我正在尝试查看gcc是否可以提供更多信息，我可以使用它从C源代码创建控制流图，然后使用它来检查Gcov {{1}执行的测试用例采用的路径输出。到目前为止，我最好的候选人使用.gcov和-fdump-tree-cfg选项（lineno），但由于节点信息是针对每个函数分开的，因此生成的文件仍然限制我。像这样：
-fdump-tree-cfg-lineno这意味着我的解析器需要识别;; 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;
}
函数内的每个函数调用，然后合并它们的节点信息，这对我来说是一项非常艰巨的任务。
所以我来自this documentation的main -graph选项，但我不知道如何使用它。正如医生所描述的那样：
-fdump-tree我不知道GraphViz是如何工作的，但是因为它能够将所有函数的子图渲染为单个图（我假设它意味着所有子图与主图合并为单个图），这意味着提供的信息通过‘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.
更加完整，简化了创建控制流图的过程。
由于我无法理解它的含义，我尝试使用该选项（-graph），但结果仍然与-fdump-tree-cfg-lineno-graph的结果相同。
我该如何使用它？
答案 0 :(得分：0)
解决方案很简单：
gcc -fdump-tree-all-graph main.c -o main
还可以看看：How can I dump an abstract syntax tree generated by gcc into a .dot file?