One question about function definition in C++ -

i'm reading material function pointer in c++, , come across 1 function definition not understand.
standard function definition have form:

type name (param...) 

but following definition seems little strange me. can explain me ? thanks.

float (*getptr1(const char opcode)) (float, float)<br> {     if(opcode == '+')         return &plus;     else         return &minus; // default if invalid operator passed } 

note: plus , minus 2 functions param (float, float) , return float.

the rule reading hairy declarations start leftmost identifier , work way out, remembering () , [] bind before * (i.e., *a[] array of pointers, (*a)[] pointer array, *f() function returning pointer, , (*f)() pointer function):

        getptr1                                       -- getptr1         getptr1(                 )                    -- function          getptr1(           opcode)                    -- taking single parameter named opcode         getptr1(const char opcode)                    -- of type const char        *getptr1(const char opcode)                    -- , returning pointer       (*getptr1(const char opcode)) (            )    -- function       (*getptr1(const char opcode)) (float, float)    -- taking 2 parameters of type float float (*getptr1(const char opcode)) (float, float)    -- , returning float 

so, if opcode equal '+', getptr1 return pointer function plus, , if it's '-', return pointer function minus.

c , c++ declaration syntax expression-centric (much bjarne pretend otherwise); form of declaration should match form of expression used in code.

if have function f returns pointer int , want access value being pointed to, execute function , dereference result:

x = *f(); 

the type of expression *f() int, declaration/definition function is

int *f() { ... } 

now suppose have function f1 returns pointer function f defined above, , want access integer value calling f1. need call f1, derefence result (which function f), , execute it, , dereference that result (since f returns pointer):

x = *(*f1())(); // *f1() == f, (*f1())() == f() , *(*f1())() == *f() 

the type of expression *(*f1())() int, decaration/definition f1 needs be

int *(*f1())() { return f; }