execl() + redirecting output to text files

JamesGoh · #1 (**permalink**) 02-13-2008

Im currently using execl() to run the ls command and redirect the output to a text file. Unfortunately, when I check the text file, no information has been written to it.

I am calling execl() with the ls command like this

Code:

execl( "/bin/ls" , "-al" , '>' , "dirlist.txt" ,(char *) 0 );

What are some possible causes of my above problem ? cheers

HPAVC · #2 (**permalink**) 02-13-2008

Is that the right syntax for execl?

Quote:

Originally Posted by man execl

execl(<shell path>, arg0, file, arg1, ..., (char *)0);

where <shell path> is an unspecified pathname for the sh utility, file is the process image file, and for execvp(), where arg0, arg1, and so on correspond to the values passed to execvp() in argv[0], argv[1], and so on.

The arguments represented by arg0,... are pointers to null-terminated character strings. These strings shall constitute the argument list available to the new process image. The list is terminated by a null pointer. The argument arg0 should point to a filename that is associated with the process being started by one of the exec functions.

The argument argv is an array of character pointers to null-terminated strings. The application shall ensure that the last member of this array is a null pointer. These strings shall constitute the argument list available to the new process image. The value in argv[0] should point to a filename that is associated with the process being started by one of the exec functions.

The argument envp is an array of character pointers to null-terminated strings. These strings shall constitute the environment for the new process image. The envp array is terminated by a null pointer.

For those forms not containing an envp pointer ( execl(), execv(), execlp(), and execvp()), the environment for the new process image shall be taken from the external variable environ in the calling process.

The number of bytes available for the new process' combined argument and environment lists is {ARG_MAX}. It is implementation-defined whether null terminators, pointers, and/or any alignment bytes are included in this total.

pflynn · #3 (**permalink**) 02-16-2008

Quote:

Originally Posted by JamesGoh

What are some possible causes of my above problem ? cheers

The problem is: you are giving ls something that actually is the shell's responsability, and that thing is output redirection. Since there is no shell involved in your program, you should perform the redirection of the output by yourself, that is, your program should do the shells's work, first by openning a new file and them redirecting its standard output and standard error to the new file descriptor using the dup2(2) function.

PHP Code:


			
#include <unistd.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>


int main(int argc, char *argv)
{
int fd; /*file descriptor to the file we will redirect ls's output*/

if((fd = open("dirlist.txt", O_RDWR | O_CREAT))==-1){ /*open the file */
  perror("open");
  return 1;
}

dup2(fd,STDOUT_FILENO); /*copy the file descriptor fd into standard output*/
dup2(fd,STDERR_FILENO); /* same, for the standard error */
close(fd); /* close the file descriptor as we don't need it more  */

/*execl ls */
execl( "/bin/ls" , "ls" , "-la" , (char *) 0 );

return 0;
}

Hope you got the point

JamesGoh · #4 (**permalink**) 02-17-2008

Hi guys

I ended up using system() instead and it resolved my probs (although I had to use a lot of strcmp() calls to create the command

) of redirecting shell output to a file. Another reason why I chose to reject execl() in favour of system() was because I didn't really understand the manual description of it too well (apart from its purpose of course)

pflynn, thanks for your suggestion. Just a question (which may be a bit stupid so bear with me). The standard output is the place where all output of any c program is directed to (i.e. the place where all printf() calls have the text displayed ) ?

pflynn · #5 (**permalink**) 02-17-2008

Quote:

Originally Posted by JamesGoh

The standard output is the place where all output of any c program is directed to (i.e. the place where all printf() calls have the text displayed ) ?

Yes, you are right!

Quote:

Standard output (stdout)

Standard output is the stream where a program writes its output data. The program requests data transfer with the write operation. Not all programs generate output. For example the file rename command (variously called mv, move, ren) is silent on success.

Unless redirected, standard output is the text terminal which initiated the program.

The file descriptor for standard output is 1 (one); the corresponding <stdio.h> variable is FILE* stdout; similarly, the <iostream> variable is std::cout.

Pay attention to this part:

Quote:

Unless redirected, standard output is the text terminal which initiated the program.

which is exactly what you want to do: you want to open a file so the ls program would write its output to this file instead of writing it to the terminal that's originally associated to the standard output.

More information on the standard streams:

Standard streams
Standard streams - Wikipedia, the free encyclopedia

JamesGoh · #6 (**permalink**) 02-18-2008

thanks heaps pflynn !

More UNIX and Linux Forum Topics You Might Find Helpful
Thread	Thread Starter	Forum	Replies	Last Post
Redirecting OUTPUT	Chanakya.m	Shell Programming and Scripting	1	08-13-2007 04:27 PM
redirecting the output of aspell	leekb	UNIX for Advanced & Expert Users	6	07-10-2006 05:50 AM
Redirecting the startup output	blakmk	UNIX for Dummies Questions & Answers	3	10-28-2003 07:22 AM
Redirecting output to TCP port	mscomms	UNIX for Dummies Questions & Answers	1	09-30-2003 10:51 PM
Redirecting output to multiple log files?	darthur	UNIX for Dummies Questions & Answers	3	01-15-2002 12:54 PM

Thread Tools	Search this Thread
Show Printable Version Email this Page	Search this Thread: Advanced Search
Display Modes	Rate This Thread
Switch to Linear Mode Hybrid Mode Switch to Threaded Mode	Rate This Thread:


	Powered by