c - while循环中的分支优化，为什么更少的指令花费更多的运行时间？

Question

我有一个用 C 实现的数据结构项目，它为其他程序导出各种 API。最近想对grofile的Hot函数做一些优化。这是供您引用的项目。

https://github.com/Incarnation-p-lee/libds有一个热门函数binary_search_tree_node_insert，如下:

/*
 * RETURN the pointer of inserted node of the binary search tree
 *        If root is NULL or node is NULL, RETURN NULL.
 */
struct binary_search_tree *
binary_search_tree_node_insert(struct binary_search_tree *root,
    struct binary_search_tree *node)
{
    register struct binary_search_tree **iter;

    if (!node || !root) {
        pr_log_warn("Attempt to access NULL pointer.\n");
    } else {
        iter = &root;
        while (*iter) {
            if (node->chain.nice == (*iter)->chain.nice) {
                if (*iter == node) {
                    pr_log_info("Insert node exist, nothing will be done.\n");
                } else {
                    doubly_linked_list_merge((*iter)->chain.link, node->chain.link);
                }
                return *iter;
#ifndef OPT_HOT
            } else if (node->chain.nice > (*iter)->chain.nice) {
                    iter = &(*iter)->right;
            } else if (node->chain.nice < (*iter)->chain.nice) {
                    iter = &(*iter)->left;
#else
            } else {
                binary_search_tree_insert_path_go_through(node, iter);
#endif
            }
        }
        return *iter = node;
    }

    return NULL;
}

我想优化两个 else-if 部分，因为它是一半到一半的分支，这可能会对管道产生很大影响。所以我写了一个宏 binary_search_tree_insert_path_go_through 来替换这两个分支。实现如下:

/*
 * 1. node->nice => rbx, *iter => rcx.
 * 2. compare rbx, and 0x8(rcx).
 * 3. update iter.
 */
#define binary_search_tree_insert_path_go_through(node, iter) \
    asm volatile (                                            \
        "mov $0x18, %%rax\n\t"                                \
        "mov $0x20, %%rdx\n\t"                                \
        "mov 0x8(%1), %%rbx\n\t"                              \
        "mov (%0), %%rcx\n\t"                                 \
        "cmp 0x8(%%rcx), %%rbx\n\t"                           \
        "cmovg %%rdx, %%rax\n\t"                              \
        "lea (%%rcx, %%rax), %0\n\t"                          \
        :"+r"(iter)                                           \
        :"r"(node)                                            \
        :"rax", "rbx", "rcx", "rdx")

但是这个功能的单元测试关于这个变化下降了大约6-8%。从objdump代码(右手优化代码)来看，指令少，我很难理解为什么优化前要花更多时间。

还有一些细节供大家引用:

struct collision_chain {
    struct doubly_linked_list *link;
    sint64                    nice;
};
/*
 * binary search tree
 */
struct binary_search_tree {
    struct collision_chain chain;
    sint32                 height;  /* reserved for avl */
    /* root node has height 0, NULL node has height -1 */
    union {
        struct binary_search_tree *left;
        struct avl_tree           *avl_left;    /* reserved for avl   */
        struct splay_tree         *splay_left;  /* reserved for splay */
    };
    union {
        struct binary_search_tree *right;
        struct avl_tree           *avl_right;    /* reserved for avl   */
        struct splay_tree         *splay_right;  /* reserved for splay */
    };
};
struct doubly_linked_list {
    uint32                    sid;
    void                      *val;
    struct doubly_linked_list *next;
    struct doubly_linked_list *previous;
};

