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# Spread syntax (unpacking arguments) —JS: Functions

The spread statement in function calls is syntactically identical to the rest statement in definitions, but does the opposite. Let's take the `sum()` function as an example:

``````const sum = (...numbers) => {
let result = 0;
for (let num of numbers) {
result += num;
}
return result;
};
``````

Let's call `sum()` by applying the spread statement to an array of arguments:

``````const numbers = [1, 7, 4];
sum(...numbers); // 12
``````

The spread statement decomposes the array into arguments. The number of arguments received by the spread statement is equal to the number of array elements. In essence, the code above is converted into a call:

``````sum(numbers, numbers, numbers);
// sum(1, 7, 4);
``````

As with function definitions, the spread operator can be used in conjunction with positional arguments:

``````const numbers = [1, 7, 4];
sum(8, ...numbers); // 20

sum(8, 10, ...numbers);     // 30
sum(8, 10, 70, ...numbers); // 100
``````

Unlike the rest-operator in the definition of functions, the spread-operator does not have to be last, it can be in any position:

``````const numbers = [1, 7, 4];

sum(8, 10, ...numbers); // 30
sum(8, ...numbers, 10); // 30
sum(...numbers, 8, 10); // 30
``````

Moreover, there can be any number of spread operators and in any order:

``````const numbers1 = [1, 7, 4];
const numbers2 = [5, 5];

// sum(1, 7, 4, 5, 5);
sum(...numbers1, ...numbers2); // 22

// sum(5, 5, 1, 7, 4);
sum(...numbers2, ...numbers1); // 22

// sum(8, 1, 7, 4, 10, 5, 5);
sum(8, ...numbers1, 10, ...numbers2); // 40
`````` ## Are there any more questions? Ask them in the Discussion section.

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