英文:
Testing for deep equality with json marshaling in golang
问题
给定这两个测试用例：
func TestEqualWhat(t *testing.T) {
testMarshalUnmarshal(t, map[string]interface{}{"a":"b"})
testMarshalUnmarshal(t, map[string]interface{}{"a":5})
}
其中testMarshalUnmarshal辅助函数只是将对象编组为JSON，然后再解组出来：
func testMarshalUnmarshal(t *testing.T, in map[string]interface{}) {
//将对象编组为字符串
jsb, err := json.Marshal(in);
if err != nil {
log.Printf("无法编组消息")
t.FailNow()
}
//解组为一个映射
res := make(map[string]interface{})
if err := json.Unmarshal(jsb, &res); err != nil { t.FailNow() }
if !reflect.DeepEqual(in, res) {
log.Printf("\n期望值 %#v\n但得到了 %#v", in, res)
t.FailNow()
}
}
为什么第一个测试用例通过了，而第二个测试用例失败了？测试的输出如下：
期望值 map[string]interface {}{"a":5}
但得到了 map[string]interface {}{"a":5}
--- FAIL: TestEqualWhat (0.00 秒)
这里是类似的代码在Go Playground上，这样你可以轻松地尝试一下。
英文:
Given these two test cases:
func TestEqualWhat(t *testing.T) {
testMarshalUnmarshal(t, map[string]interface{}{"a":"b"})
testMarshalUnmarshal(t, map[string]interface{}{"a":5})
}
Where the testMarshalUnmarshal helper just marshals to json and back out:
func testMarshalUnmarshal(t *testing.T, in map[string]interface{}) {
//marshal the object to a string
jsb, err := json.Marshal(in);
if err != nil {
log.Printf("Unable to marshal msg")
t.FailNow()
}
//unmarshal to a map
res := make(map[string]interface{})
if err := json.Unmarshal(jsb, &res); err != nil { t.FailNow() }
if !reflect.DeepEqual(in, res) {
log.Printf("\nExpected %#v\nbut got %#v", in, res)
t.FailNow()
}
}
Why does the first test case pass and the second fail? The output of the test is this:
Expected map[string]interface {}{"a":5}
but got map[string]interface {}{"a":5}
--- FAIL: TestEqualWhat (0.00 seconds)
Here is similar code on the go playground so you can have a hack at it easily.
答案1
得分: 16
我弄清楚了！JSON只有一种数值类型，即浮点数，所以在编组/解组过程中，所有整数都会转换为Float64。因此，在res映射中，5是一个float64而不是一个int。
这里是一个Go Playground，提供了我所说的上下文和证据。
英文:
I figured it out! JSON only has one numerical type, which is floating point, so all integers are converted to Float64 in the marshal/unmarshal process. So, in the res map, the 5 is a float64 instead of an int.
Here is a go playground that provides context and evidence of what I'm talking about.
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