How to receive a arg from command line in C

I want to write a program to receive a argument from command line. It's like a kind of atof().

There my program goes:

  9     char s[] = "3.1415e-4";
 10     if (argc == 1) {
 11         printf("%e\n",atof(s));
 12     }
 13     else if (argc == 2) {
 14         //strcpy(s, argv[1]);
 15         printf("%e\n",atof(argv[1]));
 16     }

1.should I just use argv[1] for the string to pass to my atof(), or, put i开发者_如何学Pythont into s[]?

2.If I'd better put it in s[], is there some build-in function to do this "put" work? maybe some function like strcpy()??

thanks.

I personally would do it this way:

#include <stdlib.h>
#include <stdio.h>

int main(int argc, char **argv)
{
  const char defaultStr[]="3.1415e-4";
  const char *arg = NULL;
  if(argc <= 1)
  {
     arg = defaultStr;
  }
  else
  {
     arg = argv[1];
  }

  printf("%e\n", atof(arg));

  return 0;
}

There is no need for put it into s. Just use argv[1].

Having said this, is you are putting it into s anyway (which is not that bad) make sure you change your code to:

char s[20];
strcpy(s, argv[1]);

Using strcpy with a constant char array is not a good idea.

You can simply pass it as argv[1], no need to copy it. Should you want to copy it, you need to ensure that s can actually hold the full string contained in argv[1], and allocate a bigger array if not... so it is messier.