C程序不执行主程序外部的功能

Question

If I execute the exec() function in another C program as a main function it works perfectly, while if I put it as a function called in the main menu it gives me some warning and the function does not run.

My code:

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h> /* for fork */
#include <sys/types.h> /* for pid_t */
#include <sys/wait.h> /* for wait */

int exec (void) {

        char array[100];
        char character;
        int i = 0;
        char* point;
        int j = 0;

        printf ("Digita una stringa");
        printf ("\n");
        do {
            character = getchar();
            array[i] = character;
            i++;
        }
        while (character != '\n');
        array[i-1] = '\0';
        i = 0;

        char* string[100];

        char *word = strtok(array, " .");
        j = 0;
        while (word != NULL) {
            printf("%s\n", word);
            string[j++] = word; // Increment j
            word = strtok(NULL, " .");
        }
        string[j] = NULL; // Make sure the array is NULL term

        printf ("\n");  

    pid_t  pid;
    pid = fork();
    int status;

    if (pid == -1) {
        perror("");
    }else if (pid == 0) {
        execvp(string[0], string);     /* execute the command  */
                fprintf(stderr, "Failed to exec");  
                exit(1);
            }
    else {

        //.. wait until the child ends
        waitpid(-1, &status, 0);
      }

return;
}

int read_input (void) {
    int choice;
    printf("Seleziona una voce dal menu");
    do {    
        printf("\n");
        scanf ("%i", &choice);
        if (choice > 8 || choice < 1)
            printf ("Attenzione, inserisci un valore contemplato dal menu");
    }
    while ( choice > 8 || choice < 1);

return choice;
}

void main (int argc, char *argv[]) {

    printf ("------------------------\n");
    printf ("          MENU         \n");
    printf ("------------------------\n");
    printf ("  \n");
    printf ("1) Esecuzione in foreground di un processo\n");
    printf ("2) Ctrl -C\n");
    printf ("3) Exit\n");
    printf ("4) Background\n");
    printf ("5) Pipe\n");
    printf ("6) Jobs\n");
    printf ("7) fg\n");
    printf ("8) kill\n");
    int menu = read_input();
    switch (menu) {

        case '1' :  
                exec ();
                break; 
        case '2' :
                //ctrl();
                break;
        case '3' :
                //exit_();
                break; 
        case '4' : 
                //background();
                break;
        case '5' :
                //pipe();
                break;
        case '6' : 
                //jobs();
                break;
        case '7' : 
                //fg();
                break;
        case '8' : 
                //kill();
                break;
    }
}

this is the warning:

elaborato.c:31:16: warning: initialization makes pointer from integer without a cast [enabled by default] char *word = strtok(array, " ."); 

Answer 1

Regarding the problem related to input,

do {    
    printf("\n");
    scanf ("%i", &choice);
    if (choice > 8 || choice < 1)
        printf ("Attenzione, inserisci un valore contemplato dal menu");
}
while ( choice > 8 || choice < 1);

Once you type an integer and press enter, the scanf() consumes the number and a newline is left in stdin. Next time the loop goes around (assuming input <1 or >8 or something else) scanf gets that newline and it goes on. add a getchar() after the scanf() .

Answer 2

The answer is in the warnings, you should move them from the comment into the question.

elaborato.c:31:16: warning: initialization makes pointer from integer without a cast [enabled by default] char *word = strtok(array, " ."); 

That means that word , which is a char pointer is being initialized from an integer. Therefore, it seems strtok() is returning an integer... that doesn't sound right.

From the strtok() man page:

#include <string.h> 

char *strtok(char *str, const char *delim);

char *strtok_r(char *str, const char *delim, char **saveptr);

That seems right, it returns a char * .... but it also says it's declared in <string.h> ... which you aren't including. Since it's not defined, the compiler assumes it as int strtok() .

Fix: add the #include <string.h> line.

Answer 3

The problem you are seeing is that of scanf getting skipped.

For more details you can refer here

The call to scanf() after the printf consumes the new-line character and continues without the user having to enter anything.

It reads the very next character from standard in, which is probably a newline character and thus not prompting you for any input.

Answer 4

I'm too tired to comment on the entire code (it is really bad), so I'll just write about the issue in question and how trivially it could be debugged.

First thing to do is to check if we get the number and return to main, hence:

int menu = read_input();
printf("got [%d]\n", menu);

Running this:

[snip]
1
got [1]

So we indeed get to this point.

As such now we check what is this compared to.

int menu = read_input();
printf("got [%d] '1'=[%d]\n", menu, '1');

Let's run:

1
got [1] '1'=[49]

So, menu stores an integer 1, while '1' is a character code for 1 in ascii, which, well, is not an integer 1.

In general I don't see what was the problem with narrowing it down.