What is the best way to parse a string into a null terminated array?

Question

I am writing a simple command line interpreter. My code reads a string using scanf and parses it using the function getArgs() shown below, and then uses that array as an argument to execvp to perform a command such as ls. It works if I call only 'ls' but when I call 'ls -la', it gives the same result as 'ls'.

void getArgs(char* command, char* args[]){

    int i = 0;
    char* p = strtok(command, " "); 
    args[i] = p;

    while(p != NULL){
        i++; 
        p = strtok(NULL, " ");
        args[i] = p;
}
}

Here is my main function which includes the initialization of the arguments given:

int main(){
char *args[1024];
char example[30]; 
char exit[5] = {'q', 'u', 'i', 't', '\0'};
int f1; 
int status;
size_t n = sizeof(args)/sizeof(args[0]);

while(strncmp(example, exit, 30) !=0){

    printf(">>>");
    scanf("%s", example);
    getArgs(example, args);
    int x = strncmp(args[0], exit, 30);

    if (x != 0){
        f1 = fork(); 
        if (f1 != 0){
            /* wait  for child process to terminate */
            waitpid(f1, &status, 0);
        }
        else{myExec(args);}}         
    else{
        return 0;}}
    return 0; 
}

My guess as to the problem is that my argument array, args, is not null terminated and so when I attempt to use it in myExec():

void myExec(char* args[]){
    execvp(args[0], args);
}

this does not work. So my question is, can I set the item after the last non-empty part of my array to null to try to get this to work? If so, how can I do that? Is there a better way to solve this?

Answer 1

The -la is being ignored because scanf("%s", example); will stop at the first space. I suggest

scanf(" %29[^\n]", example);

Which will

• Ignore whitespace left in the buffer from the previous command.
• Restrict the string input from overflowing.
• Allow space seperators in the command.

Note too that in the first execution of while(strncmp(example, exit, 30) !=0) the example is an uninitialised variable, so that needs to be

char example[30] = "";

Answer 2

The %s directive stops scanning at the first whitespace character, so it won't properly capture any commands with spaces (such as ls -la ). You should use fgets to get user input if you want to preserve any whitespace:

if ( fgets( example, sizeof example, stdin ) )
{
  getArgs( example, args);
  ...
}

fgets will read up to sizeof example - 1 characters into example (including the newline!) and 0-terminate the string. You may want to take that newline into account with your strtok call.