I would like to ask you guys a question. You see, I know what a for-loop is for but can someone please maybe explain how one works, just to help me get my head around it, an example is:
for(int i = 0; i < 10; i++) {
System.out.println("hello");
}
Now obviously that will just print Hello 10 times into the console but that's besides the point, I want to know how the for-loop works.
Sorry if i have confused anyone asking this - Shaun
The for loop in your example is more or less equivalent to this:
int i = 0;
while (i < 10) {
System.out.println("hello");
i++;
}
The only difference is that with your for loop, the variable i exists only within the scope of the loop.
Every for loop can be transformed into a while loop using this same pattern.
for (init; test; continuation) {
// loop body
}
becomes:
init;
while (test) {
// loop body
continuation;
}
Again, the only difference will be with the scope of any variables declared in init.
The for Statement
The for statement provides a compact way to iterate over a range of values. Programmers often refer to it as the "for loop" because of the way in which it repeatedly loops until a particular condition is satisfied. The general form of the for statement can be expressed as follows:
for (initialization; termination; increment) {
statement(s)
}
When using this version of the for statement, keep in mind that:
The initialization expression initializes the loop; it's executed once, as the loop begins.
When the termination expression evaluates to false, the loop terminates.
The increment expression is invoked after each iteration through the loop; it is perfectly acceptable for this expression to increment or decrement a value.
Well, this is how it is set up:
for (a; b; c)
"A" is something that is done at the beginning of the loop. It can actually be left out if necessary, like this:
for (; b; c)
"B" must be a true or false statement (like i<10, it either is or it isn't). Once "b" is no longer true, the loop stops.
"C" is something that is done at the end of the loop.
Related
I don't know interpret this code:
t:
while (true) {
break t;
}
Can you help me?
This construct called a "labeled break" and can be used to simultaneously break out of multiple nested loops.
To quote an example from an Oracle tutorial:
search:
for (i = 0; i < arrayOfInts.length; i++) {
for (j = 0; j < arrayOfInts[i].length;
j++) {
if (arrayOfInts[i][j] == searchfor) {
foundIt = true;
break search;
}
}
}
Here, an unlabeled break (i.e. simply break) would only terminate the inner loop, whereas break search terminates both loops at once.
See Is using a labeled break a good practice in Java? for a relevant debate.
It's just a label that you place anywhere and then you can "break" or "continue" depending on your conditions.It can be used in nested if-else with for loopings in order to break several loops.
Here break t; simply means break out from the while loop that is labelled as t.
It's useful for breaking out of nested loops
This is called a 'level', the alternative to 'goto' in other languages. Although 'goto' is a reserved word in Java, it is not used in the language; Java has no goto. However, it does have something that looks a bit like a jump tied in with the break and continue keywords: a level.
A label is an identifier followed by a colon, like this: label1:
The only place a label is used in Java is right before an iteration statement. The reason to put a label before an iteration is if you’re going to nest another iteration or a switch inside it. That’s because the break and continue keywords will normally interrupt only the current loop, but when used with a label, they’ll interrupt the loops up to where the label exists:
label1:
outer-iteration {
inner-iteration {
//...
break; // (1)
//...
continue; // (2)
//...
continue label1; // (3)
//...
break label1; // (4)
}
}
In (1), the break breaks out of the inner iteration and you end up in the outer iteration. In (2), the continue moves back to the beginning of the inner iteration. But in (3), the continue label1 breaks out of the inner iteration and the outer iteration, all the way back to label1. Then it does in fact continue the iteration, but starting at the outer iteration. In (4), the break label1 also breaks all the way out to label1, but it does not re-enter the iteration. It actually does break out of both iterations.
Hope this help!!!
int a = 0;
int b = 0;
int c = 0;
Scanner sc = new Scanner(System.in);
for (int i = 0; i < 6; i++) {
b = sc.nextInt();
a =+ b;
c =+ (a + 1);
if (c < 20) {
i = 2;
}
}
if I have lines numbered from 0 to 6 inside the loop, the loop would be
so if c is less than 20, it repeats the operation "c=+(a+1);" until it breaks out of the loop by c>=20.
this is a simplified code from my program, mine is GUI. every time I run the code, it freezes.
use c+= instead of c=+. try that, cheers!
and b+= instead ofb=+.
You can tag a loop and do break or continue instructions, but you need design the flow, it is not possible to go into specified line, because java don't use goto instruction. You can only switch the flow inside loops by those instructions.
myloopTag:
for (...; ...; ...) {
// and you can break current loop by:
break;
// or specific (outer) loop by
break myloopTag;
// you can also use 'continue' to go to the start of the loop and increment again
continue;
// or to 'continue' at a label:
continue myloopTag;
}
You're probably very new to the language. Welcome!
