Parameterless Constructor - Struct

Referring to - http://www.codeproject.com/csharp/structs_in_csharp.asp

Although the CLR allows it, C# does not allow structs to have a default parameterless constructor. The reason is that, for a value type, compilers by default neither generate a default constructor, nor do they generate a call to the default constructor. So, even if you happened to define a default constructor, it will not be called and that will only confuse you. To avoid such problems, the C# compiler disallows definition of a default constructor by the user.

this is a C# tutorial segment, but since C++ too supports structs, so I was wondering if the same thing holds for C++

i.e. does C++ also does not provide default parameterless constructor? and are these constructors, if i explicitly define them in my definition be ever called?

Why would you expect a rule in C# to apply to C++? Sure, in C++/CLI (which is a .NET language) you can define a value class that has these semantics, but in standard C++, the only difference between a structure and a class is the default access (public for structures, private for classes).

If you do not declare/define any constructor, the compiler will create one for you and call it when an object of that type is created.

If you declare/define a parameterized constructor, no default (parameterless) one will be created and any creation of the object must be made through the parameterized one defined. Though it doesn't prohibit a parameterless one being defined also.

Structs in C++ are stored in Stack/Heap?

What will be the default value of my struct type?

Assignment of struct types on to another struct type will copy values or reference?

yes
types don't have values.
values

Quote:

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Originally Posted by robwhit

yes

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so, are they stored in Heap or Stack???

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Originally Posted by vb.bajpai

Structs in C++ are stored in Stack/Heap?

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As with everyting in C++, depends how you create it. You can allocates structs on either the heap or the stack.

Quote:

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What will be the default value of my struct type?

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Uninitialized stack variables will contain garbage values. Initialized stucts will defaultly call the default constructors of each of there members. The default constructors for primitive types cause zero initialization.

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Assignment of struct types on to another struct type will copy values or reference?

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Assigning to a value will make a copy. Initializing a reference will make a reference. It is also possible to assign pointers, which are special types for manipulating data by reference. Again, structs are not special in this respect -- all types can be manipulated by all three ways.

stack in scope, heap with new.

Code:

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void func()
{
    df gb; //stack
    df *dfp = new df; //heap
    delete dfp;
    return;
}

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Quote:

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Originally Posted by King Mir

Uninitialized stack variables will contain garbage values. Initialized stucts will defaultly call the default constructors of each of there members. The default constructors for primitive types cause zero initialization.

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Not true. The initialization rules for primitives are confusing and hard to describe. An explicit call to a primitive's default constructor (e.g. int()) causes zero-initialization, but the compiler-generated default constructor doesn't initialize primitives at all.

If I define a vector of ints as follows:

Code:

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  std::vector<int> v(100);

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are the elements of v initialized to zero, or not? I've read that in general the default constructor is called for each element, but couldn't figure out whether that applied to a primitive type.

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are the elements of v initialized to zero, or not? I've read that in general the default constructor is called for each element, but couldn't figure out whether that applied to a primitive type.

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They are initialised to zero since the default constructor is called explicitly.

Quote:

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Originally Posted by robatino

If I define a vector of ints as follows:

Code:

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  std::vector<int> v(100);

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are the elements of v initialized to zero, or not? I've read that in general the default constructor is called for each element, but couldn't figure out whether that applied to a primitive type.

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Yes, they will be zero

** - too slow :-(

They will be zero, but that's because the line

Code:

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std::vector<int> v(100)

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is equivalent (by making explicit the default argument) to

Code:

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std::vector<int> v(100, int())

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This effectively passes 0 as the second argument to the constructor, and all elements are initialized to this value.


A vector is a much higher-level case anyway.