PHP如何工作:
$myClass = 'App\MyClass';
$object = new $myClass;
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但这会导致错误:
$myClass = 'MyClass';
$object = new 'App\\'.$myClass;
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在第二个例子中,unexpected T_CONSTANT_ENCAPSED_STRING抛出了一个.
事实证明,上面的例子是由于运算符优先级,因为new它具有最高优先级,但......
同样,我可以尝试使用一个字符串进行实例化,如:
$object = new 'App\MyClass';
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并抛出同样的错误.为什么是这样?
PHP 语法在zend_language_parser.y中定义,并且它根本没有为运算符定义任何更复杂的表达式new:
new_expr:
T_NEW class_name_reference ctor_arguments
{ $$ = zend_ast_create(ZEND_AST_NEW, $2, }
| T_NEW anonymous_class
{ $$ = $2; }
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class_name_reference:
class_name { $$ = $1; }
| new_variable { $$ = $1; }
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并new_variable允许一组有限的变量表达式:
new_variable:
simple_variable
{ $$ = zend_ast_create(ZEND_AST_VAR, $1); }
| new_variable '[' optional_expr ']'
{ $$ = zend_ast_create(ZEND_AST_DIM, $1, $3); }
| new_variable '{' expr '}'
{ $$ = zend_ast_create(ZEND_AST_DIM, $1, $3); }
| new_variable T_OBJECT_OPERATOR property_name
{ $$ = zend_ast_create(ZEND_AST_PROP, $1, $3); }
| class_name T_PAAMAYIM_NEKUDOTAYIM simple_variable
{ $$ = zend_ast_create(ZEND_AST_STATIC_PROP, $1, $3); }
| new_variable T_PAAMAYIM_NEKUDOTAYIM simple_variable
{ $$ = zend_ast_create(ZEND_AST_STATIC_PROP, $1, $3); }
// Copyright (c) 1998-2015 Zend Technologies Ltd., the Zend license 2.00
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这就是为什么那里不能有字符串表达式的原因。(它从未被扩展,因为简单的实例化和 $classvarnames 通常就足够了。所以允许的语法与 PHP3 引入时基本相同。)