两种不同数据类型之间的php算术运算

Question

hi first of all I'm new to here.

i need all of your help

below is my coding

<?php


 $a=1; 
 $b="hello";

$c=$a+$b."<br>"; 
$d=$b+$a."<br>";

 echo $c; 
 echo $d;
?>

output

1

1

i need to know whats happening why answer is 1 in both ways is there any priority for data types in php

Answer 1

This scenario is otherwise called as Type-Juggling

The variable $b="hello"; will be cast to int (in this case it will be a 0) when you did the addition operator. If the variable had $b="2hello"; , your output would have been 3

See this code.

<?php
$foo = "0";  // $foo is string (ASCII 48)
$foo += 2;   // $foo is now an integer (2)
$foo = $foo + 1.3;  // $foo is now a float (3.3)
$foo = 5 + "10 Little Piggies"; // $foo is integer (15)
$foo = 5 + "10 Small Pigs";     // $foo is integer (15)
?>

PHP Manual

Answer 2

Your $a is an integer and $b is an string. When you add integer to string, string is converted to integer. In your case $b = "hello" is converted to integer 0 .