如何获取一个链表和一个函数指针作为输入

Question

I am new to C and trying to learn function pointer.I am supposed to complete the 'map_list'function which takes a linked list and a function pointer,and return a new list in the same order, but with all the values squared.Please point it out where I did wrong.

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


struct Link {
    struct Link *next;
    int value;
};

void print_list(struct Link *list) {
    for(struct Link *l = list; l != NULL; l = l->next) {
        printf("%d", l->value);

        if(l->next) {
            printf(", ");
        }
    }

    printf("\n");
}


struct Link *append(int x, struct Link *head) {
    struct Link *head_ = (struct Link*)malloc(sizeof(struct Link));
    head_->next = head;
    head_->value = x;

    return head_;
}

struct Link *reverse_list(struct Link *list) {
    struct Link *head = NULL;

    for(struct Link *l = list; l != NULL;) {
        struct Link *next = l->next;
        l->next = head;
        head = l;

        l = next;
    }

    return head;
}

struct Link *map_list(struct Link *link_list,int (*Square)(int)   ) {

    struct Link *new_list = NULL;
    new_list = new_list ->next;
    new_list ->value = (*Square)(link_list ->value);
    return new_list;
}

int square(int x) {
    return x * x;
}

int add3(int x) {
    return x + 3;
}



struct Link *theList() {
    struct Link *l = append(1, NULL);
    l = append(2, l);
    l = append(3, l);
    l = append(5, l);
    return l;
}

int main() {


    struct Link *l = theList();
    print_list(map_list(l, &square));
    ;
    return 0;
}

I got 'Segmentation fault (core dumped)'

Answer 1

If I have understood correctly you have some trouble with writing the function map_list . It can look the following way

struct Link * map_list( const struct Link *link_list, int operation( int )   ) 
{
    struct Link *new_list  = NULL;
    struct Link **new_node = &new_list;

    for ( const struct Link *current = link_list; current != NULL; current = current->next )
    {
        *new_node = malloc( sizeof( struct Link ) );

        ( *new_node )->next  = NULL;
        ( *new_node )->value = operation( current->value );

        new_node = &( *new_node )->next;
    } 

    return new_list;
}

And the function can be called for example like

map_list( l, square );

or

map_list( l, add3 );

The function does not check whether a memory allocation for a node was successfull. You can add such a check yourself.

As for your own function implementation

struct Link *map_list(struct Link *link_list,int (*Square)(int)   ) {

    struct Link *new_list = NULL;
    new_list = new_list ->next;
    new_list ->value = (*Square)(link_list ->value);
    return new_list;
}

then for starters it has undefined behavior

ew_list = new_list ->next;

and does not make sense relative to the assignment.