groovy - 如何从groovy中的返回函数接受多个参数

Question

我想从用 groovy 编写的函数返回多个值并接收它们，但出现错误

class org.codehaus.groovy.ast.expr.ListExpression, with its value '[a, b]', is a bad expression as the left hand side of an assignment operator

我的代码是

int a=10
int b=0
println "a is ${a} , b is ${b}"
[a,b]=f1(a)
println "a is NOW ${a} , b is NOW ${b}"

def f1(int x) {   
  return [a*10,a*20]
}

Answer 1

你几乎已经拥有它了。从概念上讲，[ a, b ] 创建一个列表，( a, b ) 解包一个列表，因此您需要 (a,b)=f1(a) 而不是 [a,b]=f1(a)。

int a=10
int b=0
println "a is ${a} , b is ${b}"
(a,b)=f1(a)
println "a is NOW ${a} , b is NOW ${b}"

def f1(int x) {
    return [x*10,x*20]
}

另一个返回对象的示例，它们不需要是相同的类型:

final Date foo
final String bar
(foo, bar) = baz()
println foo
println bar

def baz() {
    return [ new Date(0), 'Test' ]
}

此外，您可以结合声明和赋值:

final def (Date foo, String bar) = baz()
println foo
println bar

def baz() {
    return [ new Date(0), 'Test' ]
}