指针和指向 C 中的指针的指针

Question

Hey I am trying to understand pointers and I noticed something which causes undefined behavior in my program In first case when I decay pointer to q1 to NULL everything is fine but when I decay pointer *q1 to NULL the terminal does not show anything. What is going wrong?

#include<stdio.h>
#include<stdlib.h>
#include<string.h>



int main ()
{
    char *p[11]={"vasilis","loukas","vasilis","vasilis","giorgos","makis","vasilis","nikos","makis","nikos"};
    
    
    char **p1;
    char**d1;
    char**q1;
    q1=NULL;
    p1=&p[0];
    d1=&p[1];
    
    
    
    /******Count words*********/
    int count=0;
    
    for(p1=&p[0] ; *p1 ; p1++)
    {
        count++;
        }
    
    
    printf("\nthe number of words : %d" ,count);
    
    return 0;
}

second case:

#include<stdio.h>
#include<stdlib.h>
#include<string.h>



int main ()
{
    char *p[11]={"vasilis","loukas","vasilis","vasilis","giorgos","makis","vasilis","nikos","makis","nikos"};
    
    
    char **p1;
    char**d1;
    char**q1;
    *q1=NULL;
    p1=&p[0];
    d1=&p[1];
    
    
    
    /******Count words*********/
    int count=0;
    
    for(p1=&p[0] ; *p1 ; p1++)
    {
        count++;
        }
    
    
    printf("\nthe number of words : %d" ,count);
    
    return 0;
}

Answer 1

In your first code, the q1 = NULL; line is assigning a zero (null) value to the pointer, q1 , which is a perfectly valid operation (even though that pointer potentially points to another pointer). In fact, assigning a NULL value to a pointer is a common technique, used so that code can test whether or not that pointer currently contains a valid address or not (if it's NULL , then we can't use it).

In your second code, *q1 = NULL; , you are trying to assign zero to the object (itself a pointer) that is pointed-to by q1 ; however, q1 hasn't been assigned a value (an address), so it doesn't point to any valid target, and your attempt fails (because you are trying to modify a 'random' part of memory that you likely don't have access to).