如何在GCOV

Question

通过以下方法转储的工作集似乎有所不同：

1. 方法1：使用gcov（gcov转储手册页here）

   gcc -ftest-coverage -fprofile-arcs program.c

   //运行程序

   gcov -w program.c

program.gcno:    01410000:   1:BLOCKS 10 blocks
program.gcno:    01430000:   3:ARCS 1 arcs 
program.gcno:    01430000:   3:ARCS 1 arcs 
program.gcno:    01430000:   3:ARCS 1 arcs 
program.gcno:    01430000:   5:ARCS 2 arcs 
program.gcno:    01430000:   3:ARCS 1 arcs 
program.gcno:    01430000:   3:ARCS 1 arcs 
program.gcno:    01430000:   5:ARCS 2 arcs 
program.gcno:    01430000:   5:ARCS 2 arcs 
program.gcno:    01430000:   3:ARCS 1 arcs 
program.gcno:    01450000:  12:LINES
program.gcno:    01450000:  11:LINES
program.gcno:    01450000:  10:LINES
program.gcno:    01450000:  13:LINES
program.gcno:    01450000:  10:LINES
program.gcno:    01450000:  10:LINES
program.gcno:    01450000:  10:LINES
program.gcno:    01450000:  10:LINES

1. 方法2：使用附加的-fdump-ipa-all（以及-ftest-coverage -fprofile-arcs）编译程序，并通读program.c.XXXi.profile转储文件。

Counter working sets:                                           
0.78%: num counts=1, min counter=891420285
1.56%: num counts=1, min counter=891420285
2.34%: num counts=1, min counter=891420285
3.12%: num counts=1, min counter=891420285
3.90%: num counts=1, min counter=891420285
...
97.50%: num counts=2, min counter=224114269
98.28%: num counts=2, min counter=224114269
99.06%: num counts=2, min counter=224114269
99.90%: num counts=2, min counter=224114269

应该如何解释在转储文件中显示工作集的输出（方法2）？在当前情况下，“工作集”的确切含义是什么？