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Undefined Behaviour in C++ [duplicate]

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0

This question already has an answer here:

• 
  Program behaving strangely on online IDEs
  
  4 answers
  
  

I got this from a facebook post. What's happening here? See the output in ideone. Output is more than 10 lines.

Code:

#include<iostream> 
using namespace std;

int main()

 for (int i = 0; i < 10; ++i)
 cout << i*1000000000 << endl;

Ideone Link

c++ undefined overflow

share|improve this question

edited Mar 22 at 10:59

Inam

asked Mar 22 at 10:54

InamInam

83

marked as duplicate by Peter c++
Users with the  c++ badge can single-handedly close c++ questions as duplicates and reopen them as needed.

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Mar 22 at 11:11

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You even tagged it with overflow. 9*1000000000 doesn't fit into an int anymore, which causes overflow, and that's undefined behavior.
  
  – Blaze
  Mar 22 at 10:55
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Why can't you copy-paste the output into your question? And why do you think it's wrong or UB? Please read about how to ask good questions, as well as this question checklist.
  
  – Some programmer dude
  Mar 22 at 10:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Probably compiler logic: "Since 3 * 1000000000 overflows, we can assume that i < 3. Thus i < 10 is always true."
  
  – molbdnilo
  Mar 22 at 11:09
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  For a 16-bit int, 3 * 1000000000 will overflow when i == 1. So the compiler can assume i is always zero, and the condition i < 10 is always true. For a 32 bit int, the compiler can assume i < 3. For a 64-bit int, the loop will only run ten times.
  
  – Peter
  Mar 22 at 11:42
  

  
  
  
  

add a comment | 

0

This question already has an answer here:

• 
  Program behaving strangely on online IDEs
  
  4 answers
  
  

I got this from a facebook post. What's happening here? See the output in ideone. Output is more than 10 lines.

Code:

#include<iostream> 
using namespace std;

int main()

 for (int i = 0; i < 10; ++i)
 cout << i*1000000000 << endl;

Ideone Link

c++ undefined overflow

share|improve this question

edited Mar 22 at 10:59

Inam

asked Mar 22 at 10:54

InamInam

83

marked as duplicate by Peter c++
Users with the  c++ badge can single-handedly close c++ questions as duplicates and reopen them as needed.

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if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
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$hover.hover(
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$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
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dismissable: false,
relativeToBody: true
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StackExchange.helpers.removeMessages();

);
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);
Mar 22 at 11:11

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You even tagged it with overflow. 9*1000000000 doesn't fit into an int anymore, which causes overflow, and that's undefined behavior.
  
  – Blaze
  Mar 22 at 10:55
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Why can't you copy-paste the output into your question? And why do you think it's wrong or UB? Please read about how to ask good questions, as well as this question checklist.
  
  – Some programmer dude
  Mar 22 at 10:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Probably compiler logic: "Since 3 * 1000000000 overflows, we can assume that i < 3. Thus i < 10 is always true."
  
  – molbdnilo
  Mar 22 at 11:09
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  For a 16-bit int, 3 * 1000000000 will overflow when i == 1. So the compiler can assume i is always zero, and the condition i < 10 is always true. For a 32 bit int, the compiler can assume i < 3. For a 64-bit int, the loop will only run ten times.
  
  – Peter
  Mar 22 at 11:42
  

  
  
  
  

add a comment | 

This question already has an answer here:

• 
  Program behaving strangely on online IDEs
  
  4 answers
  
  

I got this from a facebook post. What's happening here? See the output in ideone. Output is more than 10 lines.

Code:

#include<iostream> 
using namespace std;

int main()

 for (int i = 0; i < 10; ++i)
 cout << i*1000000000 << endl;

Ideone Link

c++ undefined overflow

share|improve this question

edited Mar 22 at 10:59

Inam

asked Mar 22 at 10:54

InamInam

83

#include<iostream> 
using namespace std;

int main()

 for (int i = 0; i < 10; ++i)
 cout << i*1000000000 << endl;

share|improve this question

edited Mar 22 at 10:59

Inam

asked Mar 22 at 10:54

InamInam

83

share|improve this question

edited Mar 22 at 10:59

Inam

asked Mar 22 at 10:54

InamInam

83

asked Mar 22 at 10:54

InamInam

83

asked Mar 22 at 10:54

InamInam

83

1 Answer
1

active

oldest

votes

3

Your platform most likely has a 32 bit int. So 1'000'000'000 is an int, and the compiler will attempt to evaluate i * 1'000'000'000 as an int too. This results in an overflow from i being 3 onwards.

The behaviour on overflowing a signed integral type is undefined.

Note that this makes the entire program behaviour undefined, which accounts for the multiple lines of output (beyond 10) that you observe.

