C++ 打印 GUID 变量
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Print a GUID variable
提问by Jeni
I have a GUID variable and I want to write inside a text file its value. GUID definition is:
我有一个 GUID 变量，我想在一个文本文件中写入它的值。GUID定义是：
typedef struct _GUID { // size is 16
DWORD Data1;
WORD Data2;
WORD Data3;
BYTE Data4[8];
} GUID;
But I want to write its value like:
但我想写它的价值，如：
CA04046D-0000-0000-0000-504944564944
I observed that:
我观察到：
Data1holds the decimal value for CA04046DData2holds the decimal value for 0Data3holds the decimal value for next 0
Data1保存 CA04046D 的十进制值Data2保存 0 的十进制值Data3保存下一个 0 的十进制值
But what about the others?
但是其他人呢？
I have to interpret myself this values in order to get that output or is there a more direct method to print such a variable?
我必须自己解释这些值才能获得该输出，还是有更直接的方法来打印这样的变量？
回答by JustinB
Sometimes its useful to roll your own. I liked fdioff's answer but its not quite right. There are 11 elements of different sizes.
有时自己滚动很有用。我喜欢 fdioff 的回答，但不太正确。有11个不同大小的元素。
printf("Guid = {%08lX-%04hX-%04hX-%02hhX%02hhX-%02hhX%02hhX%02hhX%02hhX%02hhX%02hhX}",
guid.Data1, guid.Data2, guid.Data3,
guid.Data4[0], guid.Data4[1], guid.Data4[2], guid.Data4[3],
guid.Data4[4], guid.Data4[5], guid.Data4[6], guid.Data4[7]);
Output: "Guid = {44332211-1234-ABCD-EFEF-001122334455}"
Refer to Guiddef.h for the GUID layout.
有关 GUID 布局，请参阅 Guiddef.h。
Same, as a method:
同样，作为一种方法：
void printf_guid(GUID guid) {
printf("Guid = {%08lX-%04hX-%04hX-%02hhX%02hhX-%02hhX%02hhX%02hhX%02hhX%02hhX%02hhX}",
guid.Data1, guid.Data2, guid.Data3,
guid.Data4[0], guid.Data4[1], guid.Data4[2], guid.Data4[3],
guid.Data4[4], guid.Data4[5], guid.Data4[6], guid.Data4[7]);
}
you can also pass a CLSID to this method.
您还可以将 CLSID 传递给此方法。
回答by John Sibly
Use the StringFromCLSIDfunction to convert it to a string
使用StringFromCLSID函数将其转换为字符串
e.g.:
例如：
GUID guid;
CoCreateGuid(&guid);
OLECHAR* guidString;
StringFromCLSID(guid, &guidString);
// use guidString...
// ensure memory is freed
::CoTaskMemFree(guidString);
Also see the MSDN definition of a GUIDfor a description of data4, which is an array containing the last 8 bytes of the GUID
另请参阅MSDN 对 GUID 的定义以了解 data4，它是一个包含 GUID 的最后 8 个字节的数组
回答by Jeni
In case you prefer C++ way
如果您更喜欢 C++ 方式
std::ostream& operator<<(std::ostream& os, REFGUID guid){
os << std::uppercase;
os.width(8);
os << std::hex << guid.Data1 << '-';
os.width(4);
os << std::hex << guid.Data2 << '-';
os.width(4);
os << std::hex << guid.Data3 << '-';
os.width(2);
os << std::hex
<< static_cast<short>(guid.Data4[0])
<< static_cast<short>(guid.Data4[1])
<< '-'
<< static_cast<short>(guid.Data4[2])
<< static_cast<short>(guid.Data4[3])
<< static_cast<short>(guid.Data4[4])
<< static_cast<short>(guid.Data4[5])
<< static_cast<short>(guid.Data4[6])
<< static_cast<short>(guid.Data4[7]);
os << std::nouppercase;
return os;
}
Usage:
用法：
static const GUID guid =
{ 0xf54f83c5, 0x9724, 0x41ba, { 0xbb, 0xdb, 0x69, 0x26, 0xf7, 0xbd, 0x68, 0x13 } };
std::cout << guid << std::endl;
Output:
输出：
F54F83C5-9724-41BA-BBDB-6926F7BD6813
F54F83C5-9724-41BA-BBDB-6926F7BD6813
回答by Rost
In case when your code uses ATL/MFC you also could use CComBSTR::CComBSTR(REFGUID guid)from atlbase.h:
如果您的代码使用 ATL/MFC，您也可以使用CComBSTR::CComBSTR(REFGUID guid)from atlbase.h：
GUID guid = ...;
const CComBSTR guidBstr(guid); // Converts from binary GUID to BSTR
const CString guidStr(guidBstr); // Converts from BSTR to appropriate string, ANSI or Wide
It will make conversion & memory cleanup automatically.
