reduce in javascript using curly returns undefined

Question

Just following mdn tutorial on reduce on javascript, Can someone please explain to me why below I cannot use the curly brackets around the callback?

const array1 = [1, 2, 3, 4];
    
// 0 + 1 + 2 + 3 + 4
const initialValue = 0;
    
// expected output: 10
const sumWithInitialTest = array1.reduce( (pv, cv) => 
  { pv + cv }
)
    
console.log(sumWithInitialTest)

Obviously, without curly bracket, it works fine

const array1 = [1, 2, 3, 4];
    
// 0 + 1 + 2 + 3 + 4
const initialValue = 0;
    
// expected output: 10
const sumWithInitialTest = array1.reduce( (pv, cv) => 
   pv + cv 
)
    
console.log(sumWithInitialTest)

Please read up on basic arrow function syntax. Trivial research would find e.g. https://stackoverflow.com/q/45754957/3001761. — jonrsharpe, Apr 22 '22 at 21:09
Because `{ pv + cv }` is equivalent to `{ pv + cv; return; }` — Jake Holzinger, Apr 22 '22 at 21:13

score 1 · Answer 1 · answered Apr 22 '22 at 21:14

Add return if you want to use curly bracket

const array1 = [1, 2, 3, 4];
    
// 0 + 1 + 2 + 3 + 4
const initialValue = 0;
    
// expected output: 10
const sumWithInitialTest = array1.reduce( (pv, cv) => 
    { return pv + cv }
)
    
console.log(sumWithInitialTest)

reduce in javascript using curly returns undefined

1 Answers1