What Type Does 'auto' Deduce for a C 11 Lambda Expression?

Exploring the Type of Lambda Deduction with "auto" in C 11

Contrary to the misconception that a lambda's type is a function pointer, a recent test has revealed an unexpected result. When attempting to assign a lambda expression to both a function pointer and an "auto" variable, an assertion fails. This raises the question, what is the true type of a lambda when deduced with the "auto" keyword?

To clarify, lambda expressions are effectively syntactic shortcuts for functors. During compilation, a lambda is transformed into a functor object. Elements within the [] brackets become constructor parameters and functor member variables, while parameters within the () brackets translate into the functor's operator() parameters.

Strikingly, lambdas that capture no variables (i.e., those without elements within the []) can be converted into function pointers (although this conversion is not supported by MSVC2010). However, it is crucial to emphasize that the intrinsic type of a lambda remains an unspecified functor type.

In the provided code example, the lambda expression LAMBDA is initially assigned to a function pointer pFptr and then to an "auto" variable pAuto. The subsequent assertion verifies that the types of pFptr and pAuto are equal, which indeed passes. This indicates that the "auto" keyword successfully deduces the type of the lambda to be a function pointer, allowing for seamless assignment to pFptr.

Nevertheless, the key takeaway is that lambdas are inherently unspecified functor types, regardless of their ability to be cast into function pointers. This distinction highlights the flexibility and complexity of lambda expressions in modern C programming.

The above is the detailed content of What Type Does 'auto' Deduce for a C 11 Lambda Expression?. For more information, please follow other related articles on the PHP Chinese website!