如何在Swift中取消特定警告 [英] How to suppress a specific warning in Swift
问题描述
我有一个Swift函数,其功能如下:
I have a Swift function doing something like this:
func f() -> Int {
switch (__WORDSIZE) {
case 32: return 1
case 64: return 2
default: return 0
}
}
由于 __ WORDSIZE
是常量,因此编译器始终给出开关体内至少有一个警告。实际标记的行取决于我要建立的目标(例如,iPhone 5 vs. 6;有趣的是,iPhone 5会针对64位情况发出警告,而iPhone 6会针对32位和默认情况给出两个警告)。
Because __WORDSIZE
is a constant, the compiler always gives at least one warning in the switch body. Which lines are actually marked depends on the target I am building for (e.g. iPhone 5 vs. 6; interestingly iPhone 5 gives a warning for the 64-bit case whereas iPhone 6 gives two warnings for 32-bit and default).
我发现 #pragma
的Swift等效值为 // MARK:
,所以我尝试了
I found out that the Swift equivalent for #pragma
is // MARK:
, so I tried
// MARK: clang diagnostic push
// MARK: clang diagnostic ignored "-Wall"
func f() -> Int {
switch (__WORDSIZE) {
case 32: return 1
case 64: return 2
default: return 0
}
}
// MARK: clang diagnostic pop
但警告仍然存在, MARK
似乎没有任何作用。
but the warnings remain, the MARK
s seem to have no effect.
作为一种解决方法,我现在有这样的东西:
As a workaround, I now have something like this:
#if arch(arm) || arch(i386)
return 1
#else
#if arch(arm64) || arch(x86_64)
return 2
#else
return 0
#endif
#endif
–但是当然不一样。有什么提示吗??
– but of course this is not the same. Any hints…?
推荐答案
目前(Xcode 7.1),似乎没有办法在Swift中取消特定警告(参见例如如何快速静音警告)
At present (Xcode 7.1), there seems to be no way of suppressing a specific warning in Swift (see e.g. How to silence a warning in swift).
在您的特殊情况下,您可以用
来计算一个单词中 bytes 的数量来欺骗编译器:
In your special case, you can fool the compiler by computing the number of bytes in a word:
func f() -> Int {
switch (__WORDSIZE / CHAR_BIT) { // Or: switch (sizeof(Int.self))
case 4: return 1
case 8: return 2
default: return 0
}
}
这两个32位编译器都没有警告和64位架构。
This compiles without warnings on both 32-bit and 64-bit architectures.
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