Thread: catching from command line.

hey guys, i've got an issue with my code.. I'm trying to get the program to output names when the user types in an "@" symbol.. here's the code which isnt working.. any help would be appreciated.

Code:

 int main(void) {
   char word[20];

   scanf("%s", &word);
   char names[1] =  "@";
   printf("names = %s \n", names);
	   if (word == names){
		   printf("Joe, Dave, Larry");
	   }
	   printf("wrong answer, you typed in %s\n", word);
   }

Code:

#include "stdio.h"
#include "stdlib.h"

int main(void) 
{
   char word[20];
   int i;
   scanf("%s", &word);
   char names =  '@';
   printf("names = %c \n", names);

   for(i=0;i<20;i++)
   {
	   if (word [i] == names)
     {
		   printf("Joe, Dave, Larry");
       exit(0);
	   }        
    } 
       printf("wrong answer, you typed in %s\n", word);    
  
return 0;
  }

With the exception of the item in red which shouldn't be there.

Originally Posted by shwetha_siddu

Code:

#include "stdio.h"
#include "stdlib.h"

int main(void) 
{
   char word[20];
   int i;
   scanf("%s", &word);
   char names =  '@';
   printf("names = %c \n", names);

   for(i=0;i<20;i++)
   {
	   if (word [i] == names)
     {
		   printf("Joe, Dave, Larry");
       exit(0);
	   }        
    } 
       printf("wrong answer, you typed in %s\n", word);    
  
return 0;
  }

Originally Posted by itCbitC

With the exception of the item in red which shouldn't be there.

i am sorry i directly copied that code from original post.

Regards,
Shwetha

thanks for the answer guys, it helped a lot.

I have a somewhat related follow up question though.. if i'm trying to read from the command line, how would I be able to store the arguments as a string..

For example, say if i was doing ./a.out d03dfs, how would i be able to store "d03dfs" as a string?

i've tried

Code:

char input[50];
int i;
   for(i = 0; i < argc; i++)
    	printf("arg %d: %s\n", i, argv[i]);
input[0] = argv[1];
printf("you again typed in %s \n", input);

but it does not seem to work. Any ideas? Thanks again.

You need to use strcpy() for doing that.

tried doing that too, using

Code:

    int desti = 0;
    for(i = 0; i < argc; i++){
    strcpy(input[desti],argv[i]);
    desti++;
    }

so that the value of all the arguments would be put into input, but no luck

Originally Posted by Uss_Defiant

tried doing that too, using

Code:

    int desti = 0;
    for(i = 0; i < argc; i++){
    strcpy(input[desti],argv[i]);
    desti++;
    }

so that the value of all the arguments would be put into input, but no luck

First argument of strcpy() is a pointer; do you think input[desti] is a pointer??
Increment desti by the length of the string before starting the next iteration as incrementing by one overwrites what was there before.

Originally Posted by itCbitC

First argument of strcpy() is a pointer; do you think input[desti] is a pointer??
Increment desti by the length of the string before starting the next iteration as incrementing by one overwrites what was there before.

sorry that i'm being dense, but how would you create a pointer to input[desti]? I'm very new at this an examples would be helpfull.

thanks guys, i got it working using

Code:

     for(i = 1; i < argc; i++){
    	strcat(input,argv[i]);
    }

 	printf("argc is %d \n", argc);
    printf("you typed in %s \n", input);

my issue now is that certain things mess up the code

for example, "./a.exe 12 / 3" will yield "you typed in 12/3"
but
"a.exe12 * 3" will yield "you typed in 12a.exeasdfsrc3"

the argc value from the first is 4 (expected), but from the second is 6.

Anyone got any ideas?

Thanks again.

Ah i see, is there any way to code my program to ignore these expressions?