Add FAQs about various language details.

One of these finishes off issue #225.
This commit is contained in:
Matt Pharr
2012-04-05 15:24:26 -07:00
parent 4f8cf019ca
commit c7dc8862a5

View File

@@ -14,6 +14,12 @@ distribution.
+ `Why are there multiple versions of exported ispc functions in the assembly output?`_
+ `How can I more easily see gathers and scatters in generated assembly?`_
* Language Details
+ `What is the difference between "int *foo" and "int foo[]"?`_
+ `Why are pointed-to types "uniform" by default?`_
+ `What am I getting an error about assigning a varying lvalue to a reference type?`_
* Interoperability
+ `How can I supply an initial execution mask in the call from the application?`_
@@ -214,6 +220,125 @@ easier to understand:
jmp ___pseudo_scatter_base_offsets32_32 ## TAILCALL
Language Details
================
What is the difference between "int \*foo" and "int foo[]"?
-----------------------------------------------------------
In C and C++, declaring a function to take a parameter ``int *foo`` and
``int foo[]`` results in the same type for the parameter. Both are
pointers to integers. In ``ispc``, these are different types. The first
one is a varying pointer to a uniform integer value in memory, while the
second results in a uniform pointer to the start of an array of varying
integer values in memory.
To understand why the first is a varying pointer to a uniform integer,
first recall that types without explicit rate qualifiers (``uniform``,
``varying``, or ``soa<>``) are ``varying`` by default. Second, recall from
the `discussion of pointer types in the ispc User's Guide`_ that pointed-to
types without rate qualifiers are ``uniform`` by default. (This second
rule is discussed further below, in `Why are pointed-to types "uniform" by
default?`_.) The type of ``int *foo`` follows from these.
.. _discussion of pointer types in the ispc User's Guide: ispc.html#pointer-types
Conversely, in a function body, ``int foo[10]`` represents a declaration of
a 10-element array of varying ``int`` values. In that we'd certainly like
to be able to pass such an array to a function that takes a ``int []``
parameter, the natural type for an ``int []`` parameter is a uniform
pointer to varying integer values.
In terms of compatibility with C/C++, it's unfortunate that this
distinction exists, though any other set of rules seems to introduce more
awkwardness than this one. (Though we're interested to hear ideas to
improve these rules!).
Why are pointed-to types "uniform" by default?
----------------------------------------------
In ``ispc``, types without rate qualifiers are "varying" by default, but
types pointed to by pointers without rate qualifiers are "uniform" by
default. Why this difference?
::
int foo; // no rate qualifier, "varying int".
uniform int *foo; // pointer type has no rate qualifier, pointed-to does.
// "varying pointer to uniform int".
int *foo; // neither pointer type nor pointed-to type ("int") have
// rate qualifiers. Pointer type is varying by default,
// pointed-to is uniform. "varying pointer to uniform int".
varying int *foo; // varying pointer to varying int
The first rule, having types without rate qualifiers be varying by default,
is a default that keeps the number of "uniform" or "varying" qualifiers in
``ispc`` programs low. Most ``ispc`` programs use mostly "varying"
variables, so this rule allows most variables to be declared without also
requiring rate qualifiers.
On a related note, this rule allows many C/C++ functions to be used to
define equivalent functions in the SPMD execution model that ``ispc``
provides with little or no modification:
::
// scalar add in C/C++, SPMD/vector add in ispc
int add(int a, int b) { return a + b; }
This motivation also explains why ``uniform int *foo`` represents a varying
pointer; having pointers be varying by default if they don't have rate
qualifiers similarly helps with porting code from C/C++ to ``ispc``.
The tricker issue is why pointed-to types are "uniform" by default. In our
experience, data in memory that is accessed via pointers is most often
uniform; this generally includes all data that has been allocated and
initialized by the C/C++ application code. In practice, "varying" types are
more generally (but not exclusively) used for local data in ``ispc``
functions. Thus, making the pointed-to type uniform by default leads to
more concise code for the most common cases.
What am I getting an error about assigning a varying lvalue to a reference type?
--------------------------------------------------------------------------------
Given code like the following:
::
uniform float a[...];
int index = ...;
float &r = a[index];
``ispc`` issues the error "Initializer for reference-type variable "r" must
have a uniform lvalue type.". The underlying issue stems from how
references are represented in the code generated by ``ispc``. Recall that
``ispc`` supports both uniform and varying pointer types--a uniform pointer
points to the same location in memory for all program instances in the
gang, while a varying pointer allows each program instance to have its own
pointer value.
References are represented a pointer in the code generated by ``ispc``,
though this is generally opaque to the user; in ``ispc``, they are
specifically uniform pointers. This design decision was made so that given
code like this:
::
extern void func(float &val);
float foo = ...;
func(foo);
Then the reference would be handled efficiently as a single pointer, rather
than unnecessarily being turned into a gang-size of pointers.
However, an implication of this decision is that it's not possible for
references to refer to completely different things for each of the program
instances. (And hence the error that is issued). In cases where a unique
per-program-instance pointer is needed, a varying pointer should be used
instead of a reference.
Interoperability
================