Description of the formal and actual parameters of functions in presentationfontcache.exe PHP

PHP will issue a warning when the number of actual parameters the number of formal parameters, PHP will not report an error. It will only take the first few parameters, and the excess will be discarded.
When writing a function in PHP, generally when calling a function, the changed values ​​are the formal parameters rather than the actual parameters. But if you add an address character to the formal parameters, the value of the actual parameters will be changed. Why?
Please see Example below:

Copy code The code is as follows:

//Write a function swap() and test that the actual parameter value of the function has not changed
function swap($a,$b ) {
echo "

Before entering the swqp() function
n";
echo "Before exchange: formal parameter a=$a, formal parameter b=$b
n";
$c =$b;
$a=$b;
$b=$c;
echo "After exchange: formal parameter a=$a, formal parameter b=$b
n";
echo "Exit swap() Function

n";
}
$variablea=5;
$variableb=10;
echo "Before calling the swap() function: ";
echo "Actual parameter a=$variablea, actual Parameter b=$variableb
n";
swap($variablea,$variableb);
echo "After calling the swap() function: ";
echo "Actual parameter a=$variablea,Actual parameter b=$variableb< br>n";
?>

Copy the code The code is as follows:

//Test the value change of the swap() function parameter
function swap1(&$a, &$b) {
echo "

Enter swap1() function
n";
echo "Before exchange: formal parameter a=$a, formal parameter b=$b
n";
$c=$b;
$a=$b;
$b=$c;
echo "After exchange: formal parameter a=$a, formal parameter b=$b
n";
echo "Exit swap () function

n";
}
$variablea=5;
$variableb=10;
echo "Before calling the swap1() function: ";
echo "Actual parameter a=$variablea ,actual parameter b=$variableb
n";
swap1($variablea,$variableb);
echo "After calling the swap1() function: ";
echo "Actual parameter a=$variablea,actual parameter b=$ variableb
n";
?>

//The above two examples are just for illustration, please give me some advice~~~

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