How to include last element in loops in C? - c

Question

This is a general issue I've run into, and I've yet to find a solution to it that doesn't feel very "hack-y". Suppose I have some array of elements xs = {a_1, a_2, ..., a_n}, where I know some x is in the array. I wish to loop through the array, and do something with each element, up to and including the element x. Here's a version that does almost that, except it leaves out the very last element. Note that in this example the array happens to be a sorted list of integers, but in the general case this might not necessarily be true.
int xs[] = {1,2,3,4,5};
for (int i = 0; xs[i] != 4; i++) {
foo(xs[i]);
}
The only solutions I've seen so far are:
Just add a final foo(xs[i]); statement after the for-loop. This is first of all ugly and repetitious, especially in the case where foo is not just a function call but a list of statements. Second, it requires i to be defined outside the scope of the for-loop.
Manually break the loop, with an if-statement inside an infinite loop. This again seems ugly to me, since we're not really using the for and while constructs to their full extent. The problem is almost archetypal of what you'd use a for-loop for, the only difference is that we just want it to go through the loop one more time.
Does anyone know of a good solution to this problem?

Answer 1

In C, the for loop is a "check before body" operation, you want the "check after body" variant, a do while loop, something like:
int xs[] = {1,2,3,4,5};
{
int i = 0;
do {
foo(xs[i]);
} while (xs[i++] != 4);
}
You'll notice I've enclosed the entire chunk in its own scope (the outermost {} braces). This is just to limit the existence of i to make it conform more with the for loop behaviour.
In terms of a complete program showing this, the following code:
#include <stdio.h>
void foo(int x) {
printf("%d\n", x);
}
int main(void) {
int xs[] = {1,2,3,4,5};
{
int i = 0;
do {
foo(xs[i]);
} while (xs[i++] != 4);
}
return 0;
}
outputs:
1
2
3
4
As an aside, like you, I'm also not that keen of the two other solutions you've seen.
For the first solution, that won't actually work in this case since the lifetime of i is limited to the for loop itself (the int in the for statement initialisation section makes this so).
That means i will not have the value you expect after the loop. Either there will be no i (a compile-time error) or there will be an i which was hidden within the for loop and therefore unlikely to have the value you expect, leading to insidious bugs.
For the second, I will sometimes break loops within the body but generally only at the start of the body so that the control logic is still visible in a single area. I tend to do that if the for condition would be otherwise very lengthy but there are other ways to do this.

Answer 2

Try processing the loop as long as the previous element (if available) is not 4:
int xs[] = {1,2,3,4,5};
for (int i = 0; i == 0 || xs[i - 1] != 4; i++) {
foo(xs[i]);
}

Answer 3

This may not be a direct answer to the original question, but I would strongly suggest against making a habit of parsing arrays like that (it's like a ticking bomb waiting to explode at a random point in time).
I know you said you already know x is a member of xs, but when it is not (and this can accidentally happen for a variety of reasons) then your loop will crash your program if you are lucky, or it will corrupt irrelevant data if you are not lucky.
In my opinion, it is neither ugly nor "hacky" to be defensive with an extra check.
If the hurdle is the seemingly unknown length of xs, it is not. Static arrays have a known length, either by declaration or by initialization (like your example). In the latter case, the length can be calc'ed on demand within the scope of the declared array, by sizeof(arr) / sizeof(*arr) - you can even make it a reusable macro.
#define ARRLEN(a) (sizeof(a)/sizeof(*(a)))
...
int xs[] = {1,2,3,4,5};
/* way later... */
size_t xslen = ARRLEN(xs);
for (size_t i=0; i < xslen; i++) {
if (xs[i] == 4) {
foo( xs[i] );
break;
};
}
This will not overrun xs, even when 4 is not present in the array.
EDIT:
"within the scope of the declared array" means that the macro (or its direct code) will not work on an array passed as a function parameter.
// This does NOT work, because sizeof(arr) returns the size of an int-pointer
size_t foo( int arr[] ) {
return( sizeof(arr)/sizeof(*arr) );
}
If you need the length of an array inside a function, you can pass it too as a parameter along with the array (which actually is just a pointer to the 1st element).
Or if performance is not an issue, you may use the sentinel approach, explained below.
[end of EDIT]
An alternative could be to manually mark the end of your array with a sentinel value (a value you intentionally consider invalid). For example, for integers it could be INT_MAX:
#include <limits.h>
...
int xs[] = {1,2,3,4,5, INT_MAX};
for (size_t i=0; xs[i] != INT_MAX; i++) {
if (xs[i] == 4) {
foo( xs[i] );
break;
};
}
Sentinels are quite common for parsing unknown-length dynamically-allocated arrays of pointers, with the sentinel being NULL.
Anyway, my main point is that preventing accidental buffer overruns probably has a higher priority compared to code prettiness :)