If I understand your description of your intent properly, you want your code to exit the loop when c>=20. Based on your description of numbering your lines and the fact that you have the statement:
if(c<20){
i=2;
}
it seems that you think that the iterator i in the for loop is related to the line that will be executed*. This is not the case. The iterator i is a variable that simply holds an integer (just like a, b, and c in your code).
I suggest you take a look at a tutorial on for loops. It might be helpful for you to review other language basics as well, like how control flow works (this may be a better one to start with, actually).
*This guess at your intent is further supported by you counting that there are 6 lines and that your loop goes up to 6.
I am trying to learn recursion in Java and have an array that takes in continuous input until the Scanner reads in a 0.
From there I have a method that (attempts) to calculate the number of positive integers in the array using recursion. This is the first recursive function I have ever written and I keep getting a stackoverflow error.
I have read tutorials and I still can't wrap my head around the basic understanding of recursion.
public class reuncF {
private static int start = 0;
private static int end = 98;
public static void main(String[] args) {
input = input.nextDouble();
list[i] = numInput;
computeSumPositive(numList, count);
}
}
return positives += solve(numbers, count++);
}
}
You forgot to stop your recursion!
There has to be some case where computeSumPositive returns without calling itself again. Otherwise it'll just keep going forever, never getting back to you.
If you did it with a loop, the loop would look like this:
int positives = 0;
for (int i = 0; i < numList.length; ++i) {
if (numList[i] > 0) {
positives++;
}
}
To do that recursively, you just find out what are the variables used in the loop. They are i, numList and positives.
computeSumPositive(int i, double[] numList, int positives)
Then we take a look at what the loop does. First, it checks whether we went too far,
so our recursive function should do that too. It'll have to return instead of just falling through like the loop does. And obviously, it must return the result:
{
if (! (i < numList.length))
return positives;
The loop then does the test and maybe increments positives, so the recursive function should also do that:
if (numList[i] > 0) {
positives++;
}
At the end of the loop, i is updated:
i++;
The loop just starts over, but the recursive function will have to call itself. Of course, we want it to use the new value of i and positives, but fortunately we updated those, so now we can just do:
return computeSumPositives (i, numList, positives);
}
The tricky bit is that the values i, numList, and are local to each call. Each invocation of computeSumPositives can see only the arguments it were given. If it changes them, none of the other invocation can see that change.
EDIT: So if we, for reasons we can only speculate about, wanted desperately for computeSumPositive to take only 2 parameters, we would have to "split up" positives across each invocation. Each invocation knows whether or not its number was positive or not; all we have to do is add them. Then it looks like this:
computeSumPositive(int i, double[] numList)
{
if (! (i < numList.length))
return 0; // I didn't find any at index i
if (numList[i] > 0) {
// Theres one I found + however many my later
// invocations will find.
return 1 + computeSumPositive (i+1, numList);
} else {
// I didn't find any, but my later invocations might.
return computeSumPositive (i+1, numList);
}
}
I find it helpful, when dealing with recursion, to figure out the termination case first.
It looks like you are treating 'count' as an index. So you could check if your at the last index in the array, if so and if the value is positive return a 1, if the value is non-positive return a 0 - dont recurse anymore.
If your not at the last index, and the value is positive return a 1 + the recursive function call, or if the value is non-positive just continue to recurse.
This will still cause a stack overflow for large arrays.
The value of count++ is the same as the value of count; the program uses the value and then increments it. But the result is that computeSumPositive keeps calling itself with the same value of count, which leads to infinite recursion. Note that each time computeSumPositive calls another computeSumPositive, each call has its own copy of the parameters (like count) and the local variables; so incrementing one computeSumPositive's copy of count has no effect on the value of count used by other recursive calls.
Change count++ to count + 1, and also add a way to halt the recursion. (At some point, you will be calling computeSumPositive to look at zero integers, and at that point, it should just return 0 and not call itself. You need to think about: how do you test whether you've reached that point?)
I've tried several programs that involve printing to the console in for loops, but none of them have printed anything. I can't work out the problem, and I've boiled it down as simply as possible here:
for (int x=0; x==10; x++)
{
System.out.print("Test");
}
Like I said, absolutely nothing is printed to the console. Things outside of the for loop will print, and things affected by the for loop will print.
Perhaps it's something very simple, but I wouldn't know considering I'm relatively new to programming and Eclipse gives me no errors. Any help would be much appreciated, as this is plaguing my class files at the moment.
