Why does this Golang code produce deadlock?

I'm new to Golang and I have hard time figuring out why exactly the following code produces deadlock. Also, how can I fix it so it works?

    package main
    
    import "fmt"
    
    func main() {
        m := make(map[int]chan string)
        go func() {
            m[0] = make(chan string)
            m[0] <- "abab"
        }()
        fmt.Println(<-m[0])
    }

EDIT:

Thanks for your answers! Unfortunately, initializing m[0] with

m[0] = make(chan string)

before launching a new goroutine is not exactly what I want. My problem is: is there a way to create channels "dynamically"? E.g. I have a map m of type map[int]chan string and I receive requests that contain something like id of type int. I would like to send a message via channel map[id], but initializing channels for every int would be too costly. How do I solve/work around this?

So, in other words, I would like to have a separate job queue for every id and initialize each queue lazily.

Solution

Updated answer after OP updated the question

You can just loop on all the keys in your map, maybe have another goroutine that keeps looping on all the keys. Obviously if a key hasnt been initialized, then it wont come up in the for range loop. For each key, you can then start a goroutine that listens so it doesnt block, or you can use a buffered channels so they wont block up to the buffer limit. You can also preferably use a waitGroup, rather than the time.Sleep(), these are only for this trivial example.

package main

import (
"fmt"
    "time"
)
func main() {
    m := make(map[int]chan string)

    go func() {
        m[0] = make(chan string)
        m[0] <- "abab"
    }()

    time.Sleep(time.Second * 1)  //sleep so the above goroutine initializes the key 0 channel

    for key := range m{      //loop on all non-nil keys
        fmt.Println(key)
        go func(k int){        // goroutine to listen on this channel
            fmt.Println(<- m[k])
        }(key)
    }
    time.Sleep(time.Second * 1) //sleep so u can see the effects of the channel recievers


}

Old answer

This is how the flow is. The main goroutine starts. The map is created. The main goroutine encounters another goroutine. It spawns said goroutine and goes on with its life. Then it meets this line, fmt.Println(<-m[0]), which is a problem, since the map is indeed initialized, but the channel in the map itself isnt initialized! By the time the main goroutine has reached fmt.Println(<-m[0]), the other goroutine hadn't yet initialized the channel! So its a simple fix, just initialize the channel before spawning the goroutine and you're good to go!

package main

import "fmt"

func main() {
    m := make(map[int]chan string)
    m[0] = make(chan string)

    go func() {
        m[0] <- "abab"
    }()
    fmt.Println(<-m[0])
}

Edit: Note that fmt.Println(<-m[0]) is blocking, which means that if in that other goroutine, you dont send on the channel, you will also go into a deadlock, since you are trying to recieve on the channel when no one is actually sending.