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Why Does ` []{}` Enable Lambda Reassignment in C ?

Linda Hamilton
Linda HamiltonOriginal
2024-12-28 05:34:09284browse

Why Does ` []{}` Enable Lambda Reassignment in C  ?

Decoding the Magical " " Operator: Understanding the " []{}" Lambda

In a perplexing observation, a lone " " prefix to a lambda expression in C miraculously enables its reassignment. This enigmatic sorcery prompted the question: why does the following code compile successfully?

int main() {
    auto test = +[]{}; // The "+" operator casts a lambda to a function pointer
    test = []{};
}

The answer lies in the obscure depths of the C standard. The " " triggers a conversion to a plain old function pointer for the lambda. The lambda, being non-capturing, inherently possesses a conversion function to a function pointer with an identical signature. This conversion function, as defined in the C standard, returns the address of a function that behaves identically to the lambda's function call operator.

The unary " " operator, when applied to the closure object generated by the lambda, engages a set of built-in overloads. One of these overloads accepts any pointer type and converts it to a function pointer. The closure type's conversion to a function pointer, the only candidate overloaded function, thus takes precedence.

Consequently, the type of "test" in "auto test = []{};" is deduced to be "void(*)()". This function pointer compatibility enables the assignment of a second lambda/closure object to "test" even though their closure types differ.

This knowledge unveils the secret behind the " []{}" lambda's success. By casting the lambda to a function pointer, it bypasses the restriction on redefining lambdas, unlocking the ability to assign a new lambda object.

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