shell 脚本实现乘法口诀表的两种方法——shell与C语言教程

shell 脚本实现乘法口诀表的两种方法——shell与C语言

话不多说直接给出代码：

  1 #!/bin/bash                                         
  2 if [ $# -eq 0 ]         //用于判断输入的参数个数为0
  3 then
  4         echo "welcome you!"
  5         echo "this is a test with 2 methods to output arbitrarily mux table!"
  6 else
  7         echo "sorry you input invliad argc!"
  8         echo "you input other argc before!"
  9         exit 0         //退出
 10 fi
 11 
 12 while ((1))            //创建死循环防止输错给出的命令而退出
 13 do
 14         echo "you can choose 'shell' 'gcc' or 'quit' command"
 15         read -p "please input your choice:" choice
 16 
 17         if  [ $choice == "shell" ]              //判断是shell方法实现
 18         then
 19                 echo "will do show shell"
 20                 read -p "please input a num you want:" num
 21                 touch 6.sh
 22                 echo "" > 6.sh
 23                 echo '#!/bin/bash                //将shell脚本实现的方法写入某个脚本文件中，这里是6.sh
 24 for ((j=1;j<=$1;j++))
 25 do 
 26         for((i=1;i<=j;i++))
 27         do  
 28                 echo -ne "$i*$j=$[$i*$j]\t"
 29         done
 30                 echo -e "\r"
 31 done
 32 '   >> 6.sh
 33                 #chmod 777 6.sh          //看自己是什么用户权限选择这个命令
 34                 cat 6.sh
 35                 source 6.sh $num         //执行
 36                 exit 0
 37         elif [ $choice == "gcc" ]        //判断是C语言实现方法
 38         then
 39                 echo "will do show c"
 40                 touch 6.c
 41                 echo "" > 6.c
 42                 echo '#include<stdio.h>   //将C语言的方法写入到.c文件
 43 
 44 
 45 
 46 
 47 int main(int argc,char **argv)
 48 {       if(argc<2)
 49                 perror("argc num is not correct！please do it again!");
 50         printf("%s\n",argv[1]);
 51         int num;
 52         num=atoi(argv[1]);
 53         int i,j;
 54         for (j=1;j<=num;j++)
 55         {    
 56                 for(i=1;i<=j;i++)
 57                 {   
 58                         printf("%d*%d=%d\t",i,j,i*j);
 59                 }   
 60                 printf("\n");   
 61         }   
 62 
 63         return 0;
 64 }' >> 6.c
 65                 cat 6.c                             
 66                 gcc -o 6 6.c                       //编译过程
 67                 read -p "please input a number you want:" number
 68                 ./6 $number                        //程序执行
 69                 exit 0
 70         elif [ $choice == "quit" ]                  //您选择直接退出
 71         then
 72                 exit 0
 73         else
 74                 echo "you don't choose a correct choice!"  //表示输错命令可以重新输入       
 75         fi
 76 done