docs/reference: Update constrained docs now that tuples can be const.

Signed-off-by: Damien George <damien@micropython.org>

docs/reference/constrained.rst

@@ -126,8 +126,9 @@ that the objects remain in flash memory rather than being copied to RAM. The
 Python built-in types.
 
 When considering the implications of frozen bytecode, note that in Python
-strings, floats, bytes, integers and complex numbers are immutable. Accordingly
-these will be frozen into flash. Thus, in the line
+strings, floats, bytes, integers, complex numbers and tuples are immutable.
+Accordingly these will be frozen into flash (for tuples, only if all their
+elements are immutable). Thus, in the line
 
 .. code::
 
@@ -146,16 +147,16 @@ As in the string example, at runtime a reference to the arbitrarily large
 integer is assigned to the variable ``bar``. That reference occupies a
 single machine word.
 
-It might be expected that tuples of integers could be employed for the purpose
-of storing constant data with minimal RAM use. With the current compiler this
-is ineffective (the code works, but RAM is not saved).
+Tuples of constant objects are themselves constant. Such constant tuples are
+optimised by the compiler so they do not need to be created at runtime each time
+they are used. For example:
 
 .. code::
 
-    foo = (1, 2, 3, 4, 5, 6, 100000)
+    foo = (1, 2, 3, 4, 5, 6, 100000, ("string", b"bytes", False, True))
 
-At runtime the tuple will be located in RAM. This may be subject to future
-improvement.
+This entire tuple will exist as a single object (potentially in flash if the
+code is frozen) and referenced each time it is needed.
 
 **Needless object creation**
 
@@ -214,20 +215,24 @@ buffer; this is both faster and more efficient in terms of memory fragmentation.
 
 **Bytes are smaller than ints**
 
-On most platforms an integer consumes four bytes. Consider the two calls to the
+On most platforms an integer consumes four bytes. Consider the three calls to the
 function ``foo()``:
 
 .. code::
 
     def foo(bar):
         for x in bar:
             print(x)
+    foo([1, 2, 0xff])
     foo((1, 2, 0xff))
     foo(b'\1\2\xff')
 
-In the first call a tuple of integers is created in RAM. The second efficiently
+In the first call a `list` of integers is created in RAM each time the code is
+executed. The second call creates a constant `tuple` object (a `tuple` containing
+only constant objects) as part of the compilation phase, so it is only created
+once and is more efficient than the `list`. The third call efficiently
 creates a `bytes` object consuming the minimum amount of RAM. If the module
-were frozen as bytecode, the `bytes` object would reside in flash.
+were frozen as bytecode, both the `tuple` and `bytes` object would reside in flash.
 
 **Strings Versus Bytes**