Echo声明没有出来，因为我想要它[重复]

Question

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• Not getting the specific output without braces [closed] 3 answers

This code only prints out '17' instead of "10 + 7 = 17". Why is this? And how can i solve this.

<?

$x = 10;
$y = 7;

echo $x . '+' . $y . '=' . $x+$y;

?>

</div>

Answer 1

What you should understand is that echo won't do anything until the result of the whole expression given to it is evaluated. This expression contains both . and + operators - the first one concatenates its operands (glues two strings together), the second adds numbers.

Now, you might think that + operator has a higher precedence than of . one - in other words, that result of $x + $y will be calculated before (eventually) glued to the rest of the string. But it's not. So we can say this statement is actually treated as...

echo ($x . '+' . $y . '=' . $x) + $y;

In other words, the result of all the string concatenation is added (converted to number) to $y, and only the result of this operation is printed.

The result of $x . '+' . $y . '=' . $x is '10+7=10', and it doesn't look like something that might be added. But guess what, PHP wants to be nice to you - and it assumes that you actually wanted to extract the first digits of a string when you tried to convert it to a number. So the whole line is treated as number 10. When added to 7, it's just 17 - that's why you got 17 echoed.

One possible workaround is getting rid of ., using , operator instead (as its precedence is lower - in fact, it's the lowest among operators):

echo $x, '+', $y, '=', $x + $y;

It could be simplified if one remember about such a handy feature of PHP as string interpolation:

echo "$x + $y = ", $x + $y;