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Home/ Questions/Q 8475191
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Editorial Team
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Editorial Team
Asked: June 10, 20262026-06-10T17:49:45+00:00 2026-06-10T17:49:45+00:00

I thought that in GO language, slices are passed by reference. But why the

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I thought that in GO language, slices are passed by reference. But why the following code doesn’t change the content of slice c? Am I missing something? Thank you.

package main

import (
    "fmt"
)


func call(c []int) {
    c = append(c, 1)
    fmt.Println(c)
}

func main() {
    c := make([]int, 1, 5)
    fmt.Println(c)
    call(c)
    fmt.Println(c)
}

The result printed is:

[0]
[0 1]
[0]

while I was expecting

[0]
[0 1]
[0 1]

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  1. Editorial Team
    Editorial Team
    2026-06-10T17:49:47+00:00Added an answer on June 10, 2026 at 5:49 pm

    The length of the slice is kept in the slice header which is not passed by reference. You can think of a slice as a struct containing a pointer to the array, a length, and a capacity.

    When you appended to the slice, you modified index 1 in the data array and then incremented the length in the slice header. When you returned, c in the main function had a length of 1 and so printed the same data.

    The reason slices work this way is so you can have multiple slices pointing to the same data. For example:

    x := []int{1,2,3}
    y := x[:2] // [1 2]
    z := x[1:] // [2 3]
    

    All three of those slices point to overlapping data in the same underlying array.

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