php 函数引用调用问题
function &test(){
static $b = 1;
$b += 2;
return $b;
}
$a = &test();
$a =8;
$c = &test(); // 这里再次调用test() 函数,为什么不再执行static $b = 1;这个语句。因为结果表明此时$c=10.
// 如果将static去掉,结果是3. 这个怎么解释?谢谢!
echo $c;
回复讨论(解决方案)
static修饰了变量$b,那么它就是一直存在的,去掉static,那么每次调用都会初始化一次
因为$a = &test();(加了&,地址传递)所以$a被改变,那么静态变量$b的值也被改变,所以在执行$c = &test();时,此时$b的值是8,返回10
static修饰了变量$b,那么它就是一直存在的,去掉static,那么每次调用都会初始化一次
因为$a = &test();(加了&,地址传递)所以$a被改变,那么静态变量$b的值也被改变,所以在执行$c = &test();时,此时$b的值是8,返回10
把语句改成
function &test() {
static $b = 1;
static $b = 100;
static $b = 1000;
// $b = 1000;
$b += 2;
return $b;
} 又测试了下,原来静态存在的不能再用静态方式去修改它。但是又不报错,有点受不了,呵呵~。
静态变量只在声明时赋初值,静态变量具有全局变量的性质,只是作用域不同
静态变量只存在于函数作用域内,也就是说,静态变量只存活在栈中。一般的函数内变量在函数结束后会释放,比如局部变量,但是静态变量却不会。就是说,下次再调用这个函数的时候,该变量的值会保留下来。
静态变量只在声明时赋初值,静态变量具有全局变量的性质,只是作用域不同 +1

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