检查宏 c 中的 void 指针大小

Question

我需要检查 void 指针是否适合 8 个字节，所以我检查它的长度是 4 还是 8。我知道这些值是唯一的，我可以使用 _W64，只是一个好奇的检查。

#include <windows.h>
#if (sizeof(void *) == 4)
#define IS64 0
#elif (sizeof(void *) == 8)
#define IS64 1
#else
#error "Pointer size 4 nor 8, make changes in app"
#endif
int APIENTRY WinMain(HINSTANCE hInstance,
        HINSTANCE hPrevInstance, 
        LPTSTR    lpCmdLine, 
        int       cmdShow)
{
    if (IS64) MessageBoxA(NULL, "is 64", "info", MB_OK);
    else MessageBoxA(NULL, "is 32", "info", MB_OK);
    return(0);
}

此代码无效，我收到错误 app.c(2) : fatal error C1017: invalid integer constant expression

我将宏更改为:

#if (sizeof(void *) == sizeof(int32_t))
#define IS64 0
#elif (sizeof(void *) == sizeof(int64_t))
#define IS64 1
#else
#error "Pointer size 4 nor 8, make changes in app"
#endif

同样的错误。这里有任何解决方法吗？

Answer 1

GCC 带有一堆预处理器定义；

为了解决这个特定问题，它提供了 __SIZEOF_POINTER__，其值是 8 或 4，具体取决于架构。

我不是 MS 编译器的专家，但我猜他们提供了等效的定义；

对参数的快速搜索(来源:https://learn.microsoft.com/it-it/cpp/preprocessor/predefined-macros?view=vs-2019)显示以下内容:

_M_AMD64 Defined as the integer literal value 100 for compilations that target x64 processors. Otherwise, undefined.

_M_ARM Defined as the integer literal value 7 for compilations that target ARM processors. Otherwise, undefined.

_M_ARM_ARMV7VE Defined as 1 when the /arch:ARMv7VE compiler option is set for compilations that target ARM processors. Otherwise, undefined.

_M_ARM_FP Defined as an integer literal value that indicates which /arch compiler option was set for ARM processor targets. Otherwise, undefined.

_M_ARM64 Defined as 1 for compilations that target 64-bit ARM processors. Otherwise, undefined.

使用这些信息，您可以为您的定义分配一个值。