Go declared and not used logic [duplicate]

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This question already has an answer here:

Short variable declaration and “variable declared and not used” error

2 answers

Why does golang compiler think the variable is declared but not used?

1 answer

Go compiler says “declared and not used” but they are being used

2 answers

using := gives unused error but using = don't in Go

3 answers

“declared and not used” Error

2 answers

Take this incredibly simple example. which shows variable assignment in and out of a block.

On compilation this results in: u declared and not used

var u string

{

    u, err := url.Parse("http://bing.com/search?q=dotnet")

    if err != nil {

        log.Fatal(err)

    }

}

log.Debug(u)

This simulates a logic block during which we might assess several things and set a var to the value we like depending on logic evaluation. How is this possible?

asked yesterday

David

9,75383458

marked as duplicate by JimB go
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yesterday

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2

Exactly as is says, the u declared inside the block is not used (the actual error will tell you the exact line). There's no magic here, and the error is probably saving you from a bug in your code.
– JimB
yesterday

1

The short variable declaration will create new u and err variables, because u is not declared in the same block. Redeclaration can only happen if the variable was declared in the same block.
– icza
yesterday

Can't believe nobody used the one sentence that clears everything up: u, err := is a declare and assign of both u and err inside a block, masking the var u string declared in a higher scope.
– Elias Van Ootegem
yesterday

add a comment |

up vote
-3
down vote

favorite

This question already has an answer here:

Short variable declaration and “variable declared and not used” error

2 answers

Why does golang compiler think the variable is declared but not used?

1 answer

Go compiler says “declared and not used” but they are being used

2 answers

using := gives unused error but using = don't in Go

3 answers

“declared and not used” Error

2 answers

Take this incredibly simple example. which shows variable assignment in and out of a block.

On compilation this results in: u declared and not used

var u string

{

    u, err := url.Parse("http://bing.com/search?q=dotnet")

    if err != nil {

        log.Fatal(err)

    }

}

log.Debug(u)

This simulates a logic block during which we might assess several things and set a var to the value we like depending on logic evaluation. How is this possible?

asked yesterday

David

9,75383458

marked as duplicate by JimB go
Users with the go badge can single-handedly close go questions as duplicates and reopen them as needed.

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yesterday

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2

Exactly as is says, the u declared inside the block is not used (the actual error will tell you the exact line). There's no magic here, and the error is probably saving you from a bug in your code.
– JimB
yesterday

1

The short variable declaration will create new u and err variables, because u is not declared in the same block. Redeclaration can only happen if the variable was declared in the same block.
– icza
yesterday

Can't believe nobody used the one sentence that clears everything up: u, err := is a declare and assign of both u and err inside a block, masking the var u string declared in a higher scope.
– Elias Van Ootegem
yesterday

add a comment |

up vote
-3
down vote

favorite

This question already has an answer here:

Short variable declaration and “variable declared and not used” error

2 answers

Why does golang compiler think the variable is declared but not used?

1 answer

Go compiler says “declared and not used” but they are being used

2 answers

using := gives unused error but using = don't in Go

3 answers

“declared and not used” Error

2 answers

Take this incredibly simple example. which shows variable assignment in and out of a block.

On compilation this results in: u declared and not used

var u string

{

    u, err := url.Parse("http://bing.com/search?q=dotnet")

    if err != nil {

        log.Fatal(err)

    }

}

log.Debug(u)

This simulates a logic block during which we might assess several things and set a var to the value we like depending on logic evaluation. How is this possible?

asked yesterday

David

9,75383458

This question already has an answer here:

Short variable declaration and “variable declared and not used” error

2 answers

Why does golang compiler think the variable is declared but not used?

1 answer

Go compiler says “declared and not used” but they are being used

2 answers

using := gives unused error but using = don't in Go

3 answers

“declared and not used” Error

2 answers

Take this incredibly simple example. which shows variable assignment in and out of a block.

On compilation this results in: u declared and not used

var u string

{

    u, err := url.Parse("http://bing.com/search?q=dotnet")

    if err != nil {

        log.Fatal(err)

    }

}

log.Debug(u)

This simulates a logic block during which we might assess several things and set a var to the value we like depending on logic evaluation. How is this possible?

