How does append() work internally in Go? When does it allocate new memory?

append adds elements to a slice. When there is spare capacity in the backing array, it writes directly into it and increments the length — no allocation. When capacity is exhausted, it allocates a new, larger array, copies all elements, and returns a new slice header. This is why you must always assign the return value of append.

s := make([]int, 3, 5)     // len=3, cap=5, backing array has 2 spare slots
fmt.Println(len(s), cap(s)) // 3 5

s = append(s, 10)           // len=4, cap=5 — no allocation (has room)
s = append(s, 20)           // len=5, cap=5 — no allocation (fills last slot)
s = append(s, 30)           // len=6, cap>=10 — REALLOCATES! new backing array
fmt.Println(len(s), cap(s)) // 6  10  (capacity grew to ~2x)

// Sub-slice shares backing array — subtle bug territory
a := []int{1, 2, 3, 4, 5}
b := a[1:3]                 // [2 3] — shares backing array with a
fmt.Println(cap(b))          // 4 (from position 1 to end of a's array)

b = append(b, 99)           // writes to a[3]! No reallocation needed
fmt.Println(a)               // [1 2 3 99 5] — a was modified!

// FIX: use copy to get an independent slice
b2 := make([]int, len(a[1:3]))
copy(b2, a[1:3])             // independent backing array
b2 = append(b2, 99)         // safe — doesn't affect a

// Pre-allocate when final length is known — avoids repeated reallocation
result := make([]int, 0, len(input))  // cap=len(input), no mid-loop alloc
for _, v := range input {
    result = append(result, v*2)
}

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