Thread: use pointers instead of arrays

Do you understand how an array is mapped out in memory? Do you know what the []'s are actually doing? When you pass an array to a function why don't you have to specify the size in the function prototype?

How can you mimick this behavior with pointers?

Hint: You're actually doing the opposite here with a pointer when you write a function this way. In many cases pointer notation and array notation are interchangable as are pointers to first elements of arrays and arrays themselves. This is not always true, but is for a good number of cases.

An array is allocated in contiguous memory. That means if you have an array of 5 elements, element 1 will be right before element 2.

When you have an array and you use the []'s to select an element, you're really derefencing the address of the (address of array's first element + (the index * size of one element)).

You can do that manually with pointers.

i read over some of my c programming book and this seems to be correct, but it doesnt work what is wrong with it?

Code:

int maxarray(int s, int *d){
int lo = 0;
int i = 0;
for(; i< s;i++)
{
if(*(d+i)>*(d+lo))
{lo = i;}
}
return (*(d+lo));
}

also i posted this before you made your message, but it appears as if what you said supports the change

Indeed it does support what you did, but you missed something important. Note the formula I gave you.

On your machine, an int is probably 32-bits, which means it spans 4 bytes. Remember that memory is also most likely mapped out in bytes. This means that if you have an array of 5 elements at address 1000, this is what it might look like:

Address 1000 : array[0]
Address 1004 : array[1]
Address 1008 : array[2]
Address 1012 : array[3]
Address 1016 : array[4]

The forumula to convert array[i] is (address of array's first element + (the index * size of one element)).

This can be realized by this: (array + (i*4)).

So why the multiplication by 4? Because each int is 4 bytes. Now this might not be true for all archutectures. So you shouldn't assume it. This is why we have the sizeof operator. It'll let you get the size of a variable in bytes.

This means instead of 4, you can put sizeof(int) and it will return 4. To be even more portable and allow your code to work for any data type, you can write sizeof(*array). This will take the size of whatever type the array contains.

Hope this makes sense.

Works for me

Code:

#include <stdio.h>
#include <string.h>

int maxarray(int s, int d[])
{
    int lo = 0;
    int i = 0;
    for (; i < s; i++) {
        if (d[i] > d[lo]) {
            lo = i;
        }
    }
    return (d[lo]);
}

int maxptr(int s, int *d)
{
    int lo = 0;
    int i = 0;
    for (; i < s; i++) {
        if (*(d + i) > *(d + lo)) {
            lo = i;
        }
    }
    return (*(d + lo));
}

int main()
{
    int test[] = { 1, 45, 23, 244, 11, 34, 2 };
    printf("%d\n", maxarray(7, test));
    printf("%d\n", maxptr(7, test));
    return 0;
}

$ gcc -W -Wall -ansi -pedantic foo.c
$ ./a.exe
244
244

> The forumula to convert array[i] is (address of array's first element + (the index * size of one element)).
Except C already takes this into account based on the type of the pointer.
arr[ i ] is just *( arr + i )

Yes, you're right. I'm an idiot today....

/me slaps self.

Been confusing languages lol....