C++ handles an uninitialized field in a strange way - c++

Question

The sample below shows reading an uninitialized field a.i1 in two ways. The first call does not compile. However, calling a.donothing() that does not do anything, the original call compiles fine and prints the default value 0. Why is this inconsistency?
I am using Visual Studio Community 2015, the compiler output is as follows:
Severity Code Description Project File Line Suppression State
Error C4700 uninitialized local variable 'a' used
class A {
public:
int i1;
void donothing() {}
};
int main() {
A a;
cout << "main1: " << a.i1 << endl; // compile fails
a.donothing();
cout << "main2: " << a.i1 << endl; // prints 0 (default)
return 0;
}

Answer 1

Compiler is doing what it ought to do. You can fix it like this (as one solution out of many):
class A {
public:
A(int i = 0) : i1(i) {}
int i1;
void donothing() {}
};

Answer 2

In both cases a warning must be issued, at most. The fact that calling donothing cancels the error is a clear indication that this is a bug. You can report it at Microsoft Connect.
A simple workaround for this problem is to change the declaration to A a{};.
You can test your code on different compilers at Compiler Explorer.
[EDIT] The warning message C4700 is treated as an error if Security Development Lifecycle is turned on (/sdl).

Answer 3

Depends on the compiler, the compiler should supply a default constructor that will initialize your members with a default value. But this behavior is not dependable. Since C++11 you can say ClassName()=default; The best practice is to prove your own default constructor.
Your code never had any compiler errors with g++ 5.4.0
#include <iostream>
using namespace std;
class A {
public:
//A() : i1(0) { } // compiler will provide this if you don't write anything
// since C++ 11 you can also say A() = default;
A() = default;
int i1;
void donothing() {}
void writeMember() const { cout << "i1 value: " << i1 << endl; }
};
// better provide a signature for the main function
int main(int argc, char* argv[]) {
A a;
a.writeMember();
cout << "main1: " << a.i1 << endl; // compile fails
a.donothing();
cout << "main2: " << a.i1 << endl; // prints 0 (default)
return 0;
}
To compile the above code stored in testclass.cpp
g++ -std=c++11 -o testclass testclass.cpp
By using the C++11 default I got
i1 value: 4196976
main1: 4196976
main2: 4196976
If you comment out A()=default; this will rely on the compiler provided initializer, or the compiler may be lazy and not doing anything for performance reasons. You get
i1 value: 4196944
main1: 4196944
main2: 4196944
If you uncomment the line after public: you should consistently get 0
This illustrates the importance of adhering to good conventions of alway provide your own default constructor. The compiler maybe doing the right thing by not assigning any particular value to your member because you may assign another value in the next operation. This can save one operation. The member will be simply allocated on the stack, in that case the member got a random value. If you run this code on a different computer you will for sure get a different value.