javascript reduce sum
console.log(
[1, 2, 3, 4].reduce((a, b) => a + b, 0)
)
console.log(
[].reduce((a, b) => a + b, 0)
)const arrSum = arr => arr.reduce((a,b) => a + b, 0)arrSum = function(arr){ return arr.reduce(function(a,b){ return a + b }, 0);}let nums = [1, 2, 3];
nums.reduce((curr, next) => curr + next);[1,2,3,4,5].reduce((acc, current)=>acc+current, 0)
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