Thread: is this code a list?

it simply allocates n bytes of the memory but, does this means that these bytes are connected each other and finally make a list? thanks...

Code:

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

typedef struct Person{
	int age;
	struct Person *next;
}ListPerson;

int main(){
	int n;
	ListPerson *A;
		printf("Type the size of the array:");
		scanf("%d", &n);
		A = malloc(n*sizeof(struct Person));
		if(A == NULL) exit(1);
		
return 0;
}

This struct has a pointer to another struct.
Yes, with this you can have a list. Is up to you to fill it correctly.

Probablly this code is correct right? Can any tell me a way to check if it is a list ?

Code:

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

typedef struct Person{
	int age;
	struct Person *next;
}ListPerson;

int main(){
	int n;
	ListPerson *A;
		printf("Type the size of the array:");
		scanf("%d", &n);
		A = malloc(n*sizeof(struct Person));
		if(A == NULL) exit(1);

		for(i = 0; i < (n-1); i++){
			A[i].next = &A[i+1]
		}
return 0;
}

Code:

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

typedef struct Person{
	int age;
	struct Person *next;
}ListPerson;

int main(){
	int n;
	ListPerson *A;
		printf("Type the size of the array:");
		scanf("%d", &n);
		A = malloc(n*sizeof(struct Person));
		if(A == NULL) exit(1);

		for(i = 0; i < (n-1); i++){
			A[i].next = &A[i+1]
		}
                A[n-1].next=NULL; // end of the list
return 0;
}

Originally Posted by brack

Probablly this code is correct right? Can any tell me a way to check if it is a list ?

It is not a list, it's an array of struct person. Assigning the address of the next element is not necessary here since you can access the elements directly.

can anyone tell me for sure what the above code is ? it is a list or not?