Is it possible to write a function deffinition which includeds a function as an argument ?
for ex.
typedef struct item item;
struct item{
......... }
item *add(item *p1, item* newnode()){.........}
item *newnode() {.....}
-----------------------------------------------------------------
-----------------------------------------------------------------
or maybe like this:
typedef item*newnode(){.....} node;
item *add(item*p1, node){.....}
Of course! Here, this should work:
Code:int All( char str[], int Func() ) { int i; for( i = 0; i < strlen(str); i++ ) { if( !Func( str[i] ) ) return false; } return true; } void print(char msg[]) { clrscr(); printf("\n\n\n\n %s \n", msg); getch(); } int main() { char string[] = "1234567"; if( All( string, isdigit ) ) print("The string is all numbers."); else print("The string is NOT all numbers."); if( All( string, isalpha )) print("The string is all characters."); else print("The string is NOT all characters."); return 0; }
Code:#include <cmath> #include <complex> bool euler_flip(bool value) { return std::pow ( std::complex<float>(std::exp(1.0)), std::complex<float>(0, 1) * std::complex<float>(std::atan(1.0) *(1 << (value + 2))) ).real() < 0; }
And another example:
Code:int add(int a, int b) { return a+b; } int multiply(int a, int b) { return a*b; } int calculate(int (*f)(int,int), int a, int b) { return f(a, b); } int main() { int a = 4, b = 5; printf("%d\n", calculate(add, a, b) ); printf("%d\n", calculate(multiply, a, b) ); return 0; }
Another (my preferred) method:
Function Pointers.
----
Code:typedef int (*MULTIPLYFUNC)(int,int); typedef struct { int val_1; int val_2; MULTIPLYFUNC theFuncPtr; }myStruct; int main(void); int MyMultiplyFunc(int,int); int main(void) { myStruct infoStruct; MULTIPLYFUNC fPtr; int result; fPtr = NULL; infoStruct.val_1 = 5; intoStruct.val_2 = 10; infoStruct.theFuncPtr = (MULTIPLYFUNC*)&MyMultiplyFunc; fPtr = infoStruct.theFuncPtr; result = fPtr(infoStruct.val_1,infoStruct.val2); return(0); } int MyMultilyFunc(int arg1,int arg2) { return(arg1*arg2); }
Thanks for the help.
Actually, there is no real reason to even have 'fPtr' here. I suppose it was really for just an example, but why bother having a function pointer as a stucture member if you don't do anything with it?infoStruct.theFuncPtr = (MULTIPLYFUNC*)&MyMultiplyFunc;
fPtr = infoStruct.theFuncPtr;
result = fPtr(infoStruct.val_1,infoStruct.val2);
In either case, nothing in your example actually addressed the original question, which was "Can I have a function as a parameter in another function?"Code:myStruct m; m.theFuncPtr = &someFunc; result = m.theFuncPtr( m.val_1, m.val_2 );
The answer, is yes. The simplest example of this is qsort.
Quzah.
Last edited by quzah; 06-18-2002 at 03:33 AM.
Hope is the first step on the road to disappointment.
Yes, quzah, it was to illustrate using functions pointers, which can be passed in other functions:
Code:typedef int (*MULTIPLYFUNC)(int,int); int MultiplyIntsFunc(int,int); int AMathFunc(MULTIPLYFUNC,int,int); /* <-- note */ int main(void); int main(void) { int result; result = AMathFunc(&MultiplyIntsFunc,5,5); return(0); } int AMathFunc(MULTIPLYFUNC fPtr,int a,int b) { int result; result = fPtr(a,b); return(result); } int MultiplyIntsFunc(int a, int b) { return(a*b); }
> result = AMathFunc(&MultiplyIntsFunc,5,5);
Just to point out that the & is entirely optional in this context
You get a function pointer from just the name of the function
result = AMathFunc(MultiplyIntsFunc,5,5);
Yes, that is entirely correct. I merely used it to demonstrate that an address was being supplied.
