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?

I added the rest of the code, also args[i] = strdup(p); gives 'Segmentation error: 11'. — Ars
– Ars, Commented Mar 29, 2019 at 19:46

Weather Vane · Accepted Answer · 2019-03-29 20:05:05Z

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] = "";

answered Mar 29, 2019 at 20:05

Weather Vane

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John Bode · Accepted Answer · 2019-03-29 20:44:40Z

1

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.

edited Mar 29, 2019 at 20:44

answered Mar 29, 2019 at 20:05

John Bode

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