This question already has an answer here:

Short variable declaration and “variable declared and not used” error

2 answers

Why does golang compiler think the variable is declared but not used?

1 answer

Go compiler says “declared and not used” but they are being used

2 answers

using := gives unused error but using = don't in Go

3 answers

“declared and not used” Error

2 answers

asked yesterday

David

9,75383458

asked yesterday

David

9,75383458

asked yesterday

David

9,75383458

asked yesterday

David

9,75383458

asked yesterday

David

9,75383458

marked as duplicate by JimB go
Users with the go badge can single-handedly close go questions as duplicates and reopen them as needed.

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yesterday

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by JimB go
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yesterday

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2

Exactly as is says, the u declared inside the block is not used (the actual error will tell you the exact line). There's no magic here, and the error is probably saving you from a bug in your code.
– JimB
yesterday

1

The short variable declaration will create new u and err variables, because u is not declared in the same block. Redeclaration can only happen if the variable was declared in the same block.
– icza
yesterday

Can't believe nobody used the one sentence that clears everything up: u, err := is a declare and assign of both u and err inside a block, masking the var u string declared in a higher scope.
– Elias Van Ootegem
yesterday

add a comment |

2

Exactly as is says, the u declared inside the block is not used (the actual error will tell you the exact line). There's no magic here, and the error is probably saving you from a bug in your code.
– JimB
yesterday

1

The short variable declaration will create new u and err variables, because u is not declared in the same block. Redeclaration can only happen if the variable was declared in the same block.
– icza
yesterday

Can't believe nobody used the one sentence that clears everything up: u, err := is a declare and assign of both u and err inside a block, masking the var u string declared in a higher scope.
– Elias Van Ootegem
yesterday

Exactly as is says, the u declared inside the block is not used (the actual error will tell you the exact line). There's no magic here, and the error is probably saving you from a bug in your code.
– JimB
yesterday

The short variable declaration will create new u and err variables, because u is not declared in the same block. Redeclaration can only happen if the variable was declared in the same block.
– icza
yesterday

Can't believe nobody used the one sentence that clears everything up: u, err := is a declare and assign of both u and err inside a block, masking the var u string declared in a higher scope.
– Elias Van Ootegem
yesterday

add a comment |

1 Answer
1

active

oldest

votes

up vote
0
down vote

Please pay attention that:

u, err := url.Parse("http://bing.com/search?q=dotnet")

in this line of code you have block scoped variable u which is differ from those one which declared in line: var u string, hence you have this error.

Moreover here you have shadowing for variable u from outer block scope which may lead to bugs, see this:

var u string

u = "blank"

{

    u, err := url.Parse("http://bing.com/search?q=dotnet")

    if err != nil {

        log.Fatal(err)

    }

    log.Printf("%#v", u)

}

log.Printf("%#v", u)

Result will be:

2018/11/21 19:12:33 &url.URL{Scheme:"http", Opaque:"", User:(*url.Userinfo)(nil), Host:"bing.com", Path:"/search", RawPath:"", ForceQuery:false, RawQuery:"q=dotnet", Fragment:""}

2018/11/21 19:12:33 "blank"

As you can see here you have even different data types, and variable u not changed after this block.

edited yesterday

answered yesterday

Vladimir Kovpak

10.3k43646

Thanks for the comment. What still seems to escape me a little is how to access the set value of a variable set within a logic block. Take for example a simple - if { u="bob1"}else{ u = "bob2"} log.Printf(u)
– David
17 hours ago

add a comment |

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
0
down vote

Please pay attention that:

u, err := url.Parse("http://bing.com/search?q=dotnet")

var u string

u = "blank"

{

    u, err := url.Parse("http://bing.com/search?q=dotnet")

    if err != nil {

        log.Fatal(err)

    }

    log.Printf("%#v", u)

}

log.Printf("%#v", u)

Result will be:

