Realistic usage of the C99 'restrict' keyw

I was browsing through some documentation and questions/answers and saw it mentioned. I read a brief description, stating that it would be basically a promise from the programmer that the pointer won't be used to point somewhere else.

Can anyone offer some realistic cases where its worth actually using this?

回答1:

restrict says that the pointer is the only thing that accesses the underlying object. It eliminates the potential for pointer aliasing, enabling better optimization by the compiler.

For instance, suppose I have a machine with specialized instructions that can multiply vectors of numbers in memory, and I have the following code:

void MultiplyArrays(int* dest, int* src1, int* src2, int n)
{
    for(int i = 0; i < n; i++)
    {
        dest[i] = src1[i]*src2[i];
    }
}

The compiler needs to properly handle if dest, src1, and src2 overlap, meaning it must do one multiplication at a time, from start to the end. By having restrict, the compiler is free to optimize this code by using the vector instructions.

Wikipedia has an entry on restrict, with another example, here.

回答2:

The Wikipedia example is very illuminating.

It clearly shows how it allows to save one assembly instruction.

Without restrict:

void f(int *a, int *b, int *x) {
  *a += *x;
  *b += *x;
}

Pseudo assembly:

load R1 ← *x    ; Load the value of x pointer
load R2 ← *a    ; Load the value of a pointer
add R2 += R1    ; Perform Addition
set R2 → *a     ; Update the value of a pointer
; Similarly for b, note that x is loaded twice,
; because a may be equal to x.
load R1 ← *x
load R2 ← *b
add R2 += R1
set R2 → *b

With restrict:

void fr(int *restrict a, int *restrict b, int *restrict x);

Pseudo assembly:

load R1 ← *x
load R2 ← *a
add R2 += R1
set R2 → *a
; Note that x is not reloaded,
; because the compiler knows it is unchanged
; load R1 ← *x
load R2 ← *b
add R2 += R1
set R2 → *b

Does GCC really do it?

GCC 4.8 Linux x86-64:

gcc -g -std=c99 -O0 -c main.c
objdump -S main.o

With -O0, they are the same.

With -O3:

void f(int *a, int *b, int *x) {
    *a += *x;
   0:   8b 02                   mov    (%rdx),%eax
   2:   01 07                   add    %eax,(%rdi)
    *b += *x;
   4:   8b 02                   mov    (%rdx),%eax
   6:   01 06                   add    %eax,(%rsi)  

void fr(int *restrict a, int *restrict b, int *restrict x) {
    *a += *x;
  10:   8b 02                   mov    (%rdx),%eax
  12:   01 07                   add    %eax,(%rdi)
    *b += *x;
  14:   01 06                   add    %eax,(%rsi)

For the uninitiated, the calling convention is:

rdi = first parameter
rsi = second parameter
rdx = third parameter

GCC output was even clearer than the wiki article: 4 instructions vs 3 instructions.

Arrays

So far we have single instruction savings, but if pointer represent arrays to be looped over, a common use case, then a bunch of instructions could be saved, as mentioned by supercat.

Consider for example:

void f(char *restrict p1, char *restrict p2) {
    for (int i = 0; i < 50; i++) {
        p1[i] = 4;
        p2[i] = 9;
    }
}

Because of restrict, a smart compiler (or human), could optimize that to:

memset(p1, 4, 50);
memset(p2, 9, 50);

which is potentially much more efficient as it may be assembly optimized on a decent libc implementation (like glibc): Is it better to use std::memcpy() or std::copy() in terms to performance?

Does GCC really do it?

GCC 5.2.1.Linux x86-64 Ubuntu 15.10:

gcc -g -std=c99 -O0 -c main.c
objdump -dr main.o

With -O0, both are the same.

With -O3:

with restrict:

3f0:   48 85 d2                test   %rdx,%rdx
3f3:   74 33                   je     428 <fr+0x38>
3f5:   55                      push   %rbp
3f6:   53                      push   %rbx
3f7:   48 89 f5                mov    %rsi,%rbp
3fa:   be 04 00 00 00          mov    $0x4,%esi
3ff:   48 89 d3                mov    %rdx,%rbx
402:   48 83 ec 08             sub    $0x8,%rsp
406:   e8 00 00 00 00          callq  40b <fr+0x1b>
                        407: R_X86_64_PC32      memset-0x4
40b:   48 83 c4 08             add    $0x8,%rsp
40f:   48 89 da                mov    %rbx,%rdx
412:   48 89 ef                mov    %rbp,%rdi
415:   5b                      pop    %rbx
416:   5d                      pop    %rbp
417:   be 09 00 00 00          mov    $0x9,%esi
41c:   e9 00 00 00 00          jmpq   421 <fr+0x31>
                        41d: R_X86_64_PC32      memset-0x4
421:   0f 1f 80 00 00 00 00    nopl   0x0(%rax)
428:   f3 c3                   repz retq

Two memset calls as expected.

without restrict: no stdlib calls, just a 16 iteration wide loop unrolling which I do not intend to reproduce here :-)

I haven't had the patience to benchmark them, but I believe that the restrict version will be faster.

C99

Let's look at the standard for completeness sake.

restrict says that two pointers cannot point to overlapping memory regions. The most common usage is for function arguments.

This restricts how the function can be called, but allows for more compile-time optimizations.

If the caller does not follow the restrict contract, undefined behavior.

The C99 N1256 draft 6.7.3/7 "Type qualifiers" says:

The intended use of the restrict qualifier (like the register storage class) is to promote optimization, and deleting all instances of the qualifier from all preprocessing translation units composing a conforming program does not change its meaning (i.e., observable behavior).

and 6.7.3.1 "Formal definition of restrict" gives the gory details.

Strict aliasing rule

The restrict keyword only affects pointers of compatible types (e.g. two int*) because the strict aliasing rules says that aliasing incompatible types is undefined behavior by default, and so compilers can assume it does not happen and optimize away.

See: What is the strict aliasing rule?

See also

C++14 does not yet have an analogue for restrict, but GCC has __restrict__ as an extension: What does the restrict keyword mean in C++?
Many questions that ask: according to the gory details, does this code UB or not?
- Understanding restrict qualifier by examples
- Restricted pointer questions
- Is it legal to assign a restricted pointer to another pointer, and use the second pointer to modify the value?
A "when to use" question: When to use restrict and when not to
The related GCC __attribute__((malloc)), which says that the return value of a function is not aliased to anything: GCC: __attribute__((malloc))