What happens when a constructor function calls itself in VS2013?
class IA
{
public:
virtual void Print() = 0;
}
IA* GetA();
class A : public IA
{
public:
int num = 10;
A()
{
GetA()->Print();
}
void Print()
{
std::cout << num << std::endl;
}
}
IA* GetA()
{
static A a;
return &a;
}
int main()
{
GetA();
std::cout << "End" << std::endl;
getchar();
return 0;
}
- Obviously, class A's constructor function calls itself.
- "static A a" will get stuck in a loop.
- On VS2013, this code can get out from the loop and print "End" on the console.
On VS2017, this code gets stuck in a loop.
**What does VS2013 do for this code?
c++ visual-studio-2013 visual-studio-2017
add a comment |
class IA
{
public:
virtual void Print() = 0;
}
IA* GetA();
class A : public IA
{
public:
int num = 10;
A()
{
GetA()->Print();
}
void Print()
{
std::cout << num << std::endl;
}
}
IA* GetA()
{
static A a;
return &a;
}
int main()
{
GetA();
std::cout << "End" << std::endl;
getchar();
return 0;
}
- Obviously, class A's constructor function calls itself.
- "static A a" will get stuck in a loop.
- On VS2013, this code can get out from the loop and print "End" on the console.
On VS2017, this code gets stuck in a loop.
**What does VS2013 do for this code?
c++ visual-studio-2013 visual-studio-2017
You discovered differences between two different VS versions - but the nature of the problem has wider scope (what would GCC, clang, ... do?), so I recommend dropping the two VS tags (3. and 4. remaining just examples, you even might ask in 5. for other compilers...).
– Aconcagua
2 days ago
2
Visual Studio is not a compiler. It's an IDE that invokes compilers.
– jpmc26
2 days ago
add a comment |
class IA
{
public:
virtual void Print() = 0;
}
IA* GetA();
class A : public IA
{
public:
int num = 10;
A()
{
GetA()->Print();
}
void Print()
{
std::cout << num << std::endl;
}
}
IA* GetA()
{
static A a;
return &a;
}
int main()
{
GetA();
std::cout << "End" << std::endl;
getchar();
return 0;
}
- Obviously, class A's constructor function calls itself.
- "static A a" will get stuck in a loop.
- On VS2013, this code can get out from the loop and print "End" on the console.
On VS2017, this code gets stuck in a loop.
**What does VS2013 do for this code?
c++ visual-studio-2013 visual-studio-2017
class IA
{
public:
virtual void Print() = 0;
}
IA* GetA();
class A : public IA
{
public:
int num = 10;
A()
{
GetA()->Print();
}
void Print()
{
std::cout << num << std::endl;
}
}
IA* GetA()
{
static A a;
return &a;
}
int main()
{
GetA();
std::cout << "End" << std::endl;
getchar();
return 0;
}
- Obviously, class A's constructor function calls itself.
- "static A a" will get stuck in a loop.
- On VS2013, this code can get out from the loop and print "End" on the console.
On VS2017, this code gets stuck in a loop.
**What does VS2013 do for this code?
c++ visual-studio-2013 visual-studio-2017
c++ visual-studio-2013 visual-studio-2017
edited 2 days ago
Boann
36.7k1288121
36.7k1288121
asked 2 days ago
MiCMiC
856
856
You discovered differences between two different VS versions - but the nature of the problem has wider scope (what would GCC, clang, ... do?), so I recommend dropping the two VS tags (3. and 4. remaining just examples, you even might ask in 5. for other compilers...).
– Aconcagua
2 days ago
2
Visual Studio is not a compiler. It's an IDE that invokes compilers.
– jpmc26
2 days ago
add a comment |
You discovered differences between two different VS versions - but the nature of the problem has wider scope (what would GCC, clang, ... do?), so I recommend dropping the two VS tags (3. and 4. remaining just examples, you even might ask in 5. for other compilers...).
– Aconcagua
2 days ago
2
Visual Studio is not a compiler. It's an IDE that invokes compilers.
– jpmc26
2 days ago
You discovered differences between two different VS versions - but the nature of the problem has wider scope (what would GCC, clang, ... do?), so I recommend dropping the two VS tags (3. and 4. remaining just examples, you even might ask in 5. for other compilers...).
– Aconcagua
2 days ago
You discovered differences between two different VS versions - but the nature of the problem has wider scope (what would GCC, clang, ... do?), so I recommend dropping the two VS tags (3. and 4. remaining just examples, you even might ask in 5. for other compilers...).
– Aconcagua
2 days ago
2
2
Visual Studio is not a compiler. It's an IDE that invokes compilers.
