PHP和C之间的可变范围差异：块范围不完全相同？

The following PHP code will output 3.

以下PHP代码将输出3。

function main() {
    if (1) {
        $i = 3;
    }
    echo $i;
}

main();

But the following C code will raise a compile error.

但是下面的C代码会引发编译错误。

void main() {
    if (1) {
        int i = 3;
    }

    printf("%d", i);
}

So variables in PHP are not strictly block-scoped? In PHP, variables defined in inner block can be used in outer block?

那么PHP中的变量不是严格的块范围的？在PHP中，内部块中定义的变量可以用在外部块中吗？

1 个解决方案

#1

PHP only has function scope - control structures such as if don't introduce a new scope. However, it also doesn't mind if you use variables you haven't declared. $i won't exist outside of main() or if the if statement fails, but you can still freely echo it.

PHP只有函数作用域 - 控制结构，如果不引入新的作用域。但是，如果您使用未声明的变量，它也不介意。 $ i不会存在于main（）之外，或者如果if语句失败，但你仍然可以*地回应它。

If you have PHP's error_reporting set to include notices, it will emit an E_NOTICE error at runtime if you try to use a variable which hasn't been defined. So if you had:

如果将PHP的error_reporting设置为包含通知，则在尝试使用尚未定义的变量时，它将在运行时发出E_NOTICE错误。所以如果你有：

function main() {
 if (rand(0,1) == 0) {
  $i = 3;
 }
 echo $i;
}

The code would run fine, but some executions will echo '3' (when the if succeeds), and some will raise an E_NOTICE and echo nothing, as $i won't be defined in the scope of the echo statement.

代码运行正常，但是一些执行将回显'3'（当if成功时），并且一些将引发E_NOTICE并且什么都不回显，因为$ i将不在echo语句的范围内定义。

Outside of the function, $i will never be defined (because the function has a different scope).

在函数之外，永远不会定义$ i（因为函数具有不同的范围）。

For more info: http://php.net/manual/en/language.variables.scope.php

欲了解更多信息：http：//php.net/manual/en/language.variables.scope.php

#1