Thanks,
Daniel
Your for loop condition is wrong. You want the condition to be true to continue looping, and false to stop.
Try
for (int x=0; x < 10; x++)
For more information, here's the Java tutorial on for loops.
#rgettman gave the reason your code didn't work above.
The way the for loop works is in the brackets the first variable is where the loop starts (i.e. 'x=0'), the second variable is the condition (i.e. 'x<= 10'), and the third is what to do for each loop (i.e. 'x++').
You had "x==10" for the condition, so for the first scenario where x was equal to "0", the loop ended because it was NOT equal to "10". So you want it to be "x<=10" (x is less than or equal to 10); this will go through 11 loops.
rgettman is completely correct. A for loop should be used as so:
for is a type of loop in java that will take three parameters separated by semicolons ;
The first parameter will take a variable such as int i = 0; to create a simple integer at 0.
The second parameter will take a condition such as i < 10, to say while the i integer is less than
The third and final parameter will take an incrementing value like, i++, i--, i +=5, or something to that effect.
This part should like like for(int i = 0; i < 10; i++) so far.
Now you need the brackets { and }. Inside of the brackets you will perform an action. Like you wanted to print "test" to the console.
for(int i = 0; i < 10; i++) {
System.out.println("test");
}
This would print "test" 10 times into the console. If you wanted to see what number i was at, you could simply say,
for(int i = 0; i < 10; i++) {
System.out.println(i); // Current value of i
}
Hope this was of use to you!
let say my code look like below
for(..)
for(..)
for(..){
break; //this will break out from the most inner loop OR all 3 iterated loops?
}
Your example will break out of the innermost loop only. However, using a labeled break statement, you can do this:
outer:
for(..)
for(..)
for(..){
break outer; //this will break out from all three loops
}
This will only break out from the inner loop. You can also define a scope to break out from. More from the language specs:
A break statement with no label
attempts to transfer control to the
innermost enclosing switch, while, do,
or for statement of the immediately
enclosing method or initializer block;
this statement, which is called the
break target, then immediately
completes normally.
Yes, without labels it will break only the most inner loop.
Instead of using labels you can put your loops in a seperated function and return from the function.
class Loop {
public void loopForXx() {
untilXx();
}
private void untilXx() {
for()
for()
for()
if(xx)
return;
}
}
From the most inner loop :)
int i,j,k;
for(i = 0; i < 2; i++)
for(j = 0; j < 2; j++)
for(k = 0; k < 2; k++)
{
printf("%d %d %d\n", i, j, k);
break;
}
Will produce :
0 0 0
0 1 0
1 0 0
1 1 0
You should take a look here: http://java.sun.com/docs/books/tutorial/java/nutsandbolts/branch.html
as often mentioned i don't like to break with a label eather. so while in a for loop most of the time i'm adding a boolean varible to simple exit the loop.. (only if i want to break it of cause;))
boolean exit = false;
for (int i = 0; i < 10 && !exit; i++) {
for (int j = 0; j < 10 && !exit; j++) {
exit = true;
}
}
this is in my opinion more elegant than a break..
Many people here don't like labels and breaking. This technique can be compared to using a 'goto' statement, a flow control statement which allows jumping out of a block of code in a non-standard way, obliviating use of pre- and post conditions. Edsger Dijkstra published a famous article in Communications of the ACM, march 1968, 'Goto statement considered harmful' (it's a short read).
Using the same reasoning presented in the article, returning from inside an iteration as suggested by TimW is also bad practice. If one is strict, to create readable code, with predictable entry- and exit points, one should initialize the variable which will hold the return value (if any) at the beginning of the method and return only at the end of a mehod.
This poses a challenge when using an iteration to perform a lookup. To avoid using break or return one inevitably ends up with a while-loop with a regular stop condition and some boolean variable to indicate that the lookup has succeeded:
boolean targetFound = false;
int i = 0;
while (i < values.size() && ! targetFound ) {
if (values.get(i).equals(targetValue)) {
targetFound = true;
}
}
if (!targetFound) {
// handle lookup failure
}
Ok, this works, but it seems a bit clunky to me. First of all I have to introduce a boolean to detect lookup success. Secondly I have to explicitly check targetFound after the loop to handle lookup failure.
I sometimes use this solution, which I think is more concise and readable:
lookup: {
for(Value value : values) {
if (value.equals(targetValue)) {
break lookup;
}
}
// handle lookup failure here
}
I think breaking (no pun intended) the rule here results in better code.
it will breake from most inner loop,
if you want to break from all, you can hold a variable and change its value when you want to break, then control it at the beginning of each for loop