从指向具有相同签名的函数的指针转换为函数指针,但参数的附加限定除外

converting to function pointer from pointer to function with same signature, except additional qualification of parameter

void ( * pFunc_pcInt ) ( int const * ) = nullptr ; 
void ( * pFunc_pInt ) ( int * ) = reinterpret_cast< void ( * ) ( int * ) >( pFunc_pcInt ) ;

这样的转换会导致未定义的行为吗?

standard (工作草案,强调我的):

A function pointer can be explicitly converted to a function pointer of a different type. [ Note: The effect of calling a function through a pointer to a function type ([dcl.fct]) that is not the same as the type used in the definition of the function is undefined.  — end note ] [...]

当然,void(int const *)void(int *) 是不同的类型。

类似的东西来自最著名的online reference(强调我的)之一:

Any pointer to function can be converted to a pointer to a different function type. Calling the function through a pointer to a different function type is undefined, but converting such pointer back to pointer to the original function type yields the pointer to the original function.

在您的特定情况下,这无关紧要,因为您正在将 nullptr 分配给函数指针。在任何情况下调用它都会导致错误。
话虽这么说,如果您已将有效函数指针分配给 pFunc_pcInt,通过转换后的指针 pFunc_pInt 调用它将会是一个 UB。