C法通过一个结构的成员进行迭代像一个数组? [英] C method for iterating through a struct's members like an array?
问题描述
让我们说我有一个向量类:
Let's say I have a vector class:
typedef struct vec3_s
{
float x, y, z;
}
vec3;
不过,我希望能够通过它来遍历无需将其转换为float数组。虽然投在这种情况下是可以接受的,我很好奇,看看是否沿着C的行++类的东西的功能是可行的直C.例如,在C ++中,因为的std ::矢量< T>
具有标 []
操作符重载,我可以通过它的第一个索引到一个函数采取无效*
。
But, I would like to be able to iterate through it without converting it to an array of floats. While a cast is acceptable in this case, I'm curious to see if anything along the lines of C++ like functionality is doable in straight C. For example, in C++, since std::vector< T >
has the subscript []
operator overloaded, I can pass the address of its first index to a function taking a void*
.
即,
void do_something_with_pointer_to_mem( void* mem )
{
// do stuff
}
int main( void )
{
std::vector< float > v;
// fill v with values here
// pass to do_something_with_pointer_to_mem
do_some_with_pointer_to_mem( &v[ 0 ] );
return;
}
另一个更具体的例子是,当向 glBufferData呼叫(... )在OpenGL(使用C ++时)由:
Another, more concrete example is when calls to glBufferData(...) are made in OpenGL (when using C++):
glBufferData(GL_ARRAY_BUFFER,sizeof的(somevector),放大器; somevector [0],GL_STREAM_DRAW);
那么,是不是可以使用下标操作符来实现用C类似的东西?如果没有,我只好写一个函数(例如,浮动vec3_value_at(无符号整型我)
),这将是有意义的只是静态内联
它在头文件,它的定义?
So, is it possible to accomplish something similar in C using the subscript operator? If not, and I had to write a function (e.g., float vec3_value_at( unsigned int i )
), would it make sense to just static inline
it in the header file it's defined in?
推荐答案
如果您所有的结构化字段都是同一类型的,你可以使用一个工会如下:
If all of your structure fields are of the same type, you could use a union as following:
typedef union vec3_u
{
struct vec3_s {
float x, y, z;
};
float vect3_a[3];
}
vec3;
这样您便可以独立访问到每个X,Y或Z字段或使用vect3_a阵列上进行遍历。这内存或计算,但长期的解决方案成本没有我们可能有点远从C ++类似的解决方案。
This way you could access to each x, y or z field independently or iterate over them using the vect3_a array. This solution cost nothing in term of memory or computation but we may be a bit far from a C++ like solution.
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