用于优化的LLVM静态值分析

Question

假设我有这样的函数:

int foo(int a, int b, int d, int x){
  if (c) {a = 1; b = 1; d = a;}
  else {a = 2; b = 2; d = 1;}
  if (a == b) {x = d;} else {x = 0;}
  return x;
}

这个简单的函数总是返回1.使用clang -O2选项进行编译并查看反汇编代码LLVM正确编译此函数return 1;.

我的问题是:llvm如何进行静态值分析？最弱的前提技术？价值传播？霍尔的技术？

Answer 1

LLVM做各种各样的事情:见这里.

您可以在每次优化传递后获得中间表示的转储,如下所示:

clang -c -mllvm -print-after-all -O2 foo.c

确定哪个阶段正在做什么.

事实上,这个特殊的例子并不是非常神奇!

如果将其转换为SSA格式,它看起来像这样:

  // block 1
  if (c == 0) goto L1;
  // block 2
  a_0 = 1;
  b_0 = 1;
  d_0 = a_0;
  goto L2;
L1:
  // block 3
  a_1 = 2;
  b_1 = 2;
  d_1 = 1;
  goto L2;

L2:
  // block 4 (has two possible predecessors: block 2 or block 3)
  a_2 = PHI(a_0, a_1); // i.e. a_0 if we came from block 2, a_1 if we came from block 3
  b_2 = PHI(b_0, b_1); // i.e. b_0 if we came from block 2, b_1 if we came from block 3
  d_2 = PHI(d_0, d_1); // i.e. d_0 if we came from block 2, d_1 if we came from block 3
  if (a_2 != b_2) goto L3;
  // block 5
  x_0 = d_2;
  goto L4:
L3:
  // block 6
  x_1 = 0;
  goto L4;

L4:
  // block 7 (has two possible predecessors: block 5 or block 6)
  return PHI(x_0, x_1); // i.e. x_0 if we came from block 5, x_1 if we came from block 6

简单地传播常数值会导致:

  // block 1
  if (c == 0) goto L1;
  // block 2
  a_0 = 1;
  b_0 = 1;
  d_0 = 1;
  goto L2;
L1:
  // block 3
  a_1 = 2;
  b_1 = 2;
  d_1 = 1;
  goto L2;

L2:
  // block 4
  a_2 = PHI(1, 2); // i.e. 1 if we came from block 2, 2 if we came from block 3
  b_2 = PHI(1, 2); // i.e. 1 if we came from block 2, 2 if we came from block 3
  d_2 = 1;         // PHI(d_0, d_1) == PHI(1, 1) i.e. 1 regardless of where we came from
  if (a_2 != b_2) goto L3;
  // block 5
  x_0 = 1;  // (we've deduced that d_2 == 1 regardless of control flow)
  goto L4:
L3:
  // block 6
  x_1 = 0;
  goto L4;

L4:
  // block 7
  return PHI(1, 0); // i.e. 1 if we came from block 5, 0 if we came from block 6

简化删除不再用于其他任何内容的赋值给出了:

  // block 1
  if (c == 0) goto L1;
  // block 2
  goto L2;
L1:
  // block 3
  goto L2;

L2:
  // block 4
  a_2 = PHI(1, 2); // i.e. 1 if we came from block 2, 2 if we came from block 3
  b_2 = PHI(1, 2); // i.e. 1 if we came from block 2, 2 if we came from block 3
  if (a_2 != b_2) goto L3;
  // block 5
  goto L4:
L3:
  // block 6
  goto L4;

L4:
  // block 7
  return PHI(1, 0); // i.e. 1 if we came from block 5, 0 if we came from block 6

......现在第一个条件显然是无操作; 并且第二个必须始终为真(块5路径)因为a_2并且b_2是相同的表达式.结果是

  return 1;