6 replies to this topic

[rfx]

Hi everybody,

I'm new to C and I'm fighting with pointers and argv's right now... I wrote:

#include <stdio.h>

#include <stdlib.h>


void main(int argc,char *argv[]){

	while(*argv){

		printf("%s",*argv);

		if(*argv == "show"){

			printf("Great");

		}

		*argv++;

	}

}

and the script should show "great" if it will be started with ./myscript show
What's wrong there?

[Ancient Dragon]

The array of pointers is NOT terminated with a NULL pointer. You have to test argc to find out how may strings are in argv. You also have to use strcmp() to compare two character arrays.

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

{

   int i;

  for(i = 0; i < argc; i++)

  {

      printf("%s\n", argv[i]);

      if( strcmp( argv[i], "SHOW") == 0)

        printf("Great\n");

}

[Zer033]

Hi, I understand the code here, but what exactly is the TC trying to accomplish with ./myscript show?

[Ancient Dragon]

"./myscript show" is a common *nix console way to execute a program located in the current working directory. The "show" part is the parameter to the program "myscript".

[Warrior]

Quote

The array of pointers is NOT terminated with a NULL pointer.

The command line parameters are terminated by a NULL pointer as by the standard.

Warrior

[rfx]

Thanks for all your help!! This works fine:

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

void main(int argc,char *argv[]){
	while(*argv){
		printf("%s\n",*argv);
		if(strcmp(*argv,"show") == 0){
			printf("Great\n");
		}
		argv++;
	}
}

[manux]

Warrior said:

The command line parameters are terminated by a NULL pointer as by the standard.

Warrior

Still it's not clean and/or clear to write code that uses such premises, especially when argc is provided :)