– jpmc26
2 days ago
Visual Studio is not a compiler. It's an IDE that invokes compilers.
– jpmc26
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
Nothing in particular has to happen. According to the C++ standard:
[stmt.dcl] (emphasis mine)
4 Dynamic initialization of a block-scope variable with static storage
duration or thread storage duration is performed the first time
control passes through its declaration; such a variable is considered
initialized upon the completion of its initialization. If the
initialization exits by throwing an exception, the initialization is
not complete, so it will be tried again the next time control enters
the declaration. If control enters the declaration concurrently while
the variable is being initialized, the concurrent execution shall wait
for completion of the initialization. If control re-enters the
declaration recursively while the variable is being initialized, the
behavior is undefined. [ Example:
int foo(int i) {
static int s = foo(2*i); // recursive call - undefined
return i+1;
}
— end example ]
The statement I emboldened is exactly what happens in your program. It's also what the standard's example shows as undefined. The language specification says an implementation can do whatever it deems appropriate. So it could cause an infinite loop, or it may not, depending on the synchronization primitives your implementation uses to prevent concurrent reentry into the block (the initialization has to be thread safe).
Even before C++11, the behavior of recursive reentry was undefined. An implementation could do anything to make sure an object is initialized only once, which in turn could produce different results.
But you can't expect anything specific to happen portably. Not to mention undefined behavior always leaves room for a small chance of nasal demons.
'infinite recursion' would be the better term - and at least in theory, it offers another interesting outcome: crash due to stack overflow (if TCO is not applied)...
– Aconcagua
2 days ago
@Aconcagua - You say to-may-to, I say to-mah-to. Both are Turing equivalent after all ;). Just sticking to the OP's way of phrasing and thinking about it.
– StoryTeller
2 days ago
2
And don't forget the option of a deadlock should VS be using a non recursive synchronization primitive.
– StoryTeller
2 days ago
add a comment |
The behaviour is undefined. The reason it "worked" in Visual Studio 2013 is that it didn't implement thread safe initialisation of function statics. What is probably happening is that the first call to GetA()
creates a
and calls the constructor. The second call to GetA()
then just returns the partially constructed a
. As the body of your constructor doesn't initialise anything calling Print()
doesn't crash.
Visual Studio 2017 does implement thread safe initialisation. and presumably locks some mutex on entry to GetA()
if a
is not initialised, the second call to GetA()
then encounters the locked mutex and deadlocks.
Note in both cases this is just my guess from the observed behaviour, the actual behaviour is undefined, for example GetA()
may end up creating 2 instances of A
.
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f54177087%2fwhat-happens-when-a-constructor-function-calls-itself-in-vs2013%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Nothing in particular has to happen. According to the C++ standard:
[stmt.dcl] (emphasis mine)
4 Dynamic initialization of a block-scope variable with static storage
duration or thread storage duration is performed the first time
control passes through its declaration; such a variable is considered
initialized upon the completion of its initialization. If the
initialization exits by throwing an exception, the initialization is
not complete, so it will be tried again the next time control enters
the declaration. If control enters the declaration concurrently while
the variable is being initialized, the concurrent execution shall wait
for completion of the initialization. If control re-enters the
declaration recursively while the variable is being initialized, the
behavior is undefined. [ Example:
int foo(int i) {
static int s = foo(2*i); // recursive call - undefined
return i+1;
}
— end example ]
The statement I emboldened is exactly what happens in your program. It's also what the standard's example shows as undefined. The language specification says an implementation can do whatever it deems appropriate. So it could cause an infinite loop, or it may not, depending on the synchronization primitives your implementation uses to prevent concurrent reentry into the block (the initialization has to be thread safe).
Even before C++11, the behavior of recursive reentry was undefined. An implementation could do anything to make sure an object is initialized only once, which in turn could produce different results.
But you can't expect anything specific to happen portably. Not to mention undefined behavior always leaves room for a small chance of nasal demons.
'infinite recursion' would be the better term - and at least in theory, it offers another interesting outcome: crash due to stack overflow (if TCO is not applied)...
– Aconcagua
2 days ago
@Aconcagua - You say to-may-to, I say to-mah-to. Both are Turing equivalent after all ;). Just sticking to the OP's way of phrasing and thinking about it.
– StoryTeller
2 days ago
2
And don't forget the option of a deadlock should VS be using a non recursive synchronization primitive.
