75 lines
		
	
	
		
			1.7 KiB
		
	
	
	
		
			Go
		
	
	
	
	
	
			
		
		
	
	
			75 lines
		
	
	
		
			1.7 KiB
		
	
	
	
		
			Go
		
	
	
	
	
	
| /*
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| Copyright 2019 The Kubernetes Authors.
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| 
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| Licensed under the Apache License, Version 2.0 (the "License");
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| you may not use this file except in compliance with the License.
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| You may obtain a copy of the License at
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| 
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|     http://www.apache.org/licenses/LICENSE-2.0
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| 
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| Unless required by applicable law or agreed to in writing, software
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| distributed under the License is distributed on an "AS IS" BASIS,
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| WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
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| See the License for the specific language governing permissions and
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| limitations under the License.
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| */
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| 
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| package value
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| 
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| type listUnstructured []interface{}
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| 
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| func (l listUnstructured) Length() int {
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| 	return len(l)
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| }
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| 
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| func (l listUnstructured) At(i int) Value {
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| 	return NewValueInterface(l[i])
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| }
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| 
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| func (l listUnstructured) AtUsing(a Allocator, i int) Value {
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| 	return a.allocValueUnstructured().reuse(l[i])
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| }
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| 
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| func (l listUnstructured) Equals(other List) bool {
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| 	return l.EqualsUsing(HeapAllocator, other)
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| }
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| 
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| func (l listUnstructured) EqualsUsing(a Allocator, other List) bool {
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| 	return ListEqualsUsing(a, &l, other)
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| }
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| 
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| func (l listUnstructured) Range() ListRange {
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| 	return l.RangeUsing(HeapAllocator)
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| }
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| 
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| func (l listUnstructured) RangeUsing(a Allocator) ListRange {
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| 	if len(l) == 0 {
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| 		return EmptyRange
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| 	}
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| 	r := a.allocListUnstructuredRange()
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| 	r.list = l
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| 	r.i = -1
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| 	return r
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| }
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| 
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| type listUnstructuredRange struct {
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| 	list listUnstructured
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| 	vv   *valueUnstructured
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| 	i    int
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| }
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| 
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| func (r *listUnstructuredRange) Next() bool {
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| 	r.i += 1
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| 	return r.i < len(r.list)
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| }
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| 
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| func (r *listUnstructuredRange) Item() (index int, value Value) {
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| 	if r.i < 0 {
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| 		panic("Item() called before first calling Next()")
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| 	}
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| 	if r.i >= len(r.list) {
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| 		panic("Item() called on ListRange with no more items")
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| 	}
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| 	return r.i, r.vv.reuse(r.list[r.i])
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| }
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