(If you had chosen 10'000'000'000 say instead then the multiplication would have been evaluated with long long types and the behaviour would be well-defined!)

share|improve this answer

edited Mar 22 at 11:16

answered Mar 22 at 10:56

BathshebaBathsheba

182k27257385

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I know what UB is, but somehow it is always explainable what actually happens. Could you come up with a reason why this specific platform (ideone) causes it to run further than 10 ?
  
  – Bart Friederichs
  Mar 22 at 11:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So due to the undefined behaviour somehow i < 10 condition is true when i >= 10?
  
  – Inam
  Mar 22 at 11:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Inam: My wildest guess is that the compiler is assuming that i * 1000000000 can fit into an int, so therefore i must be small, and therefore less than 10.
  
  – Bathsheba
  Mar 22 at 11:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Inam the point is that due to undefined behaviour; the whole program breaks down. The language that describes what the program should do is invalid. You've written a loop there, but the compiler may or may not decide to make it a loop anyway; but the compiler might detect that it only needs to do 3 loops before the value would overflow... and then bin the other 7
  
  – UKMonkey
  Mar 22 at 11:16
  

  
  
  
  

add a comment | 

3

Your platform most likely has a 32 bit int. So 1'000'000'000 is an int, and the compiler will attempt to evaluate i * 1'000'000'000 as an int too. This results in an overflow from i being 3 onwards.

The behaviour on overflowing a signed integral type is undefined.

Note that this makes the entire program behaviour undefined, which accounts for the multiple lines of output (beyond 10) that you observe.

(If you had chosen 10'000'000'000 say instead then the multiplication would have been evaluated with long long types and the behaviour would be well-defined!)

share|improve this answer

edited Mar 22 at 11:16

answered Mar 22 at 10:56

BathshebaBathsheba

182k27257385

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I know what UB is, but somehow it is always explainable what actually happens. Could you come up with a reason why this specific platform (ideone) causes it to run further than 10 ?
  
  – Bart Friederichs
  Mar 22 at 11:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So due to the undefined behaviour somehow i < 10 condition is true when i >= 10?
  
  – Inam
  Mar 22 at 11:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Inam: My wildest guess is that the compiler is assuming that i * 1000000000 can fit into an int, so therefore i must be small, and therefore less than 10.
  
  – Bathsheba
  Mar 22 at 11:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Inam the point is that due to undefined behaviour; the whole program breaks down. The language that describes what the program should do is invalid. You've written a loop there, but the compiler may or may not decide to make it a loop anyway; but the compiler might detect that it only needs to do 3 loops before the value would overflow... and then bin the other 7
  
  – UKMonkey
  Mar 22 at 11:16
  

  
  
  
  

add a comment | 

3

Your platform most likely has a 32 bit int. So 1'000'000'000 is an int, and the compiler will attempt to evaluate i * 1'000'000'000 as an int too. This results in an overflow from i being 3 onwards.

The behaviour on overflowing a signed integral type is undefined.

Note that this makes the entire program behaviour undefined, which accounts for the multiple lines of output (beyond 10) that you observe.

(If you had chosen 10'000'000'000 say instead then the multiplication would have been evaluated with long long types and the behaviour would be well-defined!)

share|improve this answer

edited Mar 22 at 11:16

answered Mar 22 at 10:56

BathshebaBathsheba

182k27257385

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I know what UB is, but somehow it is always explainable what actually happens. Could you come up with a reason why this specific platform (ideone) causes it to run further than 10 ?
  
  – Bart Friederichs
  Mar 22 at 11:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So due to the undefined behaviour somehow i < 10 condition is true when i >= 10?
  
  – Inam
  Mar 22 at 11:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Inam: My wildest guess is that the compiler is assuming that i * 1000000000 can fit into an int, so therefore i must be small, and therefore less than 10.
  
  – Bathsheba
  Mar 22 at 11:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Inam the point is that due to undefined behaviour; the whole program breaks down. The language that describes what the program should do is invalid. You've written a loop there, but the compiler may or may not decide to make it a loop anyway; but the compiler might detect that it only needs to do 3 loops before the value would overflow... and then bin the other 7
  
  – UKMonkey
  Mar 22 at 11:16
  

  
  
  
  

add a comment | 

Your platform most likely has a 32 bit int. So 1'000'000'000 is an int, and the compiler will attempt to evaluate i * 1'000'000'000 as an int too. This results in an overflow from i being 3 onwards.

The behaviour on overflowing a signed integral type is undefined.

Note that this makes the entire program behaviour undefined, which accounts for the multiple lines of output (beyond 10) that you observe.

(If you had chosen 10'000'000'000 say instead then the multiplication would have been evaluated with long long types and the behaviour would be well-defined!)

share|improve this answer

edited Mar 22 at 11:16

answered Mar 22 at 10:56

BathshebaBathsheba

182k27257385

share|improve this answer

edited Mar 22 at 11:16

answered Mar 22 at 10:56

BathshebaBathsheba

182k27257385

answered Mar 22 at 10:56

BathshebaBathsheba

182k27257385

answered Mar 22 at 10:56

BathshebaBathsheba

182k27257385