Exercises

For-loop

• Create a loop with the for construct. Make it loop10 times and print out the loop counter with the fmt package.

• Rewrite the loop from 1 to use goto. The keyword for may not be used.

• Rewrite the loop again so that it fills an array and then prints that array to the screen.

Answer

• There are many possibilities. One solution could be:

package main
import "fmt"
func main() {
    for i := 0; i < 10; i++ {
        fmt.Println("%d", i)
    }
}

Let’s compile this and look at the output.

    % go build for.go
    % ./for
    0
    1
    .
    .
    .
    9

• Rewriting the loop results in code that should look somethinglike this (only showing the main-function):

func main() {
    i := 0    1
Loop:            2
    if i < 10 {
        fmt.Printf("%d\n", i)
        i++
        goto Loop 3
    }
}

At 1 we define our loop variable. And at 2 we define a label and at 3 we jumpto this label.

• The following is one possible solution:

package main
import "fmt"
func main() {
    var arr [10]int 1
    for i := 0; i < 10; i++ {
        arr[i] = i 2
    }
    fmt.Printf("%v", arr) 3
}

Here 1 we create an array with 10 elements.Which we then fill 2 one by one. And finally we print it 3 with %v which letsGo to print the value for us. You could even do this in one fell swoop by using a composite literal:

fmt.Printf("%v\n", [...]int{0,1,2,3,4,5,6,7,8,9})

Average

• Write code to calculate the average of a float64 slice. Ina later exercise you will make it into a function.

Answer

• The following code calculates the average.

sum := 0.0
switch len(xs) {
case 0: 1
    avg = 0
default: 2
    for _, v := range xs {
        sum += v
    }
    avg = sum / float64(len(xs)) 3
}

Here at 1 we check if the length is zero and if so, we return 0.Otherwise we calculate the average at 2.We have to convert the value return from len to a float64to make the division work at 3.

FizzBuzz

• Solve this problem, called the Fizz-Buzz [fizzbuzz] problem:Write a program that prints the numbers from 1 to 100. But for multiplesof three print, “Fizz” instead of the number, and for multiples offive, print “Buzz”. For numbers which are multiples of both three andfive, print “FizzBuzz”.

Answer

• A possible solution to this problem is the following program.

package main
import "fmt"
func main() {
    const (
        FIZZ = 3 1
        BUZZ = 5
    )
    var p bool                 2
    for i := 1; i < 100; i++ { 3
        p = false
        if i%FIZZ == 0 { 4
            fmt.Printf("Fizz")
            p = true
        }
        if i%BUZZ == 0 { 5
            fmt.Printf("Buzz")
            p = true
        }
        if !p { 6
            fmt.Printf("%v", i)
        }
        fmt.Println()
    }
}

Here 1 we define two constants to make our code more readable, see .At 2 we define a boolean that keeps track if we already printed something.At 3 we start our for-loop, see .If the value is divisible by FIZZ - that is, 3 - , we print “Fizz” 4.And at 5 we check if the value is divisble by BUZZ – that is, 5 – if so print“Buzz”. Note that we have also taken care of the FizzBuzz case.At 6, if printed neither Fizz nor Buzz printed, we print the value.