3 ms·

How do you mean exactly? Go has an iterable for map, slices, arrays, strings, channels with the "range"[0] keyword. The values returned by range are copies of

by kyrra 12y ago

How do you mean exactly? Go has an iterable for map, slices, arrays, strings, channels with the "range"[0] keyword. The values returned by range are copies of whatever is in the structure so they should be immune to mutation (unless you're messing with pointers). [0] http://golang.org/ref/spec#RangeClause http://golang.org/ref/spec#RangeClause

rakoo 12y ago
He means that what you actually want to do is modify some state outside of the loop when you're iterating, which means that your loop doesn't provide a "safe" context where you know what to expect. Are you sure you're modifying the correct outside variable ? Are you sure you really are modifying it: bar[i] = f(foo[i]) and not creating a new one: bar[i] := f(foo[i]) ? Are you sure you're not also modifying another variable that you didn't know was "linked" somehow ? Given the direction Go has taken, it would make very little sense to introduce functional programming bits. But you must admit that the immutability/lack of side effects of the FP gives you a lot of assurance on what to expect.
- kyrra 12y ago
  I think you need to provide an example of what you're trying to say, as the syntax you showed above is not valid. the := operator is for defining a new variable, not for creating a new instance of some type. The := operator is just shorthand. foo := 5 is shorthand for: var foo int foo = 5 the "range" keyword over an array gives you the index and the value at that index. If you don't need the index or the value, you can put a "_" in it's place (to save on the allocation). http://play.golang.org/p/Z2ejUlN5An http://play.golang.org/p/Z2ejUlN5An You'll notice that foo[i] is still printing "0", because range always creates a copy of the object assigned into v. Unless you start using pointers, you'll never have mutation issues.
- NateDad 12y ago
  bar[i] := <anything> won't compile. You can only assign to slice/array indices.

by kyrra 12y ago

rakoo 12y ago

He means that what you actually want to do is modify some state outside of the loop when you're iterating, which means that your loop doesn't provide a "safe" context where you know what to expect. Are you sure you're modifying the correct outside variable ? Are you sure you really are modifying it: bar[i] = f(foo[i]) and not creating a new one: bar[i] := f(foo[i]) ? Are you sure you're not also modifying another variable that you didn't know was "linked" somehow ? Given the direction Go has taken, it would make very little sense to introduce functional programming bits. But you must admit that the immutability/lack of side effects of the FP gives you a lot of assurance on what to expect.

kyrra 12y ago
I think you need to provide an example of what you're trying to say, as the syntax you showed above is not valid. the := operator is for defining a new variable, not for creating a new instance of some type. The := operator is just shorthand. foo := 5 is shorthand for: var foo int foo = 5 the "range" keyword over an array gives you the index and the value at that index. If you don't need the index or the value, you can put a "_" in it's place (to save on the allocation). http://play.golang.org/p/Z2ejUlN5An http://play.golang.org/p/Z2ejUlN5An You'll notice that foo[i] is still printing "0", because range always creates a copy of the object assigned into v. Unless you start using pointers, you'll never have mutation issues.
NateDad 12y ago
bar[i] := <anything> won't compile. You can only assign to slice/array indices.