它运行在 virtual-box 上，2 核 i5-3xxM，cpuinfo 如下:

processor       : 0
vendor_id       : GenuineIntel
cpu family      : 6
model           : 58
model name      : Intel(R) Core(TM) i5-3230M CPU @ 2.60GHz
stepping        : 9
microcode       : 0x19
cpu MHz         : 2568.658
cache size      : 6144 KB
physical id     : 0
siblings        : 2
core id         : 0
cpu cores       : 2
apicid          : 0
initial apicid  : 0
fpu             : yes
fpu_exception   : yes
cpuid level     : 5
wp              : yes
flags           : fpu vme de pse tsc msr pae mce cx8 apic sep mtrr pge mca cmov pat pse36 clflush mmx fxsr sse sse2 ht syscall nx rdtscp lm constant_tsc rep_good nopl pni ssse3 lahf_lm
bogomips        : 5137.31
clflush size    : 64
cache_alignment : 64
address sizes   : 36 bits physical, 48 bits virtual
power management:

processor       : 1
vendor_id       : GenuineIntel
cpu family      : 6
model           : 58
model name      : Intel(R) Core(TM) i5-3230M CPU @ 2.60GHz
stepping        : 9
microcode       : 0x19
cpu MHz         : 2568.658
cache size      : 6144 KB
physical id     : 0
siblings        : 2
core id         : 1
cpu cores       : 2
apicid          : 1
initial apicid  : 1
fpu             : yes
fpu_exception   : yes
cpuid level     : 5
wp              : yes
flags           : fpu vme de pse tsc msr pae mce cx8 apic sep mtrr pge mca cmov pat pse36 clflush mmx fxsr sse sse2 ht syscall nx rdtscp lm constant_tsc rep_good nopl pni ssse3 lahf_lm
bogomips        : 5137.31
clflush size    : 64
cache_alignment : 64
address sizes   : 36 bits physical, 48 bits virtual
power management:

Answer 1

我不知道现代处理器是否相同，但是 Linus really didn't like the CMOV instruction back in '07 .

由于您正在进行微观优化，因此请将相等性检查移至最后一个位置。它几乎总是错误的，但您在每次迭代中都做到了。

此外，我会尝试不使用指针到指针模式。由于指针别名问题，间接往往会使优化器阻塞。相反，使用带有两个指针的英寸蠕虫模式:

void insert(NODE *x, NODE **root) {
  NODE *trail = NULL;
  NODE *lead = *root;
  while (lead) {
    trail = lead;
    if (x->key < lead->key)
      lead = lead->left;
    else if (x->key > lead->key)
      lead = lead->right;
    else return; // do nothing;
  }
  // lead has found null, so insert
  if (trail)
    // redo the last key comparison
    if (x->key < trail->key)
      trail->left = x;
    else
      trail->right = x;
  else 
    *root = x;
}

在我的 MacBook 上，这会编译为以下 64 位代码，其中循环仅包含 10 条指令。很难从您帖子中的微小列表中分辨出来，但看起来它要长得多:

    pushq   %rbp
    movq    %rsp, %rbp
    movq    (%rsi), %rdx
    testq   %rdx, %rdx
    je      LBB0_11
    movl    16(%rdi), %ecx
LBB0_2:                                 
    movq    %rdx, %rax     # dx=lead, ax=trail
    cmpl    16(%rax), %ecx # comparison with key
    jge     LBB0_4         # first branch
    movq    %rax, %rdx     # go left (redundant because offset(left)==0!)
    jmp     LBB0_6
LBB0_4:                                 
    jle     LBB0_12        # second branch
    leaq    8(%rax), %rdx  # go right
LBB0_6:                                 
    movq    (%rdx), %rdx   # move lead down the tree
    testq   %rdx, %rdx     # test for null
    jne     LBB0_2         # loop if not
    testq   %rax, %rax     # insertion logic
    je      LBB0_11
    movl    16(%rdi), %ecx
    cmpl    16(%rax), %ecx
    jge     LBB0_10
    movq    %rdi, (%rax)
    popq    %rbp
    retq
LBB0_11:
    movq    %rdi, (%rsi)
LBB0_12:                   # return for equal keys
    popq    %rbp
    retq
LBB0_10:
    movq    %rdi, 8(%rax)
    popq    %rbp
    retq

如果比较代价高昂(你没有显示“好”是什么)，你也可以尝试存储跟踪比较的二进制结果，以便最终检查使用它而不是重做比较。