restrict keyword is mainly used in pointer declarations as a type qualifier for pointers. It doesn't add any new functionality. It is only a way for programmer to inform about an optimization that compiler can make.
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 can occur.
If a function has two pointers pa and pb , with the same value, we say the pointers alias each other. This introduces constraints on the order of instruction execution. If two write accesses that alias occur in program order, they must happen in the same order on the processor and cannot be re-ordered.
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.
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 x may point to a (a aliased by x) thus
; the value of x will change when the value of a
; changes.
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" is no longer needed.
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 parameterrsi
= second parameterrdx
= third parameterGCC 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
restrict
, but GCC has __restrict__
as an extension: What does the restrict keyword mean in C++?
__attribute__((malloc))
, which says that the return value of a function is not aliased to anything: GCC: __attribute__((malloc))
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