How Does the ' ' Operator Work with a Non-Capturing Lambda in C ?

A Positive Lambda: ' []{}' - The Secrets Unveiled

In the realm of C coding, the behavior of the ' ' operator applied to a lambda (' []{}') has sparked curiosity. While the code seems unconventional, it surprisingly compiles successfully with major compilers. But how does this exception to the norm work?

The Conversion to Function Pointer

The key to understanding this behavior lies in the ' ' operator's overloads. When applied to the lambda, ' ' triggers a conversion to a plain old function pointer. This is possible because the lambda in question is non-capturing, which means it doesn't reference any external variables.

According to the C standard (section 5.1.2), a non-capturing lambda closure object has a public non-virtual conversion function that returns a function pointer. This pointer points to a function that has the same behavior as the lambda.

The Built-in Operator Overloads

The ' ' operator has a built-in overload that converts any type to a pointer. In the case of the lambda, the conversion function mentioned earlier is selected as the candidate overload. The result is a function pointer of the same parameter and return types as the lambda's function call operator.

So, when ' ' is applied to the first lambda, it converts the lambda closure object into a function pointer. This makes it possible to assign the second lambda closure object to the same function pointer, since both lambdas have compatible types.

Standard Conformance

Yes, the code using ' []{}' is standard conforming. The C standard allows for the conversion of non-capturing lambda closure objects to function pointers. This behavior enables some surprising but valid coding tricks.

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