运行结果：
| Starting iteration 1. Starting action Action. Action.c(11): score[0]=100 Action.c(12): *(p++)=100 Action.c(11): score[1]=98 Action.c(12): *(p++)=98 Action.c(11): score[2]=78 Action.c(12): *(p++)=78 Action.c(11): score[3]=55 Action.c(12): *(p++)=55 Action.c(11): score[4]=0 Action.c(12): *(p++)=0 Action.c(14): -------------------------- Action.c(18): score[0]=100 Action.c(19): *(p+100)=0 Action.c(18): score[1]=98 Action.c(19): *(p+98)=0 Action.c(18): score[2]=78 Action.c(19): *(p+78)=0 Action.c(18): score[3]=55 Action.c(19): *(p+55)=0 Action.c(18): score[4]=0 Action.c(19): *(p+0)=0 Action.c(21): -------------------------- Action.c(26): sixnum[0][0]=1 Action.c(27): *(*(p1+1)+0)=54385392 Action.c(26): sixnum[0][1]=2 Action.c(27): *(*(p1+2)+0)=54385392 Action.c(26): sixnum[0][2]=3 Action.c(27): *(*(p1+3)+0)=54385392 Action.c(26): sixnum[1][0]=4 Action.c(27): *(*(p1+4)+0)=54385392 Action.c(26): sixnum[1][1]=5 Action.c(27): *(*(p1+5)+0)=54385392 Action.c(26): sixnum[1][2]=6 Action.c(27): *(*(p1+6)+0)=54385392 Ending action Action. Ending iteration 1. |
本章节算做对C语言一些概念性知识的回味道吧。后面再写关于loadrunner脚本编写的内容要根据实际意义一些脚本进行分析。
相关链接：
