Typedef and function pointer

In the use of typedef, the most troublesome thing is to point to the function pointer. Without the following function, do you know the definition of the following expression and how to use it?

Int (* S_calc_func( CharOP ))( Int, Int); If you do not know, see the following Program , Which provides detailed descriptions. // Define four functions Int Add ( Int , Int ); Int Sub ( Int , Int ); Int Mul ( Int , Int ); Int Div ( Int , Int ); // Define pointer to this type of function Typedef Int (* Fp_calc )( Int , Int ); // I will not introduce it first. Can you understand the content of the next line? Int (* S_calc_func( CharOP ))( Int, Int); // The content of the next line is exactly the same as that of the previous line, // Define a function Calc_func , Which is based on the operating characters OP Returns the pointer to the corresponding computing function. Fp_calc Calc_func ( Char OP ); // According OP Returns the calculated result value. Int Calc ( Int A, Int B, Char OP ); Int Add ( Int A, Int B ){ ReturnA + B; } Int Sub ( Int A, Int B ){ ReturnA-B; } Int Mul ( Int A, Int B ){ ReturnA * B; } Int Div ( Int A, Int B ){ ReturnB? A/B:-1; } // This function is used exactly the same as the next function job and call method, // The parameter is OP Instead of the last two integers. Int (* S_calc_func ( Char OP ))( Int , Int ){ ReturnCalc_func (OP ); } Fp_calc Calc_func ( Char OP ){ Switch(OP) { Case'+ ': ReturnAdd; Case'-': ReturnSub; Case'*': ReturnMul; Case'/': ReturnDiv; Default: Return Null; } Return Null; } Int Calc ( Int A, Int B, Char OP ){ Fp_calcFp = calc_func (OP ); // The following is a similar direct definition of the pointer variable to the function. // The following line is unnecessary. Typedef To implement the pointer to the function! Int(* S_fp )( Int, Int) = S_calc_func (OP ); //Assert(FP = s_fp );// It can be asserted that the two are equal. If(FP) ReturnFP (A, B ); Else Return-1; } Void Test_fun (){ IntA = 100, B = 20; Printf ("calc (% d, % d, % C) = % d \ n", a, B, '+', Calc (A, B, '+ ')); Printf ("calc (% d, % d, % C) = % d \ n", a, B, '-', Calc (A, B, '-')); Printf ("calc (% d, % d, % C) = % d \ n", a, B, '*', Calc (A, B, '*')); Printf ("calc (% d, % d, % C) = % d \ n", a, B, '/', Calc (A, B, '/')); } Running result Calc (100, 20, +) = 120 Calc (100, 20,-) = 80 Calc (100, 20, *) = 2000 Calc (100, 20,/) = 5