Why use "(array) + count" in foreach macro?

Question

I get foreach macro from here

#define foreach(item, array) \
    for(int keep = 1, \
            count = 0,\
            size = sizeof (array) / sizeof *(array); \
        keep && count != size; \
        keep = !keep, count++) \
      for(item = (array) + count; keep; keep = !keep)

I don't understand "(array) + count", it's equal "&array[count]", but why not use "array[count]" instead of "(array) + count"

Do you mean "why not use `array[count]`" (as written) or "why not use `@array[count]`" (as per your comment on what `(array) + count` is equal to)? — rici, Oct 28 '15 at 05:50
Suggest edit/compile/runit both ways, see if there is any difference — user3629249, Oct 29 '15 at 16:25

score 4 · Accepted Answer · edited May 23 '17 at 10:27

4

but why not use "array[count]" instead of "(array) + count"

From the linked post,

foreach(int *v, values) {
    printf("value: %d\n", *v);
}

In the macro, item is of type int* and array[count] is of type int.

Hence you cannot use:

item = array[count];

But you can use

item = (array) + count;

since (array) + count evaluates to an int*.

edited May 23 '17 at 10:27

Community

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answered Oct 28 '15 at 05:14

R Sahu

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you're right, use point is more saving memory, but if i change "int *v" to "int v" and "(array) + count" to "array[count", it's more convenient. – nwaicaethi Nov 05 '15 at 15:42
@nwaicaethi, you could do that. – R Sahu Nov 05 '15 at 16:01

wrangler · Answer 2 · 2015-10-28T05:23:43.373

3

There is no difference in array+count and &array[count] or count+array.

Its just these are both addresses so you cant assign them to an integer value as you did previously:

int item=array+count;

But to an integer pointer:

int* item=array+count; or int* item= &array[count];

edited Oct 28 '15 at 05:23

answered Oct 28 '15 at 05:11

wrangler

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@chux my mistake and thanx – wrangler Oct 28 '15 at 05:24

score 1 · Answer 3 · answered Oct 28 '15 at 05:11

1

They are equivalent, which one you use is purely up to semantic style of the programmer.

Arrays decay to pointers when you pass them to a function, pointers to the first array element to be exact. Thusly, you can use both.

answered Oct 28 '15 at 05:11

Magisch

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NMDV There is no function call in the post so why "Arrays decay to pointers when you pass them to a function". – chux - Reinstate Monica Oct 28 '15 at 05:17

Why use "(array) + count" in foreach macro?

3 Answers3