Rust Function: Borrowed integers sum

Last update on December 21 2024 10:21:04 (UTC/GMT +8 hours)

Write a Rust function that borrows two integers and returns their sum.

Sample Solution:

Rust Code:

// Define a function named 'sum_of_integers' that borrows two integers and returns their sum
fn sum_of_integers(a: &i32, b: &i32) -> i32 {
    *a + *b // Dereference the integers and return their sum
}

fn main() {
    let num1 = 10;
    let num2 = 20;

    // Call the 'sum_of_integers' function and pass references to 'num1' and 'num2' to borrow them
    let result = sum_of_integers(&num1, &num2);

    // Print the result of the sum
    println!("Sum of {} and {} is: {}", num1, num2, result);
}

Output:

Sum of 10 and 20 is: 30

Explanation:

Here is a brief explanation of the above Rust code:

• fn sum_of_integers(a: &i32, b: &i32) -> i32 { ... }: This is a function named sum_of_integers that borrows two integers (&i32) and returns their sum as an integer (i32). The parameters 'a' and 'b' are references to integers, indicating borrowing.
• Inside the function:
• We dereference 'a' and 'b' using the "*" operator to access the values they point to.
• We add the dereferenced integers and return their sum.
• In the main function,
• Define two integers 'num1' and 'num2'.
• Call the "sum_of_integers()" function and pass references to 'num1' and 'num2' using & to borrow them.
• Finally we print the result of the sum.

Rust Code Editor:

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