Passing a integer through command line in C?

问题

I was wondering if someone could explain how passing arguments through command line works? I'm really confused by how it works. Right now I'm trying to pass one integer into the main program. How would I go about doing this?

EDIT: keep getting the initialization makes integer from pointer without a cast [-Wint-conversion] error?

#include <stdio.h>
#define PI 3.1416
int
main (int argc, char *argv[])

{ 
  double r,area, circ;

  char a = argv[1];
  int num =  a - '0';

  printf("You have entered %d",num); 

  r= num/2;
  area = PI * r * r;
  circ= 2 * PI * r;

  printf ("A circle with a diameter of %d ", num);
  printf ("has an area of %5.3lf cm2\n", area);
  printf ("and a circumference of %4.2lf cm.\n", circ);

  return (0);

}

回答1:

The signature for the main function in C would be this:

int main(int argc, char *argv[]);

argc is the number of arguments passed to your program, including the program name its self.

argv is an array containing each argument as a string of characters.

So if you invoked your program like this:

./program 10

argc would be 2

argv[0] would be the string program

argv[1] would be the string 10

You could fix your code like this:

#include <stdio.h>
#include <stdlib.h>
#define PI 3.1416
int
main (int argc, char *argv[])

{
  double r,area, circ;

  char *a = argv[1];
  int num = atoi(a);

  printf("You have entered %d",num);

  r= num/2;
  area = PI * r * r;
  circ= 2 * PI * r;

  printf ("A circle with a diameter of %d ", num);
  printf ("has an area of %5.3lf cm2\n", area);
  printf ("and a circumference of %4.2lf cm.\n", circ);

  return (0);

}

You probably also want to add line breaks into your print statements for readability.

来源：https://stackoverflow.com/questions/43160958/passing-a-integer-through-command-line-in-c