Is it possible to get float type result from the sum of two integer type variable?

Question

Example:

1 + 2 => 3.0

I have tried using number_format($result, 1) and sprintf("%.1f", $result), but the return value type is string.

Additionally, if I type cast to float then the return value is a float of value 3 and not 3.0

PHP uses type casting anyway, so the type doesn't matter _that_ much. Can you not just do `$result = (float)(1 + 2)` — JustCarty, Dec 28 '17 at 09:16
I don't understand, what difference does it make? If the value is `3` or `3.0` it makes no difference. If you are displaying the value then use the `number_format` as you have done, and then it, again, doesn't matter what the type is. — JustCarty, Dec 28 '17 at 09:23
@RamKesavan I think what OP wants is to combine the formatting of number_format, but **retain the float type**. — JustCarty, Dec 28 '17 at 09:28
However, I am afraid that that is not possible. I have tried everything but nothing seems to let you achieve that... — JustCarty, Dec 28 '17 at 09:45

score 3 · Answer 1 · answered Dec 28 '17 at 09:49

3

I suggest you.. you can use sprintf:-

$a = 1+2;
$result = sprintf("%.2f", $a); //3.00  or  $result = sprintf("%.1f", $a); //3.0
echo $result;

Hope it helps!

answered Dec 28 '17 at 09:49

kunal

score 0 · Answer 2 · answered Dec 28 '17 at 09:28

0

use floatval(); to swith the value to float

answered Dec 28 '17 at 09:28

Kareem Essawy

score 0 · Answer 3 · answered Dec 28 '17 at 09:47

Your integer variables are in the type int

If you add two int type, the result is inevitably an integer result.

$intOne = 1;
$intTwo = 2;

$result = $intOne + $intTwo; // = (int)3

You can change the type of your result, easily, but, is not necessary good, if you store an int in your var...

$floatResult = (float) $result; // (float)3

Also, if you don't know the type of your variable you can use the function 'floatval' (official doc here) like that:

$floatOne = floatval('3.14'); // (float)3.14
$floatTwo = floatval("3.141 is a Pi number"); // (float)3.14

In this case, if you add two flaot type, the result is in float two:

$result = $floatOne + floatTwo; // (float)6.281

The best practises want to save the right type on your variable and your database ( read this for more information about types and performance)

if you want just show decimals after your integer type, you can use the number_format() function (official documentation here) like that:

$decimalResult = number_format($floatResult, 4);
echo $decimalResult; // Show '3.0000'        ^ Number of decimals

Hope I help you ;)

3 Answers3