assignment mismatch 1 variable but json.marshal returns 2 values

Unmarshalling JSON with Null Boolean Values in Go

Whilst working on Monitoring Agent 1 , which is written in Go , I added validation for the configuration file read by the Monitoring Agent process when it is started. The configuration file is formatted in JSON and unmarshalled to a struct type.

When a mandatory configuration item is added and an administrator upgrades Monitoring Agent , they should see a useful error message about the missing configuration item instead of having the process run with unpredictable or insecure results.

The program therefore needs to ensure certain values have been set. In Go , when storage is allocated for a variable without explicit initialization, the variable is given the zero value for it’s type. For a bool type, this is 0 which is also false .

That means when configuration data is read from a file and unmarshalled, the code needs to work out whether a bool value is either false or omitted.

Solution 1: Use Pointers

Instead of defining bool types for members of the struct that JSON is unmarshaled into, we could use pointers ( *bool ) instead.

When a value is not provided to json.Unmarshal() for a member of the struct , the pointer value will be nil. Thus mystruct.var == nil can be used to see the value was included in the JSON.

Go Playground example.

The downside of this solution is that interrogating or changing the value means dereferencing or indirecting the variable, for example *input.IsGoodCode = false , as opposed to input.IsGoodCode = false .

Solution 2: Use a Custom Struct Type

I don’t like smearing code with unnecessary punctuation. Fortunately, there is another way. Instead of relying on pointers, a struct type can be used:

This is a struct with two members. IsTrue is the actual value of the bool . HasValue indicates whether or not IsTrue has been explicitly set.

Instead of input.IsGoodCode != nil and *input.IsGoodCode = true we can write input.IsGoodCode.HasValue and input.IsGoodCode.IsTrue = yes .

Before the NullBool struct can be used, we have to create a struct to represent the JSON data:

The final step is changing the way that the code unmarhals JSON into the NullBool struct .

Taking advantage of Go’s UnmarshalJSON() interface , a method on NullBool is implemented with the signature UnmarshalJSON(b []byte) error :

This method is run by any call to json.Unmarshal() when populating any instance of the NullBool struct , for which data is provided .

If a key defined in the struct representing the JSON data is omitted from the JSON string which is unmarshalled into it, then NullBool ’s UnmarshalJSON() is not run.

With bool values initiatlised to 0 ( false ), both IsTrue and HasValue will be false for any instance of NullBool that has not been populated by the unmarshalling of JSON.

Conversely, we know that instances of NullBool with IsTrue:false and HasValue:true have a corresponding key in a JSON string with a value of false .

Click here for the full Go Playground example.

Go is well documented, widely supported and easy to learn. Like many languages that use zero type initialisation , extra care needs to be afforded in some situations, like unmarshalling JSON.

The UnmarshalJSON interface is very powerful because it gives developers control over how custom types are populated when unmarshalling JSON. Not only can it be used to change the way objects are hydrated, but it can be used to add additional validation.

Variables - golang.org
The Zero Value - golang.org
Interface Types - golang.org
Struct Types - golang.org

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How can I track down a function return mismatch in Go?

How can I track down a function return mismatch in Golang? In two different build environments I am seeing a difference. Both cases should be Visual Studio Code remote to a Linux box, using Go 1.12 in module mode. The broken case is where I am driving the build using the Golang:1.12 Docker image. Below is simplified from where I'm seeing the problem.

So for this sample, derived from https://github.com/satori/go.uuid :

The unexpected build error is:

In the environment where I encounter this problem, in Visual Studio Code when I hover with mouse over the call to uuid.NewV4() I see:

and hover over uuid shows:

I may switch to a different uuid package to work around this -- but I want to understand better how to figure this out.

Use Go modules to pin the version of go.uuid. – Volker Commented Apr 16, 2020 at 14:06
@Volker thanks -- I had it on 1.2.0 in go.mod, now forced to 1.2.1, seems to work as I had expected now. – learning2learn Commented Apr 16, 2020 at 14:44

I guess problem is in different github.com/satori/go.uuid module versions. You can see, that NewV4 function signature was updated to NewV4() (uuid.UUID, error) in latest version v1.2.0 . Before that it was func NewV4() UUID

Then resolve to a specific version as in this question:

How to point Go module dependency in go.mod to a latest commit in a repo?

Thanks, do you know how VSCode ( guess via gopls) picks the function help information it displays? – learning2learn Commented Apr 16, 2020 at 14:51
@learning2learn My guess is gopls, yes. But really, i don't know about VSCode internals(and don't use it) ;) – Grigoriy Mikhalkin Commented Apr 16, 2020 at 14:55

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assignment mismatch: 2 variables but uuid.NewV4 returns 1 values #100

chingiz2387 commented Sep 13, 2019

Hello everyone, I started experimenting with mod. When I initialize the modules in my project, all libraries are loaded and github.com/satori/go.uuid is loaded accordingly.. But when starting go run server.go, I get an error #0",
"severity": 8,
"message": "cannot initialize 2 variables with 1 values",
"source": "LSP",
"startLineNumber": 41,
"startColumn": 20,
"endLineNumber": 41,
"endColumn": 30
}

I am using vscode.

