Rust Function: Check vowel or consonant

Last update on December 21 2024 10:22:35 (UTC/GMT +8 hours)

Write a Rust function that takes a tuple (char, i32) and returns "Vowel" if the first element is a vowel ('a', 'e', 'i', 'o', 'u') and "Consonant" otherwise.

Sample Solution:

Rust Code:

// Define a function named 'check_char_type' that takes a tuple '(ch, num)' as input and returns a static string slice.
fn check_char_type(tuple: (char, i32)) -> &'static str {
    // Destructure the tuple into its elements.
    let (ch, _) = tuple;
    // Check if the character 'ch' is a vowel ('a', 'e', 'i', 'o', 'u').
    if ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u' {
        // If `ch` is a vowel, return "Vowel".
        "Vowel"
    } else {
        // If 'ch' is not a vowel, return "Consonant".
        "Consonant"
    }
}

fn main() {
    // Define a tuple containing a character and an integer.
    let my_tuple: (char, i32) = ('e', 10);
    // Print the result of calling 'check_char_type' function with 'my_tuple' as input.
    println!("{}", check_char_type(my_tuple)); // Output: Vowel
}

Output:

Vowel

Explanation:

In the exercise above,

• Define a function "check_char_type()" that takes a tuple (char, i32) as input and returns a static string slice (&'static str).
• Inside the function, we destructure the input tuple into its two elements using pattern matching. We only need the first element 'ch', so we use _ to ignore the second element.
• We then check if the character 'ch' is a vowel ('a', 'e', 'i', 'o', 'u'). If it matches any of these vowels, we return the string "Vowel".
• If the character is not a vowel, we return the string "Consonant".
• In the "main()" function, we define a tuple 'my_tuple' containing a character 'a' and an integer 10.
• We call the "check_char_type()" function with 'my_tuple' as input and print the result. In this case, the output will be "Vowel", indicating that the character 'a' is indeed a vowel.

Rust Code Editor:

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