C-通过函数调用丢失双指针

Question

Quick C question here. We've been playing with double, triple, even quadruple pointers recently. We though we had a grasp on things until we ran into this problem...

char ***data;
data_generator(&data);
char **temp = data[0];          
printf("printing temp[%d]: %s\n",0, temp[0]);
printf("printing temp[%d]: %s\n",1, temp[1]);
dosomething(temp);

int dosomething(char **array) { 

    printf("printing array[%d]: %s\n",0, array[0]);
    printf("printing array[%d]: %s\n",1, array[1]);
    ......
}

int data_generator(char ****char_data) {
    char *command1[2];
    char *command2[2];

    command1[0] = "right";
    command1[1] = "left";

    command2[0] = "up";
    command2[1] = "down";

    char **commandArray[2];

    commandArray[0] = command1;
    commandArray[1] = command2;

    number_of_commands = 2;

    if(number_of_commands > 1){
    *char_data = commandArray;
    }

    return number_of_commands - 1;
}

And this prints out...

printing temp[0]: right
printing temp[1]: left
Segmentation fault

Looks like I have some misconceptions about what happens to a pointer while passed through a function. Any thoughts?

Answer 1

*char_data = commandArray;

You are putting the address of a stack (automatic) array in an outside memory location. This is a recipe for disaster (undefined behavior), since commandArray 's lifetime ends as soon as data_generator returns. The same is true for the elements of commandArray , which are themselves pointers to stack array elements.

Answer 2

Change:

char *command1[2];
char *command2[2];

to:

static char *command1[2];
static char *command2[2];

That will keep command1[] and command2[] in retained memory.

That, or malloc() them, as the other poster recommends, though proper use of malloc'c memory requires more considerations than I will discuss here.