digitalmars.D.learn - pattern matching
- %u (13/13) Sep 06 2011 template factorial(int n) { const factorial = n * factorial!(n-1); }
- bearophile (5/6) Sep 06 2011 In D unfortunately there is no pattern matching on the run-time values o...
- Jonathan M Davis (28/46) Sep 06 2011 No. It works with templates, because it's generating code. Unless you're...
template factorial(int n) { const factorial = n * factorial!(n-1); } template factorial(int n : 1) { const factorial = 1; } i think this pattern matching or like it, can i do the same thing with regular function int factorial(int n) { return n* factorial(n-1); return 1 ; } int factorial(int n : 0) { return 1 ; } is that work? thanks in advance
Sep 06 2011
%u:is that work?In D unfortunately there is no pattern matching on the run-time values of function arguments. But I don't really understand what you are asking me, sorry. Bye, bearophile
Sep 06 2011
On Tuesday, September 06, 2011 21:05:54 %u wrote:template factorial(int n) { const factorial = n * factorial!(n-1); } template factorial(int n : 1) { const factorial = 1; } i think this pattern matching or like it, can i do the same thing with regular function int factorial(int n) { return n* factorial(n-1); return 1 ; } int factorial(int n : 0) { return 1 ; } is that work? thanks in advanceNo. It works with templates, because it's generating code. Unless you're dealing with class member functions, all function calls are determined at compile time, so they can't depend on the values of their arguments at all. With class member functions, they use the class' vtable to do polymorphic calls and call the function in the type which is the actual type of that object (as opposed to what the referenc is). e.g. class A { void func(int i) {writeln("A");} } class B : A { void func(int i) {writeln("B";} } void main() { A a = new B; a.func(); } will print "B". And that's not pattern matching. It's just simple polymorphism. The closest that you could get to pattern matching with function calls is with if statements. if(n == 0) return factorial0(); else return factorialn(n); - Jonathan M Davis
Sep 06 2011