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Error reading variable. Cannot create a lazy string with address

2

0

I have following code:

#include <string>
#include <vector>

class A 
public:
 std::string s = "test";
;

class B 
public:
 std::vector<A> vec;
;

int main()

 std::vector<B> vec;

 A a1 = A();
 A a2 = A();

 B b1 = B();
 b1.vec.push_back(a1);
 b1.vec.push_back(a2);

 vec.push_back(b1); // push_1 
 vec.push_back(b1);

Whenever I execute this app under debugger, and execution procecss reachs instruction with comment push_1, the execution process is stopped, and I have two followings informations in my debugger output:

__lhs s = "test"

__rhs s = "error reading variable: Cannot create a lazy string with address 0x0, and a non zero length.

The application exit code is 0.

But when I remove property s of class A, or repleace with for instance int property, this strange behaviour does not occure. Why it is happening? Why the string occurrance in the class A, causes this error?

c++

share|improve this question

edited Mar 6 at 21:19

bielu000

asked Mar 6 at 20:58

bielu000bielu000

344316

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I can't reproduce. What is your platform/compiler? BTW, for a Minimal, Complete, and Verifiable example you are missing #include <vector> and #include <string>.
  
  – Werner Henze
  Mar 6 at 21:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You're right I missed the headers. The platform is Linux Debian x64, and to compile I'm using g++ (Debian 6.3.0-18+deb9u1) 6.3.0 20170516. The C++ standard version is 14.
  
  – bielu000
  Mar 6 at 21:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Are you debugging an optimized build?
  
  – alter igel
  Mar 6 at 21:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @bielu000 I too reckon that this is some kind of an optimized build. Tried this and couldn't reproduce. onlinegdb.com/S18rQTaUE. It's not the same architecture though...
  
  – aep
  Mar 6 at 22:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I think that you're rightit could be because optimized builds, but I don't have any optimiziation flags set. It is really confusing me - I copied all files and directories of above app into another, new directory, and it works correctly... But the original source still works as I said above. I compared all files, from CmakeLists even to IDE configuration settings, but there are any differents.
  
  – bielu000
  2 days ago
  

  
  
  
  

 | 
show 1 more comment

2

0

I have following code:

#include <string>
#include <vector>

class A 
public:
 std::string s = "test";
;

class B 
public:
 std::vector<A> vec;
;

int main()

 std::vector<B> vec;

 A a1 = A();
 A a2 = A();

 B b1 = B();
 b1.vec.push_back(a1);
 b1.vec.push_back(a2);

 vec.push_back(b1); // push_1 
 vec.push_back(b1);

Whenever I execute this app under debugger, and execution procecss reachs instruction with comment push_1, the execution process is stopped, and I have two followings informations in my debugger output:

__lhs s = "test"

__rhs s = "error reading variable: Cannot create a lazy string with address 0x0, and a non zero length.

The application exit code is 0.

But when I remove property s of class A, or repleace with for instance int property, this strange behaviour does not occure. Why it is happening? Why the string occurrance in the class A, causes this error?

c++

share|improve this question

edited Mar 6 at 21:19

bielu000

asked Mar 6 at 20:58

bielu000bielu000

344316

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I can't reproduce. What is your platform/compiler? BTW, for a Minimal, Complete, and Verifiable example you are missing #include <vector> and #include <string>.
  
  – Werner Henze
  Mar 6 at 21:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You're right I missed the headers. The platform is Linux Debian x64, and to compile I'm using g++ (Debian 6.3.0-18+deb9u1) 6.3.0 20170516. The C++ standard version is 14.
  
  – bielu000
  Mar 6 at 21:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Are you debugging an optimized build?
  
  – alter igel
  Mar 6 at 21:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @bielu000 I too reckon that this is some kind of an optimized build. Tried this and couldn't reproduce. onlinegdb.com/S18rQTaUE. It's not the same architecture though...
  
  – aep
  Mar 6 at 22:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I think that you're rightit could be because optimized builds, but I don't have any optimiziation flags set. It is really confusing me - I copied all files and directories of above app into another, new directory, and it works correctly... But the original source still works as I said above. I compared all files, from CmakeLists even to IDE configuration settings, but there are any differents.
  
  – bielu000
  2 days ago
  

  
  
  
  

 | 
show 1 more comment

I have following code:

#include <string>
#include <vector>

class A 
public:
 std::string s = "test";
;

class B 
public:
 std::vector<A> vec;
;

int main()

 std::vector<B> vec;

 A a1 = A();
 A a2 = A();

 B b1 = B();
 b1.vec.push_back(a1);
 b1.vec.push_back(a2);

 vec.push_back(b1); // push_1 
 vec.push_back(b1);

Whenever I execute this app under debugger, and execution procecss reachs instruction with comment push_1, the execution process is stopped, and I have two followings informations in my debugger output:

__lhs s = "test"

__rhs s = "error reading variable: Cannot create a lazy string with address 0x0, and a non zero length.

The application exit code is 0.

But when I remove property s of class A, or repleace with for instance int property, this strange behaviour does not occure. Why it is happening? Why the string occurrance in the class A, causes this error?

c++

share|improve this question

edited Mar 6 at 21:19

bielu000

asked Mar 6 at 20:58

bielu000bielu000

344316

#include <string>
#include <vector>

class A 
public:
 std::string s = "test";
;

class B 
public:
 std::vector<A> vec;
;

int main()

 std::vector<B> vec;

 A a1 = A();
 A a2 = A();

 B b1 = B();
 b1.vec.push_back(a1);
 b1.vec.push_back(a2);

 vec.push_back(b1); // push_1 
 vec.push_back(b1);

share|improve this question

edited Mar 6 at 21:19

bielu000

asked Mar 6 at 20:58

bielu000bielu000

344316

share|improve this question

edited Mar 6 at 21:19

bielu000

asked Mar 6 at 20:58

bielu000bielu000

344316

asked Mar 6 at 20:58

bielu000bielu000

344316

asked Mar 6 at 20:58

bielu000bielu000

344316