Strtok()

[wal99d]

The following code does not work correctly:

	char * cmd="ls -l"; /* Unix command used for listing */
	char * spCmd;	 
	spCmd = strtok(cmd , " ");/* it divides the cmd in to tokens */
	execvp(spCmd[0] , spCmd);

output of this code to execute command that is written in cmd ?
:?:

[niallj]

Firstly, you're using the wrong function. The exec() family replace the current process with the one in the arguments, so it would replace your program with ls -l.

Secondly, by passing the entire spCmd array as an argument to execvp(), you are actually telling it to run the command: ls ls -l. Just pass spCmd[1] as your second argument in execvp().

[wal99d]

Do you have any good code to show me this output DUDE. :?:

[niallj]

I'm unclear about what you want to do. Do you want to write a program that will print the output of ls -l to the screen, or one that will give your program the output of ls -l?

[Nuticulus]

I haven't used execvp() before, but after a little looking around, it seems that you supply several arguments to it in a list which will result as, when executed, argv[0], argv[1] etc. So, in other words, you are supposed to split up your string by the token " " into an array, then pass the first item to the first argument of execvp(), then the rest to the second. This array must be terminated with a null pointer.

char cmd[] = "/bin/ls /home/nuticulus/ -l";
char **toks;
int c=0;
	
printf("Running %s\n",cmd);
toks = (char**) malloc(sizeof(char*));
toks[0] = (char*)strtok(cmd," ");
	
while(toks[c]!=NULL)
{
	printf("#%d %s\n",c+1,toks[c]);
	c++;
	toks = (char**) realloc(toks, (c+1)*sizeof(char*));
	toks[c] = (char*)strtok(NULL, " ");
}
	
execvp(toks[0], toks);

That would split the string cmd into an array of character pointers, delimited by a final NULL pointer.

[wal99d]

	char *all;
	char *res[80];
	int i=0;
	
	all = "ls -l";
	res[i]= strtok(all," ");
		
	while (res[i] != NULL){
		res[++i]= strtok(NULL," ");
	}
	execvp(res[0],&res[0]);
 --->	   printf("Done\n");          <-- this is not executed why ??

thankx gentelmen , but now my question is that why the last line cannot be
executed ??
:eek:

[niallj]

As I said in my first post, execvp() stops the current process, and replaces its code space in memory with that of the new process. In other words, your app dies and is replaced by ls -l.

[wal99d]

Thankx niallj