Executing command line options

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# 1  
Old 06-26-2002
Executing command line options

Can someone please tell me how to modify/add to this code so that it recognizes UNIX command options (all beginning with "-") and executes the command with options?


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

     int i;

    system("stty -echo");

    for(i=1; i<argc; i++)
        printf("The command %s is executed.\n", argv[i]);

    system("stty echo");
    return 0;

# 2  
Old 07-01-2002
Re: Executing command line options

Hi Safia,

I think you will have to prepare a string of the unix command first and then pass this string to the system function. This string manipulation should take care of the "-" sign. I think you can also try ftok.

Hope this hints you to the solution.
# 3  
Old 07-02-2002
There's a function called getopt(), check to see if this function is supported in your compiler, this will help you.
# 4  
Old 07-03-2002
try this one

Can you try this one ??
#include "stdio.h"
#include "stdlib.h"
#include "string.h"

void main(int argc, char **argv)
char str[30];
if (argv[2][0] == '-')
strcat(str," ");
system (str);

}//end of main


void main(int argc, char **argv)

for second case, you need to give input as say "ls -l" within double quotes.
You can have any combination for this.
Like let us assume "a.out" is the binary then

./a.out "who | wc -l" also works for the second option.,....

# 5  
Old 07-03-2002
>void main(int argc, char **argv)
void main() is wrong. main() should return a value.
# 6  
Old 07-03-2002
void can be there

If main is not intended to return anything.......we must have its return type "void".

You can compile the source code using ansi compliance,

# 7  
Old 07-04-2002
ANSI C says 'int main()'. Ofcourse, to avoid compiler warnings you can use 'void main()', but that is incorrect. The OS might be wanting a return value from the terminated program, and when you are not returning any value to it, a junk value might be returned to your OS, and the results could be unpredictable.

See this
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