()
{
try
{
throw err();
}
catch (const err& e)
{
std::cout << e.what() << std::endl;
}
}
And just call it from main (without the try/catch block in main), then there is no error. Am I not understanding something, as it relates to throwing exceptions from functions?
c++ c++11 exception
c++ c++11 exception
edited Jan 2 at 18:48
Ron
10.5k21834
asked Jan 2 at 18:40
HectorJ
675
edited Jan 2 at 18:48
Ron
10.5k21834
asked Jan 2 at 18:40
HectorJ
675
edited Jan 2 at 18:48
Ron
10.5k21834
edited Jan 2 at 18:48
Ron
10.5k21834
edited Jan 2 at 18:48
Ron
10.5k21834
10.5k21834
asked Jan 2 at 18:40
HectorJ
675
asked Jan 2 at 18:40
HectorJ
675
asked Jan 2 at 18:40
HectorJ
675
675
9
void f() throw()says that the function doesn't throw exceptions. And then you do.
– Neil Butterworth
Jan 2 at 18:42
2
Remove thethrow(). You are saying "this function never throws" but then you violate that promise and throw anyway. That leads tostd::terminatesince you broke the rules of the language - at that point you cannot expect anything.
– Jesper Juhl
Jan 2 at 18:57
add a comment |
9
void f() throw()says that the function doesn't throw exceptions. And then you do.
– Neil Butterworth
Jan 2 at 18:42
2
Remove thethrow(). You are saying "this function never throws" but then you violate that promise and throw anyway. That leads tostd::terminatesince you broke the rules of the language - at that point you cannot expect anything.
– Jesper Juhl
Jan 2 at 18:57
9
9
void f() throw() says that the function doesn't throw exceptions. And then you do.– Neil Butterworth
Jan 2 at 18:42
void f() throw() says that the function doesn't throw exceptions. And then you do.– Neil Butterworth
Jan 2 at 18:42
2
2
Remove the
throw(). You are saying "this function never throws" but then you violate that promise and throw anyway. That leads to std::terminate since you broke the rules of the language - at that point you cannot expect anything.– Jesper Juhl
Jan 2 at 18:57
Remove the
throw(). You are saying "this function never throws" but then you violate that promise and throw anyway. That leads to std::terminate since you broke the rules of the language - at that point you cannot expect anything.– Jesper Juhl
Jan 2 at 18:57
add a comment |
1 Answer
1
active
oldest
votes
The throw() in void f() throw() is a dynamic exception specification and is deprecated since c++11. It's was supposed to be used to list the exceptions a function could throw. An empty specification (throw()) means there are no exceptions that your function could throw. Trying to throw an exception from such a function calls std::unexpected which, by default, terminates.
Since c++11 the preferred way of specifying that a function cannot throw is to use noexcept. For example void f() noexcept.
answered Jan 2 at 18:44
François Andrieux
15.4k32647
Thank you! I had an incorrect understanding. It looks likenoexcept(false)will provide the same functionality and looks to not have been deprecated
– HectorJ
Jan 2 at 18:59
4
@HectorJnoexcept(false)is the default for most functions and usually does not need to be provided. I believe destructors arenoexcept(true)by default and as far as I know that would be the only timenoexcept(false)has meaning (though that's not recommended).
– François Andrieux
Jan 2 at 19:00
add a comment |
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1 Answer
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1 Answer
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The throw() in void f() throw() is a dynamic exception specification and is deprecated since c++11. It's was supposed to be used to list the exceptions a function could throw. An empty specification (throw()) means there are no exceptions that your function could throw. Trying to throw an exception from such a function calls std::unexpected which, by default, terminates.
Since c++11 the preferred way of specifying that a function cannot throw is to use noexcept. For example void f() noexcept.
answered Jan 2 at 18:44
François Andrieux
15.4k32647
Thank you! I had an incorrect understanding. It looks likenoexcept(false)will provide the same functionality and looks to not have been deprecated
– HectorJ
Jan 2 at 18:59
4
@HectorJnoexcept(false)is the default for most functions and usually does not need to be provided. I believe destructors arenoexcept(true)by default and as far as I know that would be the only timenoexcept(false)has meaning (though that's not recommended).
