Question

I’m coding a shell currently and for some reason I can’t get my printenv function to work. When a command is not given, it works. When two arguments are given, it also works. However, when one argument is given it does not work and prints nothing.

The code is as follows:

else if (strcmp(args[0], "printenv")==0){
        /* Previously: if (args[1] == NULL && args[0] != NULL){ */
        if (argc == 1){
            int i = 0;
            while (envp[i] != NULL){
                printf("%s\n", envp[i++]);
            }
        }
        /* Previously: else if (args[2] == NULL && args[1] != NULL){ */
        else if (argc == 2){
            char *env;
            while (args[1] = *argv++){
                env = getenv(args[1]);
                if (env != NULL){
                    printf("%s", env);              
                }           
            }
            free(env);
        }
        else {
            fprintf(stderr, "%s: Too many arguments\n", args[0]);
        }   

    }

Answer 1

else if (strcmp(args[0], "printenv")==0){
        if (args[1] == NULL && args[0] != NULL){

I stopped reading here. The strcmp call having already determined that *args[0]=='p', why are you comparing args[0] to NULL?

---

Now, that that’s cleared-up. I think you don’t need this loop at all:

        while (args[1] = *argv++){
            env = getenv(args[1]);
            if (env != NULL){
                printf("%s", env);              
            }           
        }

Just the loop body:

            env = getenv(args[1]);
            if (env != NULL){
                printf("%s", env);              
            }

And as @Jason comments, you should not free (or otherwise modify) the pointer returned by getenv (nor the data to which it points).