我正在尝试读取和打印链表中的一个值，但我的程序没有给出任何输出

Question

i am trying to read and print one value in a linked list , but my program does not give any output, i have tryed checking where the program is failing to execute , after the first scanf the code is not printing anything, what might be the reason for that? code is as followed:

#include<stdlib.h>
#include<stdio.h>
void display();

struct ll{
    int val;
    struct ll* address;
};
struct ll *new=NULL,*start=NULL,*present=NULL;


int main(void)
{
    int num;
    scanf("%d",&num);
    //reading ll
    new=(struct ll*) malloc(sizeof(struct ll));
    new->val=num;
    new->address=NULL;
    
    if(start==NULL)
    {
        start=new;
        present=new;
       
    }
    else
    {
        present->address=new;
        present=new;
        
    }
   //calling display func to display the contents of ll
    display();
    
   
}
void display()
{
    present=start;
  // displaying.
    while (present!=NULL)
    {
        printf("%d",present->val);
        present=present->address;
    }
    printf("%d",present->val);
}

Answer 1

I have encoded my comments interspersed into your code. I have commented the last statement in function display() , to make it run properly. I have also commented the cast to malloc() (for the given reasons in the code) I have also made some aesthetic changes to make the code more readable. You can add spaces to improve readability of the code, as they don't change the compiler produced code, so please, use enough spaces to make your code more readable (I've done this also to show who it is more readable now):

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

void display(void);

struct ll {
    int        val;
    struct ll *address;
};

struct ll *new     = NULL,
          *start   = NULL,
          *present = NULL;


int main(void)
{
    int num;

    scanf("%d", &num);
    //reading ll

    /* Never cast the returned value of malloc()  This allows to
     * detect if you have properly #include'd the header file and
     * avoids other dificult to find errors.  malloc() returns a
     * void * pointer, so it will be automatically converted to
     * any other pointer type without risk. */
    new          = /* (struct ll*) */ malloc(sizeof(struct ll));
    new->val     = num;
    new->address = NULL;

    if(start == NULL)
    {
        start   = new;
        present = new;

    }
    else
    {
        /* this is never executed, as start == NULL at program
         * start. */
        present->address = new;
        present          = new;

    }
    //calling display func to display the contents of ll
    display();
    /* while it is not necessary for main() it is normal for a function 
     * that is defined to return an int value, to return something, so
     * I added the following statement: */
    return 0; 
}

void display(void)
{
    present = start;
  // displaying.
    while (present != NULL)
    {
        printf("%d",present->val);
        present = present->address;
    }
    /* as you have moved present in a while loop until the while
     * condition is false, at this point you must assume the
     * condition is false (so present == NULL) and you are trying to
     * dereference a NULL pointer below */
    /* printf("%d", present->val); */
}

Now your program will run and show the only value (I recommend you to put a \n character at the end of the printf() call, to put the printed data in a line by itself.