函数调用时自动将结构体转换为指针
我们的项目中有一个看起来非常正常的 printf 样式函数，经过修改，%g 格式意味着打印 GUID 而不是普通的浮点类型。对于我们的例子，GUID 看起来像这样：
struct guid {
uint32_t Data1;
uint16_t Data2;
uint16_t Data3;
uint8_t Data4[8];
};
实际上，打印函数期望传递一个指向 GUID 的指针，而不是结构本身：
struct guid myGuid = { 0x867FD1E7, 0x9AA7, 0x472A, { 0xAA, 0x56, 0xF2, 0xDA, 0x66, 0x62, 0xCD, 0x4D } };
print("%g", &myGuid);
然而，在源代码库中有几个地方，由于某种原因，整个 guid 都被传递：
print("%g", myGuid);
这种调用风格似乎在 MSVC2003 中工作得很好 - 是否有一些 ABI 要求使编译器将该函数调用风格转换为实际上在幕后传递指针？当将此代码库移植到使用 clang/llvm 时，它肯定不会做同样的事情。
有人可以解释为什么第二个版本的调用适用于 MSVC 吗？如果能提供相应文档的指针，我们将不胜感激！
We have a pretty normal looking printf style function in our project, with the modification that the %g format means to print a GUID instead of the normal floating-point type. For our case, a GUID looks something like this:
struct guid {
uint32_t Data1;
uint16_t Data2;
uint16_t Data3;
uint8_t Data4[8];
};
In reality, the print function expects a pointer to the GUID to be passed, as opposed to the structure itself:
struct guid myGuid = { 0x867FD1E7, 0x9AA7, 0x472A, { 0xAA, 0x56, 0xF2, 0xDA, 0x66, 0x62, 0xCD, 0x4D } };
print("%g", &myGuid);
There are several places in the source base, however, where for some reason the entire guid is passed:
print("%g", myGuid);
This style of call seems to work fine with MSVC2003 - is there some ABI requirement that makes the compiler translate that function call style to actually pass a pointer behind the scenes? When porting this codebase to use clang/llvm, it certainly doesn't do the same thing.
Can somebody explain why the second version of the call works with MSVC? A pointer to the appropriate documentation would be much appreciated!
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我想我在 MSDN 上找到了一些说明：
看来是时候修复 clang 了！
I think I found some clarification on MSDN:
Looks like it's time to fix clang!