– StoryTeller
2 days ago
add a comment |
Nothing in particular has to happen. According to the C++ standard:
[stmt.dcl] (emphasis mine)
4 Dynamic initialization of a block-scope variable with static storage
duration or thread storage duration is performed the first time
control passes through its declaration; such a variable is considered
initialized upon the completion of its initialization. If the
initialization exits by throwing an exception, the initialization is
not complete, so it will be tried again the next time control enters
the declaration. If control enters the declaration concurrently while
the variable is being initialized, the concurrent execution shall wait
for completion of the initialization. If control re-enters the
declaration recursively while the variable is being initialized, the
behavior is undefined. [ Example:
int foo(int i) {
static int s = foo(2*i); // recursive call - undefined
return i+1;
}
— end example ]
The statement I emboldened is exactly what happens in your program. It's also what the standard's example shows as undefined. The language specification says an implementation can do whatever it deems appropriate. So it could cause an infinite loop, or it may not, depending on the synchronization primitives your implementation uses to prevent concurrent reentry into the block (the initialization has to be thread safe).
Even before C++11, the behavior of recursive reentry was undefined. An implementation could do anything to make sure an object is initialized only once, which in turn could produce different results.
But you can't expect anything specific to happen portably. Not to mention undefined behavior always leaves room for a small chance of nasal demons.
'infinite recursion' would be the better term - and at least in theory, it offers another interesting outcome: crash due to stack overflow (if TCO is not applied)...
– Aconcagua
2 days ago
@Aconcagua - You say to-may-to, I say to-mah-to. Both are Turing equivalent after all ;). Just sticking to the OP's way of phrasing and thinking about it.
– StoryTeller
2 days ago
2
And don't forget the option of a deadlock should VS be using a non recursive synchronization primitive.
– StoryTeller
2 days ago
add a comment |
Nothing in particular has to happen. According to the C++ standard:
[stmt.dcl] (emphasis mine)
4 Dynamic initialization of a block-scope variable with static storage
duration or thread storage duration is performed the first time
control passes through its declaration; such a variable is considered
initialized upon the completion of its initialization. If the
initialization exits by throwing an exception, the initialization is
not complete, so it will be tried again the next time control enters
the declaration. If control enters the declaration concurrently while
the variable is being initialized, the concurrent execution shall wait
for completion of the initialization. If control re-enters the
declaration recursively while the variable is being initialized, the
behavior is undefined. [ Example:
int foo(int i) {
static int s = foo(2*i); // recursive call - undefined
return i+1;
}
— end example ]
The statement I emboldened is exactly what happens in your program. It's also what the standard's example shows as undefined. The language specification says an implementation can do whatever it deems appropriate. So it could cause an infinite loop, or it may not, depending on the synchronization primitives your implementation uses to prevent concurrent reentry into the block (the initialization has to be thread safe).
Even before C++11, the behavior of recursive reentry was undefined. An implementation could do anything to make sure an object is initialized only once, which in turn could produce different results.
But you can't expect anything specific to happen portably. Not to mention undefined behavior always leaves room for a small chance of nasal demons.
Nothing in particular has to happen. According to the C++ standard:
[stmt.dcl] (emphasis mine)
4 Dynamic initialization of a block-scope variable with static storage
duration or thread storage duration is performed the first time
control passes through its declaration; such a variable is considered
initialized upon the completion of its initialization. If the
initialization exits by throwing an exception, the initialization is
not complete, so it will be tried again the next time control enters
the declaration. If control enters the declaration concurrently while
the variable is being initialized, the concurrent execution shall wait
for completion of the initialization. If control re-enters the
declaration recursively while the variable is being initialized, the
behavior is undefined. [ Example:
int foo(int i) {
static int s = foo(2*i); // recursive call - undefined
return i+1;
}
— end example ]
The statement I emboldened is exactly what happens in your program. It's also what the standard's example shows as undefined. The language specification says an implementation can do whatever it deems appropriate. So it could cause an infinite loop, or it may not, depending on the synchronization primitives your implementation uses to prevent concurrent reentry into the block (the initialization has to be thread safe).
Even before C++11, the behavior of recursive reentry was undefined. An implementation could do anything to make sure an object is initialized only once, which in turn could produce different results.
But you can't expect anything specific to happen portably. Not to mention undefined behavior always leaves room for a small chance of nasal demons.
edited 2 days ago
answered 2 days ago
StoryTellerStoryTeller
95.3k12194259
95.3k12194259
'infinite recursion' would be the better term - and at least in theory, it offers another interesting outcome: crash due to stack overflow (if TCO is not applied)...
– Aconcagua
2 days ago
@Aconcagua - You say to-may-to, I say to-mah-to. Both are Turing equivalent after all ;). Just sticking to the OP's way of phrasing and thinking about it.
– StoryTeller
2 days ago
2
And don't forget the option of a deadlock should VS be using a non recursive synchronization primitive.
– StoryTeller
2 days ago
add a comment |
'infinite recursion' would be the better term - and at least in theory, it offers another interesting outcome: crash due to stack overflow (if TCO is not applied)...
