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Data structure implementation with Linked lists.

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−3

Could someone explain this part of the coding for data structures & linked list? I actually got this code from a textbook but no matter how I read the textbook, I still don't get the concept & what it means. Thanks :")

#include<stdio.h>
#include<stdlib.h>
struct listNode{
    char data;
    struct listNode *nextPtr
};

typedef struct listNode ListNode;
typedef ListNode *ListNodePtr;

void insert(ListNodePtr *sPtr, char value);
char delete(ListNodePtr *sPtr, char value);
int isEmpty(ListNodePtr sPtr);
void printList(ListNodePtr currentPtr);
void instructions(void);

int main (void){
    //some coding
}`
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2 comment threads

Don't typedef pointers (7 comments)
Linked list fundamentals (2 comments)

1 answer

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−1
struct listNode{
    char data;
    struct listNode *nextPtr
};

Computer, when I tell you that any region of memory is a struct called listNode, that means that the region of memory contains a char, which I will read from and write to using the name data. The region of memory also contains a pointer that can point to another region of memory matching the same listNode pattern; I will call this pointer nextPtr. I'll let you, computer, figure out the details of how to concretely arrange these contents inside the region.

(What's that, computer? I'm missing a semicolon? Well, okay; I guess you understood what I meant anyway, but if it'll make you happier...)

typedef struct listNode ListNode;
typedef ListNode *ListNodePtr;

Computer, I'll use the word ListNode (note the capitalization!) as an abbreviation for ‘a struct called listNode’, and I'll use the word ListNodePtr as an abbreviation for ‘a pointer to a ListNode’.

void insert(ListNodePtr *sPtr, char value);

Having told you all that, computer, now I'm going to give you a preview of what's to come. I'm going to define a function called insert that accepts a pointer to a ListNodePtr (that's a pointer to a pointer, of course!) and a char. I'm not going to tell you what that function does yet, but a human might recognize these words and imagine that the function needs to replace the ListNodePtr to which the pointer points with another ListNodePtr that itself points to a ListNode containing the char, in addition to the rest of the original list. I'll tell you all that later though, computer!

char delete(ListNodePtr *sPtr, char value);

Similarly, here's another preview of another function that accepts a pointer to a pointer and a char; this one also returns a char. Again, a human might infer, from the fact that this function accepts a pointer to a pointer, that the function needs to replace the ListNodePtr with a new one somehow—but I'll tell you that later!

int isEmpty(ListNodePtr sPtr);
void printList(ListNodePtr currentPtr);
void instructions(void);

Here are some more previews, computer! That last one is a function that accepts nothing and returns nothing; how mysterious! I must be planning on doing some side effects in there, like printing some information to the screen.

int main (void){
    //some coding
}

Finally, here's what you need to actually do when this program is run. I, uh, haven't decided yet. For now, do nothing at all.

Oh yeah, those function previews I gave you? Well, since I haven't told you to use those functions, I guess I don't need to keep my promises to tell you what they do, at least for now! But if I change this program to use them, I'll have to make good on those promises, won't I?

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1 comment thread

https://software.codidact.com/posts/285169/285175#answer-285175 How about in this case of creating a ... (1 comment)

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