What does the restrict keyword mean in C++?

问题

I was always unsure, what does the restrict keyword mean in C++?

Does it mean the two or more pointer given to the function does not overlap? What else does it mean?

回答1:

In his paper, Memory Optimization, Christer Ericson says that while restrict is not part of the C++ standard yet, that it is supported by many compilers and he recommends it's usage when available:

restrict keyword

! New to 1999 ANSI/ISO C standard

! Not in C++ standard yet, but supported by many C++ compilers

! A hint only, so may do nothing and still be conforming

A restrict-qualified pointer (or reference)...

! ...is basically a promise to the compiler that for the scope of the pointer, the target of the pointer will only be accessed through that pointer (and pointers copied from it).

In C++ compilers that support it it should probably behave the same as in C.

See this SO post for details: Realistic usage of the C99 ‘restrict’ keyword?

Take half an hour to skim through Ericson's paper, it's interesting and worth the time.

Edit

I also found that IBM's AIX C/C++ compiler supports the __restrict__ keyword.

g++ also seems to support this as the following program compiles cleanly on g++:

#include <stdio.h>

int foo(int * __restrict__ a, int * __restrict__ b) {
    return *a + *b;
}

int main(void) {
    int a = 1, b = 1, c;

    c = foo(&a, &b);

    printf("c == %d\n", c);

    return 0;
}

I also found a nice article on the use of restrict:

Demystifying The Restrict Keyword

Edit2

I ran across an article which specifically discusses the use of restrict in C++ programs:

Load-hit-stores and the __restrict keyword

Also, Microsoft Visual C++ also supports the __restrict keyword.

回答2:

As others said, if means nothing as of C++14, so let's consider the __restrict__ GCC extension which does the same as the C99 restrict.

C99

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 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.

A possible optimization

The Wikipedia example is very illuminating.

It clearly shows how as 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?

g++ 4.8 Linux x86-64:

g++ -g -std=gnu++98 -O0 -c main.cpp
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 and michael.

Consider for example:

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

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

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

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?, possibly with SIMD instructions.

Without, restrict, this optimization could not be done, e.g. consider:

char p1[4];
char *p2 = &p1[1];
f(p1, p2, 3);

Then for version makes:

p1 == {4, 4, 4, 9}

while the memset version makes:

p1 == {4, 9, 9, 9}

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.

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?

Does it work for references?

According to the GCC docs it does: https://gcc.gnu.org/onlinedocs/gcc-5.1.0/gcc/Restricted-Pointers.html with syntax:

int &__restrict__ rref

There is even a version for this of member functions:

void T::fn () __restrict__

回答3:

Nothing. It was added to the C99 standard.

回答4:

This is the original proposal to add this keyword. As dirkgently pointed out though, this is a C99 feature; it has nothing to do with C++.

回答5:

There's no such keyword in C++. List of C++ keywords can be found in section 2.11/1 of C++ language standard. restrict is a keyword in C99 version of C language and not in C++.

回答6:

Since header files from some C libraries use the keyword, the C++ language will have to do something about it.. at the minimum, ignoring the keyword, so we don't have to #define the keyword to a blank macro to suppress the keyword.

来源：https://stackoverflow.com/questions/776283/what-does-the-restrict-keyword-mean-in-c

标签

c++

restrict-qualifier