Why Does Adding a Unary ' ' Operator Allow Lambda Reassignment in C ?

A Positive Lambda: ' []{}' - Delving into the Unexpected

In an attempt to investigate the enigmatic nature of lambda expressions, the question of "Redefining lambdas not allowed in C 11, why?" arose. A peculiar discovery by Johannes Schaub revealed that adding a unary operator before the initial lambda allows it to compile.

The Query:

Why does the following code compile without errors:

int main() {
    auto test = +[]{}; // Note the unary operator + before the lambda
    test = []{};
}

Is this behavior in accordance with the C standards?

The Unveiling:

Yes, the code conforms to the C standards. The operator remarkably triggers a conversion of the lambda into a plain function pointer.

The Unfolding:

The compiler interprets the initial lambda ([]{}) and creates a closure object. Since this lambda does not capture any variables, a conversion function exists to convert the closure object into a function pointer with identical parameters and return type as the closure's function call operator.

This conversion aligns with the requirement of the unary operator. The built-in overloads for include one that converts any type T to a pointer to T. The closure type fulfills this requirement by providing a conversion to a function pointer.

Consequently, the expression auto test = []{}; deduces the type of test to be void(*)(). This enables the subsequent assignment in the second line, where the second lambda/closure object also undergoes the conversion to a function pointer, resulting in a compatible assignment.

Significance:

This unusual behavior highlights the versatility of lambda expressions and the intricate mechanisms of the C language. It enables the reassignment of lambda expressions by converting them to function pointers, offering flexibility and code optimization possibilities.

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