Why Does `int x = x 1;` Work in C/C ?

Declarative Initialization of Variables in C/C : A Surprising Behavior

The familiar declaration and initialization syntax in C/C allows the initialization of variables upon their declaration. However, a peculiar behavior arises when the variable to be initialized appears in the initialization expression itself. Consider the code snippet:

<code class="cpp">int x = x + 1;</code>

This code compiles successfully, and surprisingly, the value of x becomes 1 after execution. This behavior may seem counterintuitive, as the variable x does not appear to have been declared before its use in the initialization.

Explanation: Point of Declaration and Undefined Behavior

In C , the point of declaration for a variable is immediately after the complete declarator (the variable's type and name) and before the initializer (if any). Thus, in the above code, x comes into existence at the equals sign, allowing it to be used on the right-hand side.

However, it's important to note that unless the variable is initialized with static storage duration (e.g., outside a function), the behavior is undefined. This is because the x that comes into existence before the initialization has an arbitrary value.

The C 03 standard specifies that for variables initialized with their own (indeterminate) value, as in the example in the question, the behavior is undefined:

int x = x; // Example from the C++03 standard
{
    int x = x; // Undefined behavior
}

Practical Implications

While this behavior might be surprising, it's important to avoid utilizing it in production code. Unless the variable is declared with static storage duration, the undefined behavior can lead to unpredictable results and errors.

However, understanding this behavior can prove useful in comprehending complex expressions involving newly declared variables or in analyzing code with unconventional initialization techniques.

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