What Happens During Static Initialization Order Fiasco in C ?

"Static Initialization Order Fiasco" Enigma Unveiled

The infamous "static initialization order fiasco" (SIOF) arises when multiple files within a C program contain static variables that depend on each other for initialization. Consider the following example:

// file1.cpp
extern int y;
int x = y + 1;

// file2.cpp
extern int x;
int y = x + 1;

Questions:

1. During compilation of file1.cpp, does the compiler:

   • Allocate storage for y?
   • Initialize x?

2. During compilation of file2.cpp, does the compiler:

   • Allocate storage for x?
   • Initialize y?

3. When linking file1.o and file2.o, if file2.o is initialized first, does:

   • x receive an initial value of 0?
   • x remain uninitialized?

Answers:

According to the C standard (3.6.2 "Initialization of non-local objects"):

1. a. The compiler does not allocate storage for y.
   b. The compiler allocates storage for x but does not initialize it.
2. a. The compiler does not allocate storage for x.
   b. The compiler allocates storage for y but does not initialize it.
3. a. x receives an initial value of 0.
   b. x does not remain uninitialized.

Explanation:

• Step 1: x and y are zero-initialized before any other initialization.
• Step 2: Either x or y is dynamically initialized (standard does not specify which). The initialized variable receives the value 1, as the other variable is zero-initialized.
• Step 3: The other variable is then dynamically initialized, receiving the value 2.

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