Simple for loop not working [closed]

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I've just started learning programming. I'm studying for loops but this program does not work as expected. I want to break the loop when $a is equal to 3 so that I get the output 1 2 but I get 3 as output :(

for($a=0;$a<开发者_StackOverflow中文版10;++$a)
{
       if($a==3)
               break
       print"$a ";
}

Please help.

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Missing semi-colon after break

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It's rather interesting to know why your program behaves the way it does.

The general syntax of break in PHP is:

break Expression;

The expression is optional, but if present its value tells how many nested enclosing structures are to be broken out of.

break 0; and break 1; are same as break;

Your code is equivalent to

if($a==3)
       break print"$a ";

Now the print function in PHP always return 1. Hence it is equivalent to

if($a==3)
       break 1;

so when $a is 3 you print its value and break.

It's advisable to use braces to enclose the body of a conditional or a loop even if the body has a single statement. In this case enclosing the if body in braces:

if($a==3) {
  break
}
print"$a ";

would have given a syntax error: PHP expects a ; but finds a }

All of the above applies to the PHP continue as well. So the program

for($a=0;$a<10;++$a)
{
       if($a==3)
               continue
       print"$a ";
}

also prints 3 for a similar reason.

---

You are missing a semicolon at the end of break. ;)

And even with the semicolon it will not work as you'd expect it to since it will count from 0 to 2. You have to write it like this to get only 1 2.

<?php
for($a=1;$a<10;++$a)
{
   if($a==3)
           break;
   print"$a ";
}
?>

Note $a is now one in the for loop initialization.

EDIT: Another thing I've noticed which you should be aware of. In your for loop control you have a pre-increment (++$a). That basically means that PHP increments the value of $a and then returns $a. Another option is the post-increment ($a++) where $a gets returned and then gets incremented by one.

In your case both ways will get you the correct output tho.

This sometimes is pretty important. Just keep that in mind.

---

As codaddict said, you are missing the semi-colon after break.

Your code should look like:

for($a=0;$a<10;++$a)
{
       if($a==3)
           break;
       echo $a, ' ';
}

---

for($a=0;$a<10;++$a)
{
       if($a==3) break;
       print $a;
}

@Downvoters: What's wrong aside from me being laconic?

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for($a=0;$a<10;$a++) {
   if($a==3) { exit; }
   else { echo $a; }
}

---

Use echo in place of print.