2018/11/21 19:12:33 &url.URL{Scheme:"http", Opaque:"", User:(*url.Userinfo)(nil), Host:"bing.com", Path:"/search", RawPath:"", ForceQuery:false, RawQuery:"q=dotnet", Fragment:""}

2018/11/21 19:12:33 "blank"

As you can see here you have even different data types, and variable u not changed after this block.

edited yesterday

answered yesterday

Vladimir Kovpak

10.3k43646

Thanks for the comment. What still seems to escape me a little is how to access the set value of a variable set within a logic block. Take for example a simple - if { u="bob1"}else{ u = "bob2"} log.Printf(u)
– David
17 hours ago

add a comment |

up vote
0
down vote

Please pay attention that:

u, err := url.Parse("http://bing.com/search?q=dotnet")

var u string

u = "blank"

{

    u, err := url.Parse("http://bing.com/search?q=dotnet")

    if err != nil {

        log.Fatal(err)

    }

    log.Printf("%#v", u)

}

log.Printf("%#v", u)

Result will be:

2018/11/21 19:12:33 &url.URL{Scheme:"http", Opaque:"", User:(*url.Userinfo)(nil), Host:"bing.com", Path:"/search", RawPath:"", ForceQuery:false, RawQuery:"q=dotnet", Fragment:""}

2018/11/21 19:12:33 "blank"

As you can see here you have even different data types, and variable u not changed after this block.

edited yesterday

answered yesterday

Vladimir Kovpak

10.3k43646

Thanks for the comment. What still seems to escape me a little is how to access the set value of a variable set within a logic block. Take for example a simple - if { u="bob1"}else{ u = "bob2"} log.Printf(u)
– David
17 hours ago

add a comment |

up vote
0
down vote

Please pay attention that:

u, err := url.Parse("http://bing.com/search?q=dotnet")

var u string

u = "blank"

{

    u, err := url.Parse("http://bing.com/search?q=dotnet")

    if err != nil {

        log.Fatal(err)

    }

    log.Printf("%#v", u)

}

log.Printf("%#v", u)

Result will be:

2018/11/21 19:12:33 &url.URL{Scheme:"http", Opaque:"", User:(*url.Userinfo)(nil), Host:"bing.com", Path:"/search", RawPath:"", ForceQuery:false, RawQuery:"q=dotnet", Fragment:""}

2018/11/21 19:12:33 "blank"

As you can see here you have even different data types, and variable u not changed after this block.

edited yesterday

answered yesterday

Vladimir Kovpak

10.3k43646

Please pay attention that:

u, err := url.Parse("http://bing.com/search?q=dotnet")

var u string

u = "blank"

{

    u, err := url.Parse("http://bing.com/search?q=dotnet")

    if err != nil {

        log.Fatal(err)

    }

    log.Printf("%#v", u)

}

log.Printf("%#v", u)

Result will be:

2018/11/21 19:12:33 &url.URL{Scheme:"http", Opaque:"", User:(*url.Userinfo)(nil), Host:"bing.com", Path:"/search", RawPath:"", ForceQuery:false, RawQuery:"q=dotnet", Fragment:""}

2018/11/21 19:12:33 "blank"

As you can see here you have even different data types, and variable u not changed after this block.

edited yesterday

answered yesterday

Vladimir Kovpak

10.3k43646

edited yesterday

answered yesterday

Vladimir Kovpak

10.3k43646

answered yesterday

Vladimir Kovpak

10.3k43646

answered yesterday

Vladimir Kovpak

10.3k43646

Thanks for the comment. What still seems to escape me a little is how to access the set value of a variable set within a logic block. Take for example a simple - if { u="bob1"}else{ u = "bob2"} log.Printf(u)
– David
17 hours ago

add a comment |

Thanks for the comment. What still seems to escape me a little is how to access the set value of a variable set within a logic block. Take for example a simple - if { u="bob1"}else{ u = "bob2"} log.Printf(u)
– David
17 hours ago

Thanks for the comment. What still seems to escape me a little is how to access the set value of a variable set within a logic block. Take for example a simple - if { u="bob1"}else{ u = "bob2"} log.Printf(u)
– David
17 hours ago

add a comment |

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