dynamic array

Sun, 22 Nov 2009 03:14:16 -0800

I don't know what's with the index part. Please help. Thanks!

#include<stdio.h>  main(){ 	int* a;	int number;	int choice;	int index = 0;	int head;  	while(choice!=4){ 		printf("\tMENU\t\n");		printf("[1] add number to array\n");		printf("[2] view array\n");		printf("[3] search using linear search\n");		printf("[4] quit program\n");		scanf("%d", &choice); 		switch(choice){			case 1: head = add(&index, a);					break; 			case 2:head = view(&index,a);					break; 			//case 3:linsearch(&index,array[5]);					//break; 			case 4: return;					break; 			default: printf("invalid input!!!\n"); 		}}}  int add(int* index, int* a){ 	int number;	int i; 		printf("enter number you want to enter in the array.\n");		scanf("%d", &number); 		*a = (int) malloc (5*sizeof(int)); 		for(i = 0; i <= *index; i ++){			a[i] = number;		} 		return *index;}  view(int* index, int* a){ 	int i; 	printf("\n"); 		for(i = 0; i <= *index; i++){			printf("index: %d, %d ", *index, a[i]);		}		printf("\n\n"); 		return a[5];}#include<stdio.h>


main(){

	int* a;
	int number;
	int choice;
	int index = 0;
	int head;
	
	
	while(choice!=4){
	
		printf("\tMENU\t\n");
		printf("[1] add number to array\n");
		printf("[2] view array\n");
		printf("[3] search using linear search\n");
		printf("[4] quit program\n");
		scanf("%d", &choice);
		
		switch(choice){
			case 1: head = add(&index, a);
					break;
			
			case 2:head = view(&index,a);
					break;
					
			//case 3:linsearch(&index,array[5]);
					//break;
					
			case 4: return;
					break;
					
			default: printf("invalid input!!!\n");
			
		}
}
}

 int add(int* index, int* a){
	
	int number;
	int i;
	
		printf("enter number you want to enter in the array.\n");
		scanf("%d", &number);
		
		*a = (int) malloc (5*sizeof(int));
		
		for(i = 0; i <= *index; i ++){
			a[i] = number;
		}
		
		return *index;
}

 view(int* index, int* a){

	int i; 
	printf("\n");
	
		for(i = 0; i <= *index; i++){
			printf("index: %d, %d ", *index, a[i]);
		}
		printf("\n\n");
		
		return a[5];
}

Re: dynamic array

Sun, 22 Nov 2009 06:50:26 -0800

This code will not even compile! What exactly are you trying to do here? Way too many problems in code. What is "return a[5];" in a function which does not declare a return value type? And what are these mallocs() inside the add()?

If you need an array with exactly 5 numbers - simply declare it: "int a[5];" and only after that begin working with menus.

If you want to return allocated pointer from a function - it is done in two ways:

1. Simply return it as a return code:

int* NewArray (int size);

int* a = NewArray (5);

int* NewArray (int size)
{
    return (int*) malloc (size * sizeof (int));
}

2. Return it using pointer to a pointer:

int NewArray (int size, int** parray);

int* a = NULL;
int ok = NewArray (5, &a);

int NewArray (int size, int** parray)
{
    *parray = (int*) malloc (size * sizeof (int));
    return (*parray != NULL);
}

'dynamic array' Thread RSS Feed

dynamic array

Re: dynamic array