Why is there no pointer in the second case?

Question

C++ / C#

Good afternoon!
I don’t understand much why compilation is allowed in the second case:

typedef int (*SIG_TYP)(int); 
typedef void (SIG_ARG_TYP)(int);
SIG_TYP signal1(int, SIG_ARG_TYP){};

Answer 1

typedef int (*SIG_TYP)(int); 
typedef void (SIG_ARG_TYP)(int);
SIG_TYP signal1(int, SIG_ARG_TYP){};

how to understand the second construction? is it a pointer?

No, this type is the type of a function (not a pointer to a function).
Those. can be written like this:
and this will be the declaration of the function foo.