C中的命令行参数

Question

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;
  }    

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

Where did that (null) come from ??

Answer 1

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).

Answer 2

argc is the total number of elements in the argv array; they are numbered from 0 to argc - 1 . You are printing five values and only the last four are valid.

Answer 3

他们教你在学校计算的方式不适用于C.在C中我们算0,1,2，......

Answer 4

Because you're printing out argv[4], argv[3], argv[2], argv[1], argv[0], instead of argv[3], argv[2], argv[1], argv[0].

Basically you've got an off by one error.

Answer 5

argc will have number of elements which can be accessed from argv[0] to argv[argc-1] . So modify your condition accordingly viz print from argv[argc-1].

Here is a command line arguments tutorial link as there are many things which you may have missed when reading it. Hence you are not able to understand the reason for that output.

Numbering for indexes is usually from 0, because of many reasons. Please check this question which will help you understand why its zero based. https://stackoverflow.com/questions/393462?tab=votes&page=1#tab-top

Answer 6

I think that the fact that the code is while(argc >= 0) shows that the you know that arrays are zero indexed. The problem is that you start at argc instead of argc-1.

or, put another way, you appear to understand that argv[0] is the name of the program, argc INCLUDES that as an argument, so when it says argc = 4, it means that there are 3 arguments in addition to the program name...

And, as Jerry Coffin pointed out, C requires the argv[argc] to be NULL, as a sentinal, in case knowing that the arguments are 1 - argc-1 isn't enough... (Belt and suspenders)

Answer 7

It is giving output as (null) 30 20 10 because you are using argv[argc], and when you pass 10,20,30 then number of arguments is 4 (as the first argument is always program name) so for the first time argv[argc] means argv[4] ie, fifth argument (as array begins with a[0] therefore argv[0] will be the first argument ...argv 1 will be the second....so on.....argv[4] will be the fifth argument passed) which is null therefore it is printing as (null) 30 20 10. For the proper result use

 printf("%s ",argv[argc-1]); 

Command Line Argument in C Programming - Concept