Little-known function variable scope (implementation steps)
The last article introduced you to "Two kinds of parameters you must fully master in PHP (with examples) Selection". This article continues to introduce to you the variable scope of the function (that is, function-related The scope of the variable) Now let's go and take a look! ! !
Local variables:
In a function, the defined variables are local variables and their scope Only the content of the function;
Formal parameters are also variables inside the function and local invariants;
<?php /****** 局部变量*/ function demo(){ $str = '找个富二代,可以少奋斗好几十年。'; } demo (); echo $str; ?>
The code demonstration results are as follows:
In local variables, we define a function, and then declare a variable inside the function. If we can output the variable of this function outside the function, according to the code demonstration, We can get that the result shows that there is no output and this variable is not defined, so we can conclude that the variable we define inside the function is what we call a local variable. (In other words, the variables inside the function will be destroyed once they are executed).
Suppose we define a $str in demo() and then output $str1 externally. Can we output the contents of the defined variable?
The code demonstration is as follows:
<?php /****** 局部变量*/ function demo($str1 = '论如何成为一个有钱人'){ $str = '找个富二代,可以少奋斗好几十年。'; } demo (); echo $str; echo $str1; ?>
The code demonstration results are as follows:
We can know from the code demonstration that the operation is still the same is wrong, it still means that $str1 has no defined variable.
So we can also conclude that our formal parameter is also used inside the function. The code demonstration is as follows:
<?php /****** 局部变量*/ function demo($str1 = '论如何成为一个有钱人'){ echo $str1; $str = '找个富二代,可以少奋斗好几十年。'; } demo (); echo $str; echo $str1; ?>
The code demonstration result is as follows:
Recommended study: "PHP Video Tutorial"
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