C++ 使用 GLuint 而不是 unsigned int 有什么好处？
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What's the advantage of using GLuint instead of unsigned int?
提问by JSeven
I like to be more standard as possible, so why should I "constrain" my classes defining it's members as OpenGL types when I can use primitive types? Is there any advantage?
我喜欢尽可能更标准，那么当我可以使用原始类型时，为什么要“约束”我的类将其成员定义为 OpenGL 类型？有什么优势吗？
回答by mah
The type "unsigned int" has a different size depending on the platform you're building on. I expect this to normally be 32 bits, however it could be 16 or 64 (or something else -- depending on the platform).
“unsigned int”类型的大小取决于您构建的平台。我希望这通常是 32 位，但也可能是 16 位或 64 位（或其他 - 取决于平台）。
Library-specific types are often created to be typedef'd according to platform-specific rules. This allows a generic application to use the right type without having to be aware of the platform it will be built for. Instead, the platform-specific knowledge is constrained to a single common header file.
特定于库的类型通常被创建为根据特定于平台的规则进行 typedef。这允许通用应用程序使用正确的类型，而不必知道将为其构建的平台。相反，特定于平台的知识仅限于单个公共头文件。
回答by JSeven
i don't think it matters in this case because the spec says they are minimum sizes, not strict sizes. have a look at gl.h ~line 149 they're just typedefs of basic C types. they are just a convenience - for example there is a boolean type, so if you're using C89 and don't use any booleans then there's one set up for you to use with GL. GLuint is just a shorter way of typing unsigned int:
我认为在这种情况下这并不重要，因为规范说它们是最小尺寸，而不是严格尺寸。看看 gl.h ~line 149 它们只是基本 C 类型的 typedef。它们只是一种方便 - 例如有一个布尔类型，所以如果您使用 C89 并且不使用任何布尔值，那么有一个设置供您与 GL 一起使用。GLuint 只是输入 unsigned int 的一种更短的方式：
typedef unsigned int GLenum;
typedef unsigned char GLboolean;
typedef unsigned int GLbitfield;
typedef void GLvoid;
typedef signed char GLbyte; /* 1-byte signed */
typedef short GLshort; /* 2-byte signed */
typedef int GLint; /* 4-byte signed */
typedef unsigned char GLubyte; /* 1-byte unsigned */
typedef unsigned short GLushort; /* 2-byte unsigned */
typedef unsigned int GLuint; /* 4-byte unsigned */
typedef int GLsizei; /* 4-byte signed */
typedef float GLfloat; /* single precision float */
typedef float GLclampf; /* single precision float in [0,1] */
typedef double GLdouble; /* double precision float */
typedef double GLclampd; /* double precision float in [0,1] */
回答by karlphillip
Better cross-platform compatibility.
更好的跨平台兼容性。
回答by user877329
The advantages has already been mentioned here. However, there is a disadvantage clear from the following examples:
这里已经提到了优点。但是，从以下示例中可以明显看出缺点：
class FileIn
{
public:
//Public interface like read
private:
void* handle;
};
The above code fits very well in a platform independent header but writing
上面的代码非常适合独立于平台的标头，但编写
#define WIN32_LEAN_AND_MEAN
#include <windows.h>
class FileIn
{
public:
//Public interface like read
private:
HANDLE handle;
};
does not.
才不是。
Though the former will require ugly typecasts like
虽然前者需要丑陋的类型转换，如
int fd=(int)( (size_t)handle );
close(fd);
i do not know any system which have sizeof(void*)< sizeof(int). Yes it will fail if open returns a negative number for a valid file handle.
我不知道任何具有sizeof(void*)< 的系统sizeof(int)。是的，如果 open 返回有效文件句柄的负数，它将失败。
What to learn about this? Avoid using typedefs in library include files. Instead use struct declarations even though C programmers need to write structa dozen times. Here, some C standard library implementations do it all wrong.
关于这个要学习什么？避免在库包含文件中使用 typedef。即使 C 程序员需要编写struct十几次，也可以使用结构声明。在这里，一些 C 标准库实现都做错了。
Right
对
In stdio.h:
在stdio.h：
struct FILE;
And in the application:
在应用程序中：
struct FILE* the_file=fopen("filename.txt","rb");
/*...*/
Wrong
错误的
In stdio.h:
在stdio.h：
typedef struct SOMENAMETHATNOONESHOULDUSE
{
/* Internal data members */
} FILE;
In application
申请中
FILE* the_file=fopen("filename.txt","rb");
When writing a C++ wrapper, this forces either #include <cstdio>or simply declare the handle as above.
在编写 C++ 包装器时，这会强制#include <cstdio>或简单地声明句柄，如上。
