出个题目：如何把颜色的十六进制值，赋值给:Color.r .g .b .a变量？要求高效。
我先来一个：
long c=0xffffffff;
printf("r:%d,g:%d,b:%d,a:%d\n%x",(char)(c>>8*3),(char)(c>>8*2),(char)(c>>8),(char)c,c);
还有没有更好的思路？

学这个，居然发现我的cpu是小端的

union C_Color{
struct {char a;
char b;
char g;
char r;
};
int color;};
可以吗

规范下重发↓↓↓↓↓↓↓↓↓
#include <stdio.h>
struct COLOR
{
char r,g,b,a;
};
int hex2color(long &hex,COLOR *color)
{
static union
{
char c[4];
unsigned long mylong;
} endian_test = {{'l', '?', '?', 'b'}};
#define ENDIANNESS ((char)endian_test.mylong)
if (ENDIANNESS == 'l')
{
struct s_c
{
char a, b, g, r;
};
s_c *tem_color = (s_c*)&hex;
color->r=tem_color->r;
color->g=tem_color->g;
color->b=tem_color->b;
color->a=tem_color->a;
return 0;
}
else
{
COLOR *tem_color=(COLOR*)&hex;
color->r=tem_color->r;
color->g=tem_color->g;
color->b=tem_color->b;
color->a=tem_color->a;
return 1;
}
}
int main()
{
long hex = 0xffffffff;
COLOR color;
hex2color(hex,&color);
printf("color:%x\nr:%x,g:%x,b:%x,a:%x",hex, color.r, color.g, color.b,
color.a);
}

#include <stdio.h>
//#include <stdlib.h>
union COLOR
{
struct
{
char r, g, b, a;
};
unsigned long val;
COLOR() {};
COLOR(const long &hex)
{
static union
{
char c[4 ];
unsigned long mylong;
} endian_test = {{'l', '?', '?', 'b'}};
#define ENDIANNESS ((char)endian_test.mylong)
switch (ENDIANNESS)
{
case 'l':
{
COLOR *color = (COLOR *) & hex;
this->r = color->a;
this->g = color->b;
this->b = color->g;
this->a = color->r;
break;
}
case 'b':
{
this->val=hex;
//memcpy(this, &hex, sizeof(hex));
break;
}
default:
this->val=hex;
break;
}
}
};