c - 如何在 gcc -fdump-tree 中使用 "graph"选项？

Question

我正在尝试查看 gcc 是否可以提供更多信息，我可以使用这些信息从 C 源代码创建控制流图，然后使用它来检查 Gcov .gcov 中执行的测试用例采用的路径。输出。到目前为止，我最好的候选人是使用 -fdump-tree-cfg与 lineno选项( -fdump-tree-cfg-lineno )，但生成的文件对我来说仍然有限制，因为每个函数的节点信息都是分开的。像这样的事情:

;; Function main (main, funcdef_no=13, decl_uid=3054, cgraph_uid=13, symbol_order=13)

Removing basic block 7
Merging blocks 5 and 6
;; 1 loops found
;;
;; Loop 0
;;  header 0, latch 1
;;  depth 0, outer -1
;;  nodes: 0 1 2 3 4 5 6
;; 2 succs { 3 4 }
;; 3 succs { 5 }
;; 4 succs { 5 }
;; 5 succs { 6 }
;; 6 succs { 1 }
main ()
{
  int number;
  int D.3066;

  <bb 2> [0.00%]:
  [evenodd.c:7:5] printf ([evenodd.c:7:12] "\nEnter a number: ");
  [evenodd.c:8:5] scanf ([evenodd.c:8:11] "%d", [evenodd.c:8:5] &number);
  [evenodd.c:10:2] number.0_1 = number;
  [evenodd.c:10:2] checkNegative (number.0_1);
  [evenodd.c:12:17] number.1_2 = number;
  [evenodd.c:12:17] number.2_3 = (unsigned int) number.1_2;
  [evenodd.c:12:17] _4 = number.2_3 & 1;
  [evenodd.c:12:7] if (_4 == 0)
    goto <bb 3>; [0.00%]
  else
    goto <bb 4>; [0.00%]

  <bb 3> [0.00%]:
  [evenodd.c:13:9] number.3_5 = number;
  [evenodd.c:13:9] printf ([evenodd.c:13:16] "%d is Even\n", number.3_5);
  [0:0] goto <bb 5>; [0.00%]

  <bb 4> [0.00%]:
  [evenodd.c:15:9] number.4_6 = number;
  [evenodd.c:15:9] printf ([evenodd.c:15:16] "%d is Odd\n", number.4_6);

  <bb 5> [0.00%]:
  [evenodd.c:18:9] D.3066 = 0;
  number = {CLOBBER};

<L4> [0.00%]:
  return D.3066;

}



;; Function checkNegative (checkNegative, funcdef_no=14, decl_uid=3057, cgraph_uid=14, symbol_order=14)

;; 1 loops found
;;
;; Loop 0
;;  header 0, latch 1
;;  depth 0, outer -1
;;  nodes: 0 1 2 3 4 5
;; 2 succs { 3 4 }
;; 3 succs { 5 }
;; 4 succs { 5 }
;; 5 succs { 1 }
checkNegative (int number)
{
  <bb 2> [0.00%]:
  [evenodd.c:23:4] if (number < 0)
    goto <bb 3>; [0.00%]
  else
    goto <bb 4>; [0.00%]

  <bb 3> [0.00%]:
  [evenodd.c:24:9] printf ([evenodd.c:24:16] "%d is Negative\n", number);
  [0:0] goto <bb 5>; [0.00%]

  <bb 4> [0.00%]:
  [evenodd.c:26:9] printf ([evenodd.c:26:16] "%d is Positive\n", number);

  <bb 5> [0.00%]:
  [evenodd.c:28:1] return;

}

这意味着我的解析器需要识别 main 内的每个函数调用函数，然后合并它们的节点信息，这对我来说是一项相当艰巨的任务。

所以我遇到了-graph -fdump-tree 的选项来自this documentation但我不知道如何使用它。正如文档所描述的:

‘graph’

    For each of the other indicated dump files (-fdump-rtl-pass),
  dump a representation of the control flow graph suitable for 
  viewing with GraphViz to file.passid.pass.dot. Each function in
  the file is pretty-printed as a subgraph, so that GraphViz can
  render them all in a single plot.

    This option currently only works for RTL dumps, and the RTL
  is always dumped in slim form.

我不知道 GraphViz 是如何工作的，但由于它能够将所有函数的子图渲染为单个图(我假设这意味着所有子图都与主图合并为单个图)，这意味着提供的信息通过-graph更加完整并简化了创建控制流图的过程。

由于我无法理解它的含义，因此我尝试按原样使用该选项( -fdump-tree-cfg-lineno-graph )，但结果仍然与 -fdump-tree-cfg-lineno 相同的结果。

如何使用它？

Answer 1

解决方案很简单:

gcc -fdump-tree-all-graph main.c -o main

另请查看:How can I dump an abstract syntax tree generated by gcc into a .dot file?