可编译器优化可变长度的循环吗？ [英] Can Compiler Optimize Loop with Variable Length?

问题描述

如果循环的最后一个索引（ a 和 b 在下面的示例中）在编译时不知道？

Can the compiler optimize loops if the last index of the loops (a and b in the following example) are not known at compile time?

未优化：

int* arr = new int[a*b];
for (i = 0; i < a; ++i){
    for(j = 0; j < b; ++j){
        arr[i*b+j] *= 8;
    }
}

//delete arr after done.

更多优化:(假设a和b很大...）

More Optimized: (assuming a and b are large...)

int c = a*b;
int* arr = new int[c];
for (i = 0; i < c; ++i){
        arr[c] *= 8;
}

//delete arr after done.

推荐答案

如果将数组视为线性空间，gcc

If you treat the array as linear space, gcc (and presumably others) will optimise even without knowing the extents at compile time.

此代码：

void by8(int* arr, int a, int b)
{
  auto extent = a * b;
  for (int i = 0; i < extent; ++i)
  {
    arr[i] *= 8;
  }
}

编译循环是向量化的）

by8(int*, int, int):
        imull   %esi, %edx
        testl   %edx, %edx
        jle     .L23
        movq    %rdi, %rax
        andl    $31, %eax
        shrq    $2, %rax
        negq    %rax
        andl    $7, %eax
        cmpl    %edx, %eax
        cmova   %edx, %eax
        cmpl    $8, %edx
        jg      .L26
        movl    %edx, %eax
.L3:
        sall    $3, (%rdi)
        cmpl    $1, %eax
        je      .L15
        sall    $3, 4(%rdi)
        cmpl    $2, %eax
        je      .L16
        sall    $3, 8(%rdi)
        cmpl    $3, %eax
        je      .L17
        sall    $3, 12(%rdi)
        cmpl    $4, %eax
        je      .L18
        sall    $3, 16(%rdi)
        cmpl    $5, %eax
        je      .L19
        sall    $3, 20(%rdi)
        cmpl    $6, %eax
        je      .L20
        sall    $3, 24(%rdi)
        cmpl    $7, %eax
        je      .L21
        sall    $3, 28(%rdi)
        movl    $8, %ecx
.L5:
        cmpl    %eax, %edx
        je      .L27
.L4:
        leal    -1(%rdx), %r8d
        movl    %edx, %r9d
        movl    %eax, %r10d
        subl    %eax, %r9d
        subl    %eax, %r8d
        leal    -8(%r9), %esi
        shrl    $3, %esi
        addl    $1, %esi
        leal    0(,%rsi,8), %r11d
        cmpl    $6, %r8d
        jbe     .L7
        leaq    (%rdi,%r10,4), %r10
        xorl    %eax, %eax
        xorl    %r8d, %r8d
.L9:
        vmovdqa (%r10,%rax), %ymm0
        addl    $1, %r8d
        vpslld  $3, %ymm0, %ymm0
        vmovdqa %ymm0, (%r10,%rax)
        addq    $32, %rax
        cmpl    %r8d, %esi
        ja      .L9
        addl    %r11d, %ecx
        cmpl    %r11d, %r9d
        je      .L22
        vzeroupper
.L7:
        movslq  %ecx, %rax
        sall    $3, (%rdi,%rax,4)
        leal    1(%rcx), %eax
        cmpl    %eax, %edx
        jle     .L23
        cltq
        sall    $3, (%rdi,%rax,4)
        leal    2(%rcx), %eax
        cmpl    %eax, %edx
        jle     .L23
        cltq
        sall    $3, (%rdi,%rax,4)
        leal    3(%rcx), %eax
        cmpl    %eax, %edx
        jle     .L23
        cltq
        sall    $3, (%rdi,%rax,4)
        leal    4(%rcx), %eax
        cmpl    %eax, %edx
        jle     .L23
        cltq
        sall    $3, (%rdi,%rax,4)
        leal    5(%rcx), %eax
        cmpl    %eax, %edx
        jle     .L23
        cltq
        addl    $6, %ecx
        sall    $3, (%rdi,%rax,4)
        cmpl    %ecx, %edx
        jle     .L28
        movslq  %ecx, %rcx
        sall    $3, (%rdi,%rcx,4)
        ret
.L22:
        vzeroupper
.L23:
        ret
.L27:
        ret
.L26:
        testl   %eax, %eax
        jne     .L3
        xorl    %ecx, %ecx
        jmp     .L4
.L28:
        ret
.L21:
        movl    $7, %ecx
        jmp     .L5
.L15:
        movl    $1, %ecx
        jmp     .L5
.L16:
        movl    $2, %ecx
        jmp     .L5
.L17:
        movl    $3, %ecx
        jmp     .L5
.L18:
        movl    $4, %ecx
        jmp     .L5
.L19:
        movl    $5, %ecx
        jmp     .L5
.L20:
        movl    $6, %ecx
        jmp     .L5

$ b b

编译器：gcc 5.4，带有命令行选项：-std = c ++ 14 -O3 -march = native

compiler : gcc 5.4 with command line options: -std=c++14 -O3 -march=native

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