Pointer arithmetic

pointer_arithmetic_example.c

// The MIT License (MIT)

// Copyright (c) 2015 Leonardo de Oliveira Lourenço

// Permission is hereby granted, free of charge, to any person obtaining a copy
// of this software and associated documentation files (the "Software"), to deal
// in the Software without restriction, including without limitation the rights
// to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
// copies of the Software, and to permit persons to whom the Software is
// furnished to do so, subject to the following conditions:

// The above copyright notice and this permission notice shall be included in all
// copies or substantial portions of the Software.

// THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
// IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
// FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
// AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
// LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
// OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
// SOFTWARE.

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

int main(void)
{
    int *elementos, *p;
    int i, total = 15;

    elementos = (int *) malloc(total * sizeof(int));

    if(elementos == NULL)
    {
        return -1;
    }

    for(i = 0; i < total; ++i)
    {
        elementos[i] = i;
    }

    printf("\nAcessando fazendo p+=1:\n");
    for(p = elementos; p < (elementos+total); p += 1)
    {
        printf("%d\n", *p);
    }

    printf("\nAcessando fazendo p+=sizeof(int):\n");
    for(p = elementos; p < (elementos+total); p += sizeof(int))
    {
        printf("%d\n", *p);
    }

    free(elementos);

    return 0;
}