它将自动进行转换和内存清理。
回答by Tim Dowty
You can eliminate the need for special string allocations/deallocations by using StringFromGUID2()
您可以使用 StringFromGUID2() 消除对特殊字符串分配/解除分配的需要
GUID guid = <some-guid>;
// note that OLECHAR is a typedef'd wchar_t
wchar_t szGUID[64] = {0};
StringFromGUID2(&guid, szGUID, 64);
回答by Marco M.
Courtesy of google's breakpadproject:
由google 的 breakpad项目提供：
std::string ToString(GUID *guid) {
char guid_string[37]; // 32 hex chars + 4 hyphens + null terminator
snprintf(
guid_string, sizeof(guid_string),
"%08x-%04x-%04x-%02x%02x-%02x%02x%02x%02x%02x%02x",
guid->Data1, guid->Data2, guid->Data3,
guid->Data4[0], guid->Data4[1], guid->Data4[2],
guid->Data4[3], guid->Data4[4], guid->Data4[5],
guid->Data4[6], guid->Data4[7]);
return guid_string;
}
UUID guid = {0};
UuidCreate(&guid);
std::cout << GUIDToString(&guid);
回答by steve
Inspired by JustinB's answer
灵感来自JustinB的回答
#define GUID_FORMAT "%08lX-%04hX-%04hX-%02hhX%02hhX-%02hhX%02hhX%02hhX%02hhX%02hhX%02hhX"
#define GUID_ARG(guid) guid.Data1, guid.Data2, guid.Data3, guid.Data4[0], guid.Data4[1], guid.Data4[2], guid.Data4[3], guid.Data4[4], guid.Data4[5], guid.Data4[6], guid.Data4[7]
and then
进而
printf("Int = %d, string = %s, GUID = {" GUID_FORMAT "}\n", myInt, myString, GUID_ARG(myGuid));
回答by Cookie
I know the question is quite old, but would this work maybe?
我知道这个问题已经很老了，但这可能有用吗？
inline std::ostream& operator <<(std::ostream& ss,GUID const& item) {
OLECHAR* bstrGuid;
::StringFromCLSID(item, &bstrGuid);
ss << bstrGuid;
::CoTaskMemFree(bstrGuid);
return ss;
}
回答by user2443842
Use UuidToStringfunction to convert GUID to string. The function accepts UUID type which is typedef of GUID.
使用UuidToString函数将 GUID 转换为字符串。该函数接受 UUID 类型，即 GUID 的 typedef。
回答by tcb
std::string
GuidToString(const GUID& guid, bool lower = false)
{
const char* hexChars = lower ? "0123456789abcdef" : "0123456789ABCDEF";
auto f = [hexChars](char* p, unsigned char v)
{
p[0] = hexChars[v >> 4];
p[1] = hexChars[v & 0xf];
};
char s[36];
f(s, static_cast<unsigned char>(guid.Data1 >> 24));
f(s + 2, static_cast<unsigned char>(guid.Data1 >> 16));
f(s + 4, static_cast<unsigned char>(guid.Data1 >> 8));
f(s + 6, static_cast<unsigned char>(guid.Data1));
s[8] = '-';
f(s + 9, static_cast<unsigned char>(guid.Data2 >> 8));
f(s + 11, static_cast<unsigned char>(guid.Data2));
s[13] = '-';
f(s + 14, static_cast<unsigned char>(guid.Data3 >> 8));
f(s + 16, static_cast<unsigned char>(guid.Data3));
s[18] = '-';
f(s + 19, guid.Data4[0]);
f(s + 21, guid.Data4[1]);
s[23] = '-';
f(s + 24, guid.Data4[2]);
f(s + 26, guid.Data4[3]);
f(s + 28, guid.Data4[4]);
f(s + 30, guid.Data4[5]);
f(s + 32, guid.Data4[6]);
f(s + 34, guid.Data4[7]);
return std::string(s, 36);
}
