referencing to array-values with pointers

[Placebo]

Hi!

Im not sure why the following output of the following code is:
2 15 6 8 10

Code being:

#include <stdio.h>





void change(int *b, int n) {
    int i;
    for(i=0; i<n; i++)
        *(b+1) = *(b+i)+5;
}


int main() {
    int i, a[] = {2, 4, 6, 8, 10};
    change(a, 5);
    for(i=0; i<=4; i++)
        printf("%d  ", a[i]);


    return 0;
}

I mean, from my understanding an array is basically a pointer of the 1st element of the array, i.e. an array is the address of the first element of an array.

So here *b should be basically a[0] and hence the value of 2 - is this right so far?

if so, then *(b + 1) should be just be basically a[1] and hence 4?
But what is happening then?
it is set to *(b + i) + 5, which should be within the first iteration
*(b + 0) + 5, and hence a[0] + 5 and finally this should be 7?

Im not sure how the calc works to be honest and would be grateful for help to understand it.

[cooper1200]

your assigning the value of a[0] + 5 to a{0+1] then your assining 6+5 to a[1] and so on which is why the 2 doesnt change or the 6, 8 and 10 and why a[1] is 15 ( 10 + 5)

[cooper1200]

by adding 1 or i to the pointer is another way of saying b[ 1 ] or b[ i ]

[Placebo]

Hi!
Thanks for the reply - Im afraid I still do not fully get it.

your assigning the value of a[0] + 5 to a{0+1]
But this should be 7 then? Since a[0] is 2?
then your assining 6+5 to a[1]
when and how?
and so on which is why the 2 doesnt change or the 6, 8 and 10 and why a[1] is 15 ( 10 + 5)
Need to think more about it, because so far I really apparently do not get it.

[laserlight]

You're only ever changing *(b + 1). You keep assigning to it, so only the last assignment has a net effect, and in that assignment i == 4, i.e., you compute *(b + 4) + 5, which is 15.

[Placebo]

You're only ever changing *(b + 1). You keep assigning to it, so only the last assignment has a net effect, and in that assignment i == 4, i.e., you compute *(b + 4) + 5, which is 15.

so basically 10 + 5, since *(b + 4) should be b[4], which is 10.
If this is right, I think I get it now with your help - thanks much^^

...............

[flp1969]

so basically 10 + 5, since *(b + 4) should be b[4], which is 10.
If this is right, I think I get it now with your help - thanks much^^

laserlight's answer is the right one... and you understood perfectly well the equivalence of *(b+n) and b[n].

[Placebo]

laserlight's answer is the right one... and you understood perfectly well the equivalence of *(b+n) and b[n].

All right, thanks