英文:
Does the golang `range` keyword botch uint type information?
问题
考虑以下这个Go语言程序：
func main() {
one := uint(1)
ones := []uint{1, 1, 1}
for x := range ones {
if x != one {
print("ERR")
}
}
}
当我尝试编译时，出现了一个意外的错误：
$ go build foo.go
# command-line-arguments
./foo.go:7: invalid operation: x != one (mismatched types int and uint)
为什么Go认为x的类型是int而不是uint？
英文:
Consider this golang program:
<!-- language: lang-go -->
func main() {
one := uint(1)
ones := []uint{1, 1, 1}
for x := range ones {
if x != one {
print("ERR")
}
}
}
When I try to compile I get an unexpected error:
$ go build foo.go
# command-line-arguments
./foo.go:7: invalid operation: x != one (mismatched types int and uint)
Why does go think x has type int instead of uint?
答案1
得分: 4
range返回的第一个值是索引，而不是值。你需要的是：
func main() {
one := uint(1)
ones := []uint{1, 1, 1}
for _, x := range ones {
if x != one {
print("ERR")
}
}
}
range返回的第一个值是索引，而不是值。你需要的是：
func main() {
one := uint(1)
ones := []uint{1, 1, 1}
for _, x := range ones {
if x != one {
print("ERR")
}
}
}
英文:
The first value returned by range is the index, not the value. What you need is:
func main() {
one := uint(1)
ones := []uint{1, 1, 1}
for _, x := range ones {
if x != one {
print("ERR")
}
}
}
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