不了解自定义C Shell的逻辑

Question

So I am trying to debug a shell programmed in C

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define MAX_NUM_ARGS 256
#define SIZE 256

//void orders(char *command[SIZE]);

int main() {

    char buffer[SIZE]= "";
    //char input_args[MAX_NUM_ARGS];
    char **input_args = NULL;
    int i = 0;// counting variable
    int j = 0;// second counting variable (thank you Nathan)
    int next_counter = 0;

    printf("Welcome to my shell.\n");
  while(1){

    // sees to it that the buffer is clean (thanks erik =) )
    memset(buffer, '\0', sizeof(buffer));
    i = 0;
    j = 0; //ensure that the counting variables are reset


    //initialize array of strings
    //first free any prevously allocated memory
    if (input_args != NULL)
    {   //memory has been allocated free it
        for (i = 0; i <MAX_NUM_ARGS; i++)
        {
            free(input_args[i]);
        }
    }   
    //free array of strings
    free(input_args);

    //new allocate memory
    input_args = (char**) malloc(sizeof(char*) * MAX_NUM_ARGS);
    //check return value for error
    if (input_args == NULL)
    {
        printf("We are out of memory. =( Can't run. Sorry!\n");

        return -1; //Thank you Erik for this idea!
    }
    //allocate memory for each string
    for (i = 0; i <MAX_NUM_ARGS; i++)
    { 
        input_args[i]= (char*)malloc(sizeof(char) * MAX_NUM_ARGS);
        if(input_args[i] == NULL)
            {//error
            printf("Error, the input is empty.");
            return -1;
            }//end of if statement
    }//end of for loop


    printf("~$: "); //prompts the user for input
    fgets(buffer, sizeof(buffer), stdin);
    //if the user types in exit, quit
    if (strcmp(buffer, "exit\n") == 0){ 
        exit(0);
    } //end of if statement
    //if user types in clear, wipe the screen and repeat the loop
    else if(strcmp(buffer, "clear\n")==0){

        system("clear");    
        continue;   

    }//end of else if
    //should the user punch in nothing, repeat the loop
    else if (strcmp(buffer, "\n") == 0) {
        continue;
    }//end of else if




    for (i = 0; i < SIZE; i++) {

        if(buffer[i] != '\n' && buffer[i] != ' ' && buffer[i] != '\t'){

            input_args[j][i] = buffer[i];
        }   //end of if statement
        else{
            input_args[j][i] = '\0';
            j++;

        }//end of else statment

    }//end of for loop

    input_args[1] = NULL;

    //block down here handles the command arugments
    int retval = 0; //return value
    int pid = 0;
    int childValue = 0;
    pid = fork();

    if (pid != 0){
    //  printf("I'm the parent, waiting on the child.\n");//debug
        pid = waitpid(-1, &childValue,0);
    //  printf("Child %d returned a value of %x in hex.\n", pid, childValue);

    }//end of if statement
    else{
    //  printf("I am the first child.\n");
        retval = execvp(input_args[0], input_args);
        //exit(2);
        if (retval != -1){
            //print error!
            printf("Invalid command!\n");
            exit(2);
        }
    }//end of else block

   } //end of while loop
    return 0;

}//end of main function

Now, I can make this shell execute one word commands like 'ls', or 'pwd', or go into vi and open up a new file. But multi-word arguments don't seem to be panning out.

I'm having trouble with the basic logic of the code. I mean, looking at the block of code at the bottom, it seems like it has been coded to take in both arguments, but right now, only the first one is getting parsed. What logic error am I making exactly? I'm interested in learning about this.

Answer 1

Maybe that's the only problem: after the loop that parses the arguments you have

input_args[1] = NULL;

so whatever you have done before, now you don't have any more arguments ( input_args[0] is the progam name). It surely should be

input_args[j] = NULL;

Edit

that could make your shell work but you would still have a memory leak as you first allocate memory for all input_args[i] (0 <= i < MAX_NUM_ARGS). So when you set input_args[j] = NULL that memory will never be freed again. A not very elegant, but working solution would be to call

free( input_args[j] );

before, but I would suggest only to allocate memory for the arguments that are actually needed.