Home/ Questions/Q 7445331
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-05-29T11:47:46+00:00

Are there any drawbacks / disadvantages using the default constructor for default initialization for

  • 0

Are there any drawbacks / disadvantages using the default constructor for default initialization for primitive data types?

For example

class MyClass
{
public:
    MyClass(); 

private:
    int     miInt;
    double  mdDouble;
    bool    mbBool;
};

Using this constructor:

MyClass::MyClass() 
  : miInt(int())
  , mdDouble(double())
  , mbBool(bool())
{}

instead of this:

MyClass::MyClass() 
  : miInt(0)
  , mdDouble(0.0)
  , mbBool(false)
{}
Share

Leave an answer

You must login to add an answer.

Need An Account,

1 Answer

  1. Editorial Team

    No, and the compiler will most probably generate the same code for both.

    With optimization off, the following code is generated:

    MyClass::MyClass() 
  : miInt(0)
  , mdDouble(0.0)
  , mbBool(false)
{}
012313A0  push        ebp  
012313A1  mov         ebp,esp 
012313A3  sub         esp,0CCh 
012313A9  push        ebx  
012313AA  push        esi  
012313AB  push        edi  
012313AC  push        ecx  
012313AD  lea         edi,[ebp-0CCh] 
012313B3  mov         ecx,33h 
012313B8  mov         eax,0CCCCCCCCh 
012313BD  rep stos    dword ptr es:[edi] 
012313BF  pop         ecx  
012313C0  mov         dword ptr [ebp-8],ecx 
012313C3  mov         eax,dword ptr [this] 
012313C6  mov         dword ptr [eax],0 
012313CC  mov         eax,dword ptr [this] 
012313CF  fldz             
012313D1  fstp        qword ptr [eax+8] 
012313D4  mov         eax,dword ptr [this] 
012313D7  mov         byte ptr [eax+10h],0 
012313DB  mov         eax,dword ptr [this] 
012313DE  pop         edi  
012313DF  pop         esi  
012313E0  pop         ebx  
012313E1  mov         esp,ebp 
012313E3  pop         ebp  
012313E4  ret

    and

    MyClass::MyClass() 
  : miInt(int())
  , mdDouble(double())
  , mbBool(bool())
{}
001513A0  push        ebp  
001513A1  mov         ebp,esp 
001513A3  sub         esp,0CCh 
001513A9  push        ebx  
001513AA  push        esi  
001513AB  push        edi  
001513AC  push        ecx  
001513AD  lea         edi,[ebp-0CCh] 
001513B3  mov         ecx,33h 
001513B8  mov         eax,0CCCCCCCCh 
001513BD  rep stos    dword ptr es:[edi] 
001513BF  pop         ecx  
001513C0  mov         dword ptr [ebp-8],ecx 
001513C3  mov         eax,dword ptr [this] 
001513C6  mov         dword ptr [eax],0 
001513CC  mov         eax,dword ptr [this] 
001513CF  fldz             
001513D1  fstp        qword ptr [eax+8] 
001513D4  mov         eax,dword ptr [this] 
001513D7  mov         byte ptr [eax+10h],0 
001513DB  mov         eax,dword ptr [this] 
001513DE  pop         edi  
001513DF  pop         esi  
001513E0  pop         ebx  
001513E1  mov         esp,ebp 
001513E3  pop         ebp  
001513E4  ret

    As you can see, it’s identical.

    • 0