将变量传递给作为可调用执行的闭包

Question

在Groovy中我们可以做到这一点......

def pool = Executors.newFixedThreadPool(1)
def defer = { c -> pool.submit(c as Callable) }   

def myfunction = {"hello"}
def deferedInvocation = defer(myfunction)

def response = deferedInvocation.get() 
println response    // outputs hello

我想做的是延迟闭包将一个已关闭的变量传递给它调用的闭包c。

所以我们有一个外变量tony

这样的事情：

def tony = "tony"
def pool = Executors.newFixedThreadPool(1)
def defer = { c -> pool.submit(c(tony) as Callable) }   // tony is closed variable

def myfunction = {"hello " + it}
def deferedInvocation = defer(myfunction)

def response = deferedInvocation.get() 
println response

但是，当我尝试时，我得到了：

java.util.concurrent.ExecutionException: org.codehaus.groovy.runtime.metaclass.MissingMethodExceptionNoStack: No signature of method: java.lang.String.call() is applicable for argument types: () values: []
Possible solutions: wait(), any(), wait(long), each(groovy.lang.Closure), any(groovy.lang.Closure), take(int)

有什么想法吗？

Answer 1

我无法重现您的示例，但似乎已经执行了as Callable部分，并且生成的"hello" + it正在提交到池中

尝试将Callable封装在一个闭包中以真正推迟执行：

def defer = { c -> pool.submit({ c(tony) } as Callable) }