String Basics and Memory Representation in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a string variable named greeting with the value "Hello".

DSA Go

var greeting [1] = "Hello"

Drag options to blanks, or click blank then click option'

Aint

Bbool

Cfloat64

Dstring

Attempts:

3 left

💡 Hint

Common Mistakes

Using numeric types like int or float64 for text.

Forgetting to specify the type.

The keyword string declares a variable of string type in Go.

2fill in blank

medium

Complete the code to get the length of the string variable name.

DSA Go

length := len([1])

Drag options to blanks, or click blank then click option'

A"name"

Bname

Clen

Dstring

Attempts:

3 left

💡 Hint

Common Mistakes

Passing the string literal instead of the variable.

Passing the function name instead of the variable.

The len function returns the number of characters in the string variable name.

3fill in blank

hard

Fix the error in the code to access the first character of the string s.

DSA Go

firstChar := s[[1]]

Drag options to blanks, or click blank then click option'

A1

B'0'

C0

D"0"

Attempts:

3 left

💡 Hint

Common Mistakes

Using quotes around the index.

Using 1 instead of 0 for the first character.

Indexing strings in Go uses integer indices starting at 0 without quotes.

4fill in blank

hard

Fill both blanks to convert a string to a byte slice and back to a string.

DSA Go

bytes := []byte([1])
newStr := string([2])

Drag options to blanks, or click blank then click option'

Aoriginal

Bbytes

Cstr

Ddata

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing variable names.

Using string in place of byte slice or vice versa.

Convert the original string to bytes, then convert bytes back to string.

5fill in blank

hard

Fill all three blanks to create a map from string to int with word lengths for words longer than 3 characters.

DSA Go

lengths := map[string]int{}
for _, [3] := range words {
    if len([3]) > 3 {
        lengths[[1]] = [2]
    }
}

Drag options to blanks, or click blank then click option'

Aword

Blen(word)

Dw

Attempts:

3 left

💡 Hint

Common Mistakes

Using inconsistent variable names.

Using the wrong variable for map keys or values.

Use word as the loop variable, map key is word, and value is len(word).

Practice

1. What does a string represent in Go language?
var s string = "hello"

easy

A. A sequence of bytes representing characters

B. A mutable array of characters

C. A pointer to an integer

D. A floating-point number

Solution

1. Step 1: Understand Go string type

   In Go, a string is a sequence of bytes that represent characters, usually UTF-8 encoded.

2. Step 2: Check mutability and type

   Strings in Go are immutable, so they cannot be changed after creation. They are not arrays or pointers to integers.

3. Final Answer:

   A sequence of bytes representing characters -> Option A

4. Quick Check:

   Go strings = immutable byte sequences [OK]

Hint: Remember: Go strings are immutable byte sequences [OK]

Common Mistakes:

• Thinking strings are mutable arrays
• Confusing strings with pointers
• Assuming strings store integers

2. Which of the following is the correct way to declare and initialize a string in Go?

easy

A. var s string = 'hello'

B. var s string = "hello"

C. var s = hello

D. string s = "hello"

Solution

1. Step 1: Check string literal syntax in Go

   Go uses double quotes "" for string literals, single quotes '' are for rune (character) literals.

2. Step 2: Validate declaration syntax

   Correct declaration uses var keyword, type string, and double quotes for value.

3. Final Answer:

   var s string = "hello" -> Option B

4. Quick Check:

   Use double quotes for strings in Go [OK]

Hint: Use double quotes "" for strings, single quotes '' for runes [OK]

Common Mistakes:

• Using single quotes for strings
• Omitting quotes around string value
• Using incorrect variable declaration syntax

3. What will be the output of the following Go code?

package main
import "fmt"
func main() {
    s := "GoLang"
    fmt.Println(len(s))
}

medium

A. 6

B. 7

C. 5

D. Error

Solution

1. Step 1: Understand the string value

   The string "GoLang" has 6 characters: G, o, L, a, n, g.

2. Step 2: Use len function on string

   len(s) returns the number of bytes in the string. Since all characters are ASCII, length is 6.

3. Final Answer:

   6 -> Option A

4. Quick Check:

   len("GoLang") = 6 [OK]

Hint: len() returns byte count for ASCII strings [OK]

Common Mistakes:

• Counting characters incorrectly
• Expecting len() to count runes instead of bytes
• Assuming error due to missing imports

4. Identify the error in the following Go code snippet:

package main
import "fmt"
func main() {
    s := "hello"
    s[0] = 'H'
    fmt.Println(s)
}

medium

A. No error; code runs fine

B. Missing import for fmt package

C. Incorrect string declaration syntax

D. Strings are immutable; cannot assign to s[0]

Solution

1. Step 1: Check string mutability in Go

   Strings in Go are immutable, so you cannot change individual bytes like s[0].

2. Step 2: Analyze the assignment statement

   Attempting s[0] = 'H' causes a compile-time error because strings cannot be modified.

3. Final Answer:

   Strings are immutable; cannot assign to s[0] -> Option D

4. Quick Check:

   Strings immutable = no s[0] assignment [OK]

Hint: Remember: Go strings cannot be changed after creation [OK]

Common Mistakes:

• Trying to modify string bytes directly
• Ignoring immutability errors
• Confusing strings with byte slices

5. Given the string s := "Go😊Lang", what will len(s) return and why?

hard

A. 7, because len counts characters

B. 8, because emoji counts as two bytes

C. 10, because emoji is 4 bytes plus other characters

D. 6, because emoji is ignored

Solution

1. Step 1: Analyze string content and encoding

   The string "Go😊Lang" has 7 characters: G, o, 😊, L, a, n, g. The emoji 😊 is a Unicode character encoded in UTF-8 using 4 bytes.

2. Step 2: Calculate total byte length

   Characters G, o, L, a, n, g are 1 byte each (6 bytes). The emoji is 4 bytes. Total bytes = 6 + 4 = 10 bytes.

3. Step 3: Verify len(s) output

   len(s) returns the number of bytes, so it returns 10, not the number of characters.

4. Final Answer:

   10, because emoji is 4 bytes plus other characters -> Option C

5. Quick Check:

   len counts bytes, emoji uses multiple bytes [OK]

Hint: len() counts bytes, not characters; emojis use multiple bytes [OK]

Common Mistakes:

• Assuming len counts characters, not bytes
• Ignoring multi-byte encoding of emojis
• Thinking emoji counts as one byte