– François Andrieux
Jan 2 at 19:00
add a comment |
The throw() in void f() throw() is a dynamic exception specification and is deprecated since c++11. It's was supposed to be used to list the exceptions a function could throw. An empty specification (throw()) means there are no exceptions that your function could throw. Trying to throw an exception from such a function calls std::unexpected which, by default, terminates.
Since c++11 the preferred way of specifying that a function cannot throw is to use noexcept. For example void f() noexcept.
answered Jan 2 at 18:44
François Andrieux
15.4k32647
Thank you! I had an incorrect understanding. It looks likenoexcept(false)will provide the same functionality and looks to not have been deprecated
– HectorJ
Jan 2 at 18:59
4
@HectorJnoexcept(false)is the default for most functions and usually does not need to be provided. I believe destructors arenoexcept(true)by default and as far as I know that would be the only timenoexcept(false)has meaning (though that's not recommended).
– François Andrieux
Jan 2 at 19:00
add a comment |
The throw() in void f() throw() is a dynamic exception specification and is deprecated since c++11. It's was supposed to be used to list the exceptions a function could throw. An empty specification (throw()) means there are no exceptions that your function could throw. Trying to throw an exception from such a function calls std::unexpected which, by default, terminates.
Since c++11 the preferred way of specifying that a function cannot throw is to use noexcept. For example void f() noexcept.
answered Jan 2 at 18:44
François Andrieux
15.4k32647
The throw() in void f() throw() is a dynamic exception specification and is deprecated since c++11. It's was supposed to be used to list the exceptions a function could throw. An empty specification (throw()) means there are no exceptions that your function could throw. Trying to throw an exception from such a function calls std::unexpected which, by default, terminates.
Since c++11 the preferred way of specifying that a function cannot throw is to use noexcept. For example void f() noexcept.
answered Jan 2 at 18:44
François Andrieux
15.4k32647
edited Jan 2 at 18:59
answered Jan 2 at 18:44
François Andrieux
15.4k32647
answered Jan 2 at 18:44
François Andrieux
15.4k32647
answered Jan 2 at 18:44
François Andrieux
15.4k32647
15.4k32647
Thank you! I had an incorrect understanding. It looks likenoexcept(false)will provide the same functionality and looks to not have been deprecated
– HectorJ
Jan 2 at 18:59
4
@HectorJnoexcept(false)is the default for most functions and usually does not need to be provided. I believe destructors arenoexcept(true)by default and as far as I know that would be the only timenoexcept(false)has meaning (though that's not recommended).
– François Andrieux
Jan 2 at 19:00
add a comment |
Thank you! I had an incorrect understanding. It looks likenoexcept(false)will provide the same functionality and looks to not have been deprecated
– HectorJ
Jan 2 at 18:59
4
@HectorJnoexcept(false)is the default for most functions and usually does not need to be provided. I believe destructors arenoexcept(true)by default and as far as I know that would be the only timenoexcept(false)has meaning (though that's not recommended).
– François Andrieux
Jan 2 at 19:00
Thank you! I had an incorrect understanding. It looks like
noexcept(false) will provide the same functionality and looks to not have been deprecated– HectorJ
Jan 2 at 18:59
Thank you! I had an incorrect understanding. It looks like
noexcept(false) will provide the same functionality and looks to not have been deprecated– HectorJ
Jan 2 at 18:59
4
4
@HectorJ
noexcept(false) is the default for most functions and usually does not need to be provided. I believe destructors are noexcept(true) by default and as far as I know that would be the only time noexcept(false) has meaning (though that's not recommended).– François Andrieux
Jan 2 at 19:00
@HectorJ
noexcept(false) is the default for most functions and usually does not need to be provided. I believe destructors are noexcept(true) by default and as far as I know that would be the only time noexcept(false) has meaning (though that's not recommended).– François Andrieux
Jan 2 at 19:00
add a comment |
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9
void f() throw()says that the function doesn't throw exceptions. And then you do.– Neil Butterworth
Jan 2 at 18:42
2
Remove the
throw(). You are saying "this function never throws" but then you violate that promise and throw anyway. That leads tostd::terminatesince you broke the rules of the language - at that point you cannot expect anything.– Jesper Juhl
Jan 2 at 18:57