– Aconcagua
2 days ago
@Aconcagua - You say to-may-to, I say to-mah-to. Both are Turing equivalent after all ;). Just sticking to the OP's way of phrasing and thinking about it.
– StoryTeller
2 days ago
2
And don't forget the option of a deadlock should VS be using a non recursive synchronization primitive.
– StoryTeller
2 days ago
'infinite recursion' would be the better term - and at least in theory, it offers another interesting outcome: crash due to stack overflow (if TCO is not applied)...
– Aconcagua
2 days ago
'infinite recursion' would be the better term - and at least in theory, it offers another interesting outcome: crash due to stack overflow (if TCO is not applied)...
– Aconcagua
2 days ago
@Aconcagua - You say to-may-to, I say to-mah-to. Both are Turing equivalent after all ;). Just sticking to the OP's way of phrasing and thinking about it.
– StoryTeller
2 days ago
@Aconcagua - You say to-may-to, I say to-mah-to. Both are Turing equivalent after all ;). Just sticking to the OP's way of phrasing and thinking about it.
– StoryTeller
2 days ago
2
2
And don't forget the option of a deadlock should VS be using a non recursive synchronization primitive.
– StoryTeller
2 days ago
And don't forget the option of a deadlock should VS be using a non recursive synchronization primitive.
– StoryTeller
2 days ago
add a comment |
The behaviour is undefined. The reason it "worked" in Visual Studio 2013 is that it didn't implement thread safe initialisation of function statics. What is probably happening is that the first call to GetA()
creates a
and calls the constructor. The second call to GetA()
then just returns the partially constructed a
. As the body of your constructor doesn't initialise anything calling Print()
doesn't crash.
Visual Studio 2017 does implement thread safe initialisation. and presumably locks some mutex on entry to GetA()
if a
is not initialised, the second call to GetA()
then encounters the locked mutex and deadlocks.
Note in both cases this is just my guess from the observed behaviour, the actual behaviour is undefined, for example GetA()
may end up creating 2 instances of A
.
add a comment |
The behaviour is undefined. The reason it "worked" in Visual Studio 2013 is that it didn't implement thread safe initialisation of function statics. What is probably happening is that the first call to GetA()
creates a
and calls the constructor. The second call to GetA()
then just returns the partially constructed a
. As the body of your constructor doesn't initialise anything calling Print()
doesn't crash.
Visual Studio 2017 does implement thread safe initialisation. and presumably locks some mutex on entry to GetA()
if a
is not initialised, the second call to GetA()
then encounters the locked mutex and deadlocks.
Note in both cases this is just my guess from the observed behaviour, the actual behaviour is undefined, for example GetA()
may end up creating 2 instances of A
.
add a comment |
The behaviour is undefined. The reason it "worked" in Visual Studio 2013 is that it didn't implement thread safe initialisation of function statics. What is probably happening is that the first call to GetA()
creates a
and calls the constructor. The second call to GetA()
then just returns the partially constructed a
. As the body of your constructor doesn't initialise anything calling Print()
doesn't crash.
Visual Studio 2017 does implement thread safe initialisation. and presumably locks some mutex on entry to GetA()
if a
is not initialised, the second call to GetA()
then encounters the locked mutex and deadlocks.
Note in both cases this is just my guess from the observed behaviour, the actual behaviour is undefined, for example GetA()
may end up creating 2 instances of A
.
The behaviour is undefined. The reason it "worked" in Visual Studio 2013 is that it didn't implement thread safe initialisation of function statics. What is probably happening is that the first call to GetA()
creates a
and calls the constructor. The second call to GetA()
then just returns the partially constructed a
. As the body of your constructor doesn't initialise anything calling Print()
doesn't crash.
Visual Studio 2017 does implement thread safe initialisation. and presumably locks some mutex on entry to GetA()
if a
is not initialised, the second call to GetA()
then encounters the locked mutex and deadlocks.
Note in both cases this is just my guess from the observed behaviour, the actual behaviour is undefined, for example GetA()
may end up creating 2 instances of A
.
answered 2 days ago
Alan BirtlesAlan Birtles
8,595933
8,595933
add a comment |
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f54177087%2fwhat-happens-when-a-constructor-function-calls-itself-in-vs2013%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
You discovered differences between two different VS versions - but the nature of the problem has wider scope (what would GCC, clang, ... do?), so I recommend dropping the two VS tags (3. and 4. remaining just examples, you even might ask in 5. for other compilers...).
– Aconcagua
2 days ago
2
Visual Studio is not a compiler. It's an IDE that invokes compilers.
– jpmc26
2 days ago