My project is outside gopath

rajmaniar commented Sep 18, 2019

I struggled with this today before I realized the api for NewV4 at the latest semver tag v1.2.0 has one return variable but the head of master branch on the repo has two.

vs

go modules will always try to collect the latest semver tag by default so you need to do something like:

👍 15 reactions

Sorry, something went wrong.

3013216027 commented Sep 24, 2019

Same to me, the default behavior of go mod confuse with the old tag.
I wonder if it's possible to add a new version tag :)

👍 3 reactions

cameracker commented Oct 16, 2019

this package is obsolete.

Recommend using

3013216027 commented Nov 5, 2019

@CameronAckermanSEL thks, I've moved to instead

👍 2 reactions
👎 1 reaction

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COMMENTS

mongodb
I've done almost everything but I have problem with json.marshal(). Can . Skip to main content. Stack Overflow. About; Products OverflowAI; ... It returns two values. The problem here is that you just give it one variable to get those two values: b := json.Marshal(result) So you just have to correct it this way: b, err := json.Marshal(result) ...
go
The string conversion you are trying to perform requires a single argument, but the json.Marshal function returns two ([]byte and error). You need to store the first return value and then do the conversion:
go 我这里的代码一个不理解的报错，麻烦会的看下谢谢
go 我这里的代码一个不理解的报错，麻烦会的看下谢谢. 凉瓜. 70 1 11 20. 发布于. 2019-04-09. 报错信息是：serviceservice.go:38:11: assignment mismatch: 2 variables but 1 values. 代码如下. package service. import (.
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In this example, we define a Person struct with two fields, Name and Age. We create a Person object with some values, and then marshal it into a JSON string using the json.Marshal function. We ...
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We can unmarshal the "value" from JSON into a variable of type "Value," as demonstrated in the snippet below: func (v ... if len (v) == 1 {return json. Marshal (v [0])} var vals [] string = [] string (v) return json. Marshal (vals)} In summary, this article provides a technique for custom marshaling and unmarshaling of a value that can either ...
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Assignment mismatch: 2 variable but returns three values?
func2() returns three values, and you assign the returned values of func2 to two variables. func2 expects three variables. If you post the definition of func2 we can point out the specific location where this is happening, but it might be more useful for the long term to go through the Go tour as Jakob above suggests.
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go:51: assignment count mismatch: 2 = 1. errors. I'm working my way through the book as well as the Udemy training. I think I get the general idea that for things like this one: Executes the GET request and is returned a byte array. Passes the byte array to ioutil and ReadAll to convert that to a slice of unsigned int8 values.
Unmarshalling JSON with Null Boolean Values in Go
Instead of defining bool types for members of the struct that JSON is unmarshaled into, we could use pointers (*bool) instead. When a value is not provided to json.Unmarshal() for a member of the struct, the pointer value will be nil. Thus mystruct.var == nil can be used to see the value was included in the JSON.
Error: (random.go) assignment mismatch: 2 variables but 1 values
Error: (random.go) assignment mismatch: 2 variables but 1 values #355. Closed gencer opened this issue Aug 1, 2018 · 3 comments Closed Error: (random.go) assignment mismatch: 2 variables but 1 values #355. gencer opened this issue Aug 1, 2018 · 3 comments Comments. Copy link Contributor.
table/table.go:357:7: assignment mismatch: 1 variable but z
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assignment mismatch: 2 variables but uuid.NewV4 returns 1 values
assignment mismatch: 2 variables but uuid.NewV4 returns 1 values #106. ... Open assignment mismatch: 2 variables but uuid.NewV4 returns 1 values #106. barry-en opened this issue Mar 5, 2020 · 6 comments Comments. Copy link barry-en commented Mar 5, 2020. I use go mod go mod init
assignment mismatch: 2 variables but uuid.NewV4 returns 1 values
assignment mismatch: 2 variables but uuid.NewV4 returns 1 values #34280. Closed chingiz2387 opened this issue Sep 13, 2019 · 3 comments Closed assignment mismatch: 2 variables but uuid.NewV4 returns 1 values #34280. chingiz2387 opened this issue Sep 13, 2019 · 3 ... "cannot initialize 2 variables with 1 values", "source": "LSP ...
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./prog.go:8:7: assignment mismatch: 1 variable but fmt.Printf returns 2 values ./prog.go:9:2: cannot refer to unexported name fmt.println Explanation: The above program will generate a syntax error, because the function fmt.Printf() returns two values but we used only one value as an lvalue. Golang Find Output Programs »
How can I track down a function return mismatch in Go?
./main.go:11:5 assignment mismatch: 2 variables but uuid.NewV4() returns 1 values In the environment where I encounter this problem, in Visual Studio Code when I hover with mouse over the call to uuid.NewV4() I see: func uuid.NewV4() (uuid.UUID, error) uuid.NewV4 on pkg.go.dev NewV4 returns random generated UUID. and hover over uuid shows:
assignment mismatch: xlsx.GetRows · Issue #75 · shenwei356/csvtk
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assignment mismatch: 2 variables but uuid.NewV4 returns 1 values
assignment mismatch: 2 variables but uuid.NewV4 returns 1 values #100. Open chingiz2387 opened this issue Sep 13, 2019 · 4 comments Open assignment mismatch: 2 variables but uuid.NewV4 returns 1 values #100. ... I struggled with this today before I realized the api for NewV4 at the latest semver tag v1.2.0 has one return variable but the head ...