pointers for getting elements of an array in C [duplicate]

First, Please don't update code to fix bugs while a question is open. It makes most of the current answers meaningless, doesn't grant credit where it is due to the person(s) that solved one or more issues in your prior code, and makes the casual reader looking for a related problem to their own completely confused by both the question and the answers therein. If you want to amend an update do so in addition to the original problem, but if it an entirely different issue, then mark as answered, give credit where it is due, and open a new question with your new code and different problem(s) (ideally, anyway).

As written now, your function is fine. But your real issue is this:

// compile with -Wall -Werror and look at the warning here
int v={10,12,20,34,45}; // <== WRONG
getelements(&v,5);      // <== Harmless, but bad form.

This should be like this instead, assuming you want to print all elements in the array:

int v[] = {10,12,20,34,45};
getelements(v, sizeof(v)/sizeof(v[0]));

Note the [] following your array. Without it, the &v was masking what would have been a big-fat warning or error from the compiler that int is being passed as an int *. Furthermore, if you compile your prior code with full warnings treated as errors (-Wall -Werror for gcc) you will get an error like the following on your v declaration line:

main.c:116:15: Excess elements in scalar initializer

In other words, everything past the first element was ignored, and thus your pointer was running off into undefined behavior land. Changing the declaration and invocation to what I have above will address this as well as ensure you don't make that mistake again, since sizeof(v[0]) won't even compile unless v is an array or pointer type. The latter can still cause headaches when you use a pointer rather than an array with such a calculation, but thats something you just have to discipline yourself against doing in C.

try this and let me know if that works.

void getelements(int *a)
{
    int *p;
    int l=5;
    for (p=a;p<a+l;p++)
    {
         printf("%d\n",*p);
    }
}

It's best to pass in the length of the array along with the array itself.

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

void get_elements(int* values, int length)
{
    int i;
    for (i = 0; i < length; ++i)
    {
        printf("%d\n", values[i]);
    }
}

int main(int argc, char** argv)
{
    int vals[3];

    vals[0] = 0;
    vals[1] = 1;
    vals[2] = 2;

    get_elements(vals, 3);

    getchar();
    return 0;
}

Using the code similar to your original post (before the addition of the array length as a method parameter), you could do the follow (which is a bit convoluted if you ask me).

void get_elements(int* values, int length)
{
    int *p;

    for (p = &values[0]; p < &values[length]; p++)
    {
        printf("%d\n", *p);
    }
}

Actually you are passing array address in

getelements(&v,5) 

in function

getelements() 

you are treating it like an array!!!

void getelements(int *a, int cl)
{
    int *p;
    for (p=a;p<a+cl;p++)
    {
         printf("%d\n",*p);
    }
}

Let me know if you are cleared conceptually!!!

As per my knowledge and seeing your code.you have hardcoded the length of array to 5. I think you can also pass array length as a parameter to function; or you can use this function and see if it gives the desired result

void getelements(int *a)
{
    int *p;
    int i = 0;
    int l=5;
    p = a
    for (i = 0;i<l;i++)
    {
         printf("%d\n",*(p + i));
    }
}