英文:
Mismatch types uint32 and bool in Go
问题
我有一个定义如下的C宏：
#define normalize(c, a) c = (a) + ((a) == 0xFFFFFFFF)
我正在用Go重新编写它，据我所知，Go中没有C宏这样的东西。因此，我创建了一个普通函数：
func normalize(a uint32, c *uint32) {
*c = a + (a == 0xFFFFFFFF)
}
问题是这给我带来了类型不匹配的错误。有什么办法可以修复它吗？
英文:
I have a C macro defined like this:
#define normalize(c, a) c = (a) + ((a) == 0xFFFFFFFF)
I was rewriting it in Go, and as long as I know there is no such things as C macros in Go. Therefore, I created a normal function:
func normalize(a uint32, c *uint32) {
*c = a + (a == 0xFFFFFFFF)
}
The problem is that this gives me a type mismatch error. Any ideas how to fix it?
答案1
得分: 2
所以你的 C 语言的 normalize 宏会将 c 赋值为 a，如果 a 不等于 0xffffffff，否则赋值为 0。我不确定这是什么样的归一化，但现在不是我的关注点。
根据你提供的 Go 函数签名，可以这样实现：
func normalize(a uint32, c *uint32) {
if a != 0xffffffff {
*c = a
} else {
*c = 0
}
}
然而，我不确定为什么不直接返回一个值，而是通过 c 指针进行写入？
func normalize(a uint32) uint32 {
if a != 0xffffffff {
return a
}
return 0
}
顺便提一下：对于你的 C 宏也是一样的。顺便说一下，这个宏会对 a 进行两次求值，如果你将带有副作用的函数作为 a 传递进去，可能会让你感到惊讶。有没有什么理由不使用（内联）函数来替代宏，或者至少使其求值为一个新值，而不是将 c 赋值为它呢？
英文:
So your C normalize macro assigns c to a if a is not equal to 0xffffffff, or to 0 otherwise. I'm not sure what kind of normalization it is, but it's not my concern now.
So given the Go function signature you provided, this would work:
func normalize(a uint32, c *uint32) {
if a != 0xffffffff {
*c = a
} else {
*c = 0
}
}
However, I'm not sure why not just return a value instead of writing it via c pointer?
func normalize(a uint32) {
if a != 0xffffffff {
return a
}
return 0
}
Side note: the same applies to your C macro. By the way, the macro evaluates a twice, this might come as a surprise if you ever pass some function with side effects as a. Any reason not to use (inline) function instead of a macro, or at least make it so that it evaluates to a new value, instead of assigning c to it?
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