如何确定是否在Xcode中编译64位iOS [英] How to determine if compiling for 64-bit iOS in Xcode

问题描述

考虑以下函数

CGSize CGSizeIntegral(CGSize size)
{
    return CGSizeMake(ceilf(size.width), ceilf(size.height));
}

CGSize 实际上包含两个 CGFloat s， CGFloat 的定义根据体系结构而变化：

CGSize actually consists of two CGFloats, and CGFloat's definition changes depending on the architecture:

typedef float CGFloat;// 32-bit
typedef double CGFloat;// 64-bit

因此，上述代码在64位系统上是错误的，需要更新类似

So, the above code is wrong on 64-bit systems, and needs to be updated with something like

CGSize CGSizeIntegral(CGSize size)
{
#if 64_bit
    return CGSizeMake(ceil(size.width), ceil(size.height));
#else
    return CGSizeMake(ceilf(size.width), ceilf(size.height));
#endif
}

肯定有编译器宏/常量（对于Mac，我们可以使用 INTEL_X86 ）但我无法在 64位转换指南。

There is surely a compiler macro/constant for this (for Mac we can use INTEL_X86 for example) but I haven't been able to find this in the 64-bit transition guide.

如何确定正在构建的体系结构？

推荐答案

确定您是否正在编译64位，使用 __ LP64 __ ：

To determine if you are compiling for 64-bit, use __LP64__:

#if __LP64__
    return CGSizeMake(ceil(size.width), ceil(size.height));
#else
    return CGSizeMake(ceilf(size.width), ceilf(size.height));
#endif

__ LP64 __ 看台因为longs and pointers是64位并且是架构中立的。

__LP64__ stands for "longs and pointers are 64-bit" and is architecture-neutral.

根据你的转换指南它也适用于iOS：

According to your transition guide it applies for iOS as well:

编译64位
运行时时，编译器定义 __ LP64 __ 宏。

但是，处理用例的首选方法是使用 CGFLOAT_IS_DOUBLE 。无法保证 __ LP64 __ 总是意味着 CGFloat 是一个双倍，但它将保证 CGFLOAT_IS_DOUBLE 。

However, the preferred way to handle your use case is to use CGFLOAT_IS_DOUBLE. There is no guarantee that __LP64__ will always mean the CGFloat is a double, but it would be guaranteed with CGFLOAT_IS_DOUBLE.

#if CGFLOAT_IS_DOUBLE
    return CGSizeMake(ceil(size.width), ceil(size.height));
#else
    return CGSizeMake(ceilf(size.width), ceilf(size.height));
#endif

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