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I want to know, in c++, when does the initialization of objects take place?
Is it at the compile time or link time?
For ex: 

//file1.cpp
extern int i;
int j=5;

//file2.cpp ( link with file1.cpp)
extern j;
int i=10;

Now, what does compiler do : according to me, it allocates storage for variables.
Now I want to know :
does it also put initialization value in that storage or is it done at link time?

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3 Answers 3

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Actually there are different cases:

  • global variables or static variables (not classes): these values are stored in an init section of the exe/dll. These values are created by the linker based on the compiled object files information. (initialization on loading + mapping the dll/exe into memory)
  • local non static variables: these values are set by the compiler by putting these values on the stack (push/pop on x86) (compiler initialization)
  • objects: memory is reserved on the stack, the actual setting of values is deferred to the call to the constructor (run-time initialization)
  • pointers to objects (not actually a new case): space is reserved for the pointer only. The object pointed to exists only after a call to new that reserves memory and calls the constructor to initialize it (run-time initialization)
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4 Comments

What about "static const" members (not classes)? Do they really exist as a 'variable'? I think that every usage of them is simply replaced by the value and the "static const variable" just disappears. Right? For the rest, good overview, +1.
Ok..So in above code..will following things happen ? 1. while compiling file1.cpp, compiler leaves i as it is i.e doesn't allocate storage for i. 2. compiler allocates storage for j, but doesn't initialize it. 3. While compiling file2.cpp, compiler leaves j as it is i.e doesn't allocate storage for it. 4. compiler allocates storage for i, but doesn't initialize it. 5. While linking file1.o and file2.o, now let file2.o is initialized first, so now: Does j gets initial value of 0? or doesn't get initialized?
@Patrick, good point, I do not know for sure. I guess you are right, as the cost of having a variable is higher than an inlined value for basic types. However not 100% sure if all compilers behave that way...
@Happy Mittal, the linker doesn't care about the value while processing the object files. It only ensures an init section is created with the proper values for i and j. The order in which the .o files are processed does not matter. For the linker j is just the content of a specific address. At run-time j will be 5 and i 10.
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As you said the compiler allocates storage for variables. I think the intialization value will also be done at compile time and not during link time.

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3 Comments

But I read in a book that compiler can't determine what would be value in a storage, that's why we can't use something like this : int a=5; int arr[a];
@Happy If the compiler did enough static analysis to realize that that could be converted to int arr[5]; it would be fine, but they generally don't (it might not even be allowed by the spec)
The reason for int a=5; int arr[a]; is not defined is because the array size is undefined. Later you can change a to 10 that time how do you think the compiler will resizze the previously declared size.
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There are no objects in your example, just ints. If by "initialization" you mean when are their values assigned, those ints will be converted to word-sized entries in a data section in the object file, which will be hard-coded with their initial values. The data section, along with the rest of the object file, is created by the compiler, so I suppose the answer to your question is compile-time

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3 Comments

if u consider OOP then we can say int also as a class.
@vis Ints aren't classes in C++, they're a primitive type; there's no constructor call or any of the other scaffolding associated with classes
Thats correct. I understand. maybe we must change the question to Initialization of variables rather than objects.

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