我似乎无法弄清楚这里发生了什么。我有以下警告图标;
K_USHORT usaAlertIcon[16] = { 0x0000, 0x0000, 0x0180, 0x03C0, 0x03C0, 0x0660, 0x0660, 0x0E70, 0x0E70, 0x1E78, 0x3E7C, 0x3FFC, 0x7E7E, 0x7E7E, 0xFFFF, 0x0000 };
现在,我想在我的代码中使用它,将数组视为 8 位数据(unsigned char);
//---------------------------------------------------------------------------
void GraphicsDriver::Stamp(DrawStamp_t *pstStamp_)
{
K_USHORT usRow;
K_USHORT usCol;
K_USHORT usShift;
K_USHORT usIndex;
DrawPoint_t stPoint;
usIndex = 0;
for (usRow = pstStamp_->usY; usRow < (pstStamp_->usY + pstStamp_->usHeight); usRow++)
{
usShift = 0x80;
for (usCol = pstStamp_->usX; usCol < (pstStamp_->usX + pstStamp_->usWidth); usCol++)
{
// If the packed bit in the bitmap is a "1", draw the color.
if (pstStamp_->pucData[usIndex] & usShift)
{
stPoint.usX = usCol;
stPoint.usY = usRow;
stPoint.uColor = pstStamp_->uColor;
DrawPixel(&stPoint);
}
// Stamps are opaque, don't fill in the BG
// Shift to the next bit in the field
usShift >>= 1;
// Rollover - next bit in the bitmap.
// This obviously works best for stamps that are multiples of 8x8
if (usShift == 0)
{
usShift = 0x80;
usIndex++;
}
}
}
}
当将数组分配给要发送到此函数的数据结构时,我将其转换为;
stStamp.pucData = (K_UCHAR*)usaAlertIcon;
但是,当我这样做时,它似乎翻转了字节。先画左半边,再画右半边;所以它把图像从中间切下来,把左边的部分换成右边的部分。
作为健全性检查,如果我自己明确地拆分数组;
K_UCHAR ucaAlertIcon[32] = { 0x00, 0x00, 0x00, 0x00, 0x01, 0x80, 0x03, 0xC0, 0x03, 0xC0, 0x06, 0x60, 0x06, 0x60, 0x0E, 0x70, 0x0E, 0x70, 0x1E, 0x78, 0x3E, 0x7C, 0x3F, 0xFC, 0x7E, 0x7E, 0x7E, 0x7E, 0xFF, 0xFF, 0x00, 0x00 };
一切都按预期工作。
有人可以解释为什么在我进行类型转换时这些字节似乎在翻转吗?
最佳答案
Can someone explain why these bytes seem to be flipping when I do a type cast?
字节顺序。您显然在一台小端机器上,这意味着最低地址的字节是最低有效字节。您需要将 unsigned short
值转换为大端格式,例如 htons
就是这样做的。
#include <arpa/inet.h>
uint16_t htons(uint16_t hostshort);
当然你也可以自己做,
array[i] = (array[i] << 8) | (array[i] >> 8);
关于c++ - 将无符号短裤数组转换为无符号字符数组,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/14150166/