C：如何做結構打包和解壓縮到char緩衝區？

Question

我試圖打包和解包的GPS座標轉換成字符緩衝區：

我有一個gpscoords結構爲這樣：