我可以在golang的for-range迭代中使索引为int64吗？

Question

According to the spec the

for idx, val range a_slice 

statement returns idx as an integer.

Since creating a large size slice is possible, is there a way to chance idx to int64?

Thank you.

Answer 1

No, the spec specifies the type of index to be int if you use a "for" statement with a "range" clause:

Range expression                          1st value          2nd value

array or slice  a  [n]E, *[n]E, or []E    index    i  int    a[i]       E
string          s  string type            index    i  int    see below  rune
map             m  map[K]V                key      k  K      m[k]       V
channel         c  chan E, <-chan E       element  e  E

Nothing you can do about it, and nothing you should do about it. The length of the slice/array will fit into int.

It is not possible to make a slice bigger than max int. Attempting to make a larger slice with a constant expression is a compile-time error:

x := make([]struct{}, 3123456789)

Compile-time error: len argument too large in make([]struct {})

Note: size of int is implementation-specific: it is either 32-bit or 64-bit. The constant expressions here to produce the errors are for 32-bit ints (the Go Playground uses 32-bit ints).

If length is a runtime expression, it panics:

i := uint(3123456789)
y := make([]struct{}, i)

Runtime error: panic: runtime error: makeslice: len out of range

And length of array types must also fit into int: Spec: Array types:

The length is part of the array's type; it must evaluate to a non-negative constant representable by a value of type int.

Attempting to use a larger length is a compile-time error:

var x [3123456789]struct{}
type t1 [3123456789]byte
type t2 [3123456789]struct{}

All compile-time error: array bound is too large