我正在关注使用列表和地图作为构造函数的this博文.
为什么以下列表无法强制反对?
class Test {
static class TestObject {
private int a = 1;
protected int b = 2;
public int c = 3;
int d = 4;
String s = "s";
}
static main(args) {
def obj = [1, 2, 3, 4, 's'] as TestObject
}
}
我得到这个例外:
Caught: org.codehaus.groovy.runtime.typehandling.GroovyCastException: Cannot cast object '[1, 2, 3, 4, s]' with class 'java.util.ArrayList' to class 'in.ksharma.Test$TestObject' due to: groovy.lang.GroovyRuntimeException: Could not find matching constructor for: in.ksharma.Test$TestObject(java.lang.Integer, java.lang.Integer, java.lang.Integer, java.lang.Integer, java.lang.String)
org.codehaus.groovy.runtime.typehandling.GroovyCastException: Cannot cast object '[1, 2, 3, 4, s]' with class 'java.util.ArrayList' to class 'in.ksharma.Test$TestObject' due to: groovy.lang.GroovyRuntimeException: Could not find matching constructor for: in.ksharma.Test$TestObject(java.lang.Integer, java.lang.Integer, java.lang.Integer, java.lang.Integer, java.lang.String)
at in.ksharma.Test.main(Test.groovy:22)
解决方法:
你可以使用map:
class Test {
static class TestObject {
private int a = 1;
protected int b = 2;
public int c = 3;
int d = 4;
String s = "s";
}
static main(args) {
def o = ['a':1,b:'2',c:'3','d':5,s:'s'] as TestObject
println o.d
}
}
马上就会考虑清单.
编辑
嗯..我不确定列表是否可行.仅当您添加适当的构造函数时.
完整样本:
class Test {
static class TestObject {
TestObject() {
}
TestObject(a,b,c,d,s) {
this.a = a
this.b = b
this.c = c
this.d = d
this.s = s
}
private int a = 1;
protected int b = 2;
public int c = 3;
int d = 4;
String s = "s";
}
static main(args) {
def obj = ['a':1,b:'2',c:'3','d':5,s:'s'] as TestObject
assert obj.d == 5
obj = [1, 2, 3, 6, 's'] as TestObject
assert obj.d == 6
}
}