Nested loop only access X elements at once

Question

I need to access all elements of a list, but only ever 10 elements at most in one go. For this I thought about a nested loop like this:

for (int i = 0; i < 50 / 10; i++) {
    for (int k = 0; k < 10; k++) {
        paramList.addParam(xyz)
    }
    sendRequest(paramList);
}

With this nested loop I can access all elements from 0-49 and only ever 10 elements in the inner loop. My problem occurs when the list only hast 49 elements instead of 50. Then I only get up to 39 elements with this logic. Also my best thought on only accessing e.g. 5 elements in the last iteration (list size = 45), is to break the loop, before it goes any further.

So my question is, how can I access all elements in a List with undefined size, but only a maximum of 10 elements in the inner loop (and less if list size doesn't allow 10)?

Why not simply go for(int i = myStartIndex; i < myStartIndex + 10 && i < myList.size(); ++i)? — Ben
– Ben, Commented Mar 21, 2018 at 10:37
@Ben you'd end up processing elements 0-10, 1-11, 2-12 etc. You need to add 10 onto i each time. — Data Man
– Data Man, Commented Mar 21, 2018 at 10:39
@user7 makes no sense because with my current loop i wont even got to e.g. 45 if that is the list size — XtremeBaumer
– XtremeBaumer, Commented Mar 21, 2018 at 10:40
OP, what do you mean by access all elements of a list, but only ever 10 elements at most in one go? Can you clarify what you mean by access, and why you are printing a modified value of i? — Robo Mop
– Robo Mop, Commented Mar 21, 2018 at 10:41

Ruslan Akhundov · Accepted Answer · 2018-03-21 10:58:58Z

3

As an option you can just iterate in outer loop using step of 10 elements:

int step = 10;
for (int i = 0; i < list.size(); i+=step) {
    for (int k = i; k < i + step && k < list.size()); k++) {
        System.out.println(list.get(k));
    }
}

edited Mar 21, 2018 at 10:58

answered Mar 21, 2018 at 10:41

Ruslan Akhundov

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4 Comments

XtremeBaumer Over a year ago

Is it possible to not use Math.min() for this?

Robo Mop Over a year ago

@XtremeBaumer You could use a ternary operator for it... (i+step)>list.size()?list.size():i+step

Ruslan Akhundov Over a year ago

@XtremeBaumer as an option you can use ternary operator, so it would be something like: k < i + step <= list.size() ? i + step : list.size() but honestly I think min here seems to be just fine, or you can just use a condition as @Kepotx mention

AxelH Over a year ago

This kind of loops are not easily understood, I suggest good comment or simply find a better approach for any production environment.

AxelH · Accepted Answer · 2018-03-21 11:22:28Z

2

I had to find a reason for your need, breaking the iteration with an inner loop like your snippet doesn't do anything useful, it's still iterate a list.

The only reason I see your logic needed is if you want to call a method after 10 element like :

for (int i = 0; i < 50 / 10; i++) {
    for (int k = 0; k < 10; k++) {
        System.out.println((i * 10) + k);
    }
    System.out.println("####");
}

That would make sense, but that second loop is not needed and also give a "complex" code to understand.

As an alternative, you can iterate normally a list and add a condition to execte that method. It gives a much more readable code

for (int i = 0; i < list.size(); i++) {
    System.out.println(list.get(i));

    if( (i + 1)%10 == 0){ //add one to not trigger this on the first iteration
        System.out.println("####");
    }
}

From your comment:

I retrieve up to 50 records from my database. For each record I need to create a request which will be send in a POST request. Each POST request can contain 10 data requests and I can only send 5 of those request per minute (5*10=50). Since I always fetch TOP 50 data from my database, it can happen that I only get 35 entries instead of 50. To create my requests, I need that data from my database and therefore I have to iterate the whole list, but only ever 10 elements in 1 go to create 1 full POST request

I can see the need here, let's assume you have an instance post that have two method addParam and sendRequest. You need to iterate the list to send String parameters :

int i = 0; 
for(String s : params){
    post.addParam(s);
    if( ++i % 10 == 0){
         post.sendRequest();
    }
}
//If the last `addParam` was not sended, we check again.
if( i % 10 != 0){
     post.sendRequest();
}

Note the condition change outside the loop

edited Mar 21, 2018 at 11:22

answered Mar 21, 2018 at 11:09

AxelH

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2 Comments

XtremeBaumer Over a year ago

Very nice approach. Somehow I was thinking that nested loops are the best (and only) way to do it and never even thought about just adding a condition to trigger what I need. Yet I found it hard to come up with a solution using nested loops

AxelH Over a year ago

@XtremeBaumer I have added some content based on your comment (the context). That give a readable solution that's easily undestood I believe. Nested-loop are sometimes a bit more difficult to understand because of the scope. On some cases, a sheet of paper is the best approch to understand one ;) or good comments !

Kepotx · Accepted Answer · 2018-03-21 10:59:57Z

2

as user7 say, you can add a condition to the nested loop

for (int i = 0; i < list.size(); i+=10) {
    for (int k = i; k < list.size() && (k-i) < 10; k++) {
        System.out.println(k);
    }
}

edited Mar 21, 2018 at 10:59

answered Mar 21, 2018 at 10:47

Kepotx

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5 Comments

XtremeBaumer Over a year ago

I dont think that the check && i < 10 is even needed in the inner loop

Ruslan Akhundov Over a year ago

@XtremeBaumer no, it's needed, otherwise you will each time iterate through the whole collection starting from the index i

XtremeBaumer Over a year ago

I tested this solution and its not working as needed at all. It only iterates to the list size once and that in the inner loop. Only became clear, when adding a System.out.println() to the outer loop

Ruslan Akhundov Over a year ago

@XtremeBaumer that's because of a small mistake in this peace of code, but the idea is good, I edited my solution according to this proposal.

Kepotx Over a year ago

just small mistake, it's not "i < 10" but "(k-i) < 10" or "k < i + 10", as you want

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