Implicit variable initialization in C++

Question

#include <iostream>

int a;
void foo();

int main() 
{
 std::cout << "a = " << a << std::endl;
 foo();
 return 0;
}

void foo(){
    int b;
    std::cout << "b = " << b << std::endl;
}

Output:

a = 0
b = 32650

I have created a function named foo that declares a int variable and prints it. It prints some junk value because b is not initialized at the time of declaration then how is a getting initialized to 0 everytime?

Why is a initialized to 0 while b being initialized to some junk value?

Answer 1

In the original version of C++ language standard, all variables with static storage duration were zero-initialized before any other initialization took place.

In modern C++ this initialization stage (aka static initialization ) is split into constant initialization (for variables with explicit constant initializers) and zero initialization (for everything else). Your a falls into the second category. So your a is zero-initialized.

Automatic variables of non-class types, like your b , begin their life with indeterminate values, unless you initialize them explicitly. Using that indeterminate value in an expression leads to undefined behavior.

Answer 2

From draft of c++17 standard (other standards states almost the same):

3.6.2 Initialization of non-local variables [basic.start.init]

...

Variables with static storage duration (3.7.1) or thread storage duration (3.7.2) shall be zero-initialized (8.5) before any other initialization takes place.

...

3.7.1 Static storage duration [basic.stc.static]

...

All variables which do not have dynamic storage duration, do not have thread storage duration, and are not local have static storage duration . The storage for these entities shall last for the duration of the program (3.6.2, 3.6.3).

If you will go deep into definitions, you will find that a is actually has static storage duration (it is global variable) and therefore it is zero-initialized.