在C命令行参数 [英] Command line arguments in C

问题描述

我有这个计划与命令行给出的值10,20,30执​​行。

I have this program execute with the values 10,20,30 given at command line.

int main(int argc , char **argv)
 { 
  printf("\n Printing the arguments of a program \n");
  printf("\n The total number of arguments in the program is %d",argc);
   while(argc>=0)
    { 
     printf("%s   ",argv[argc]);
     argc--;
     }
     return 0;
  }    

的输出是
 的参数中的节目的总数为4（空）30 20 10 ./a.out

The outputs is The total number of arguments in the program is 4(null) 30 20 10 ./a.out

如果这样做（空）来自??

Where did that (null) come from ??

推荐答案

的argv [0] 应该是能够识别该程序存在（尽可能）跑。 的argv [1] 到的argv [ARGC-1] 是实际上是在命令行中输入的参数。 ARGV [ARGC] 要求是一个空指针（§5.1.2.2.1/ 2）。

argv[0] is (to the extent possible) supposed to be something that identifies the program being run. argv[1] through argv[argc-1] are the arguments that were actually entered on the command line. argv[argc] is required to be a null pointer (§5.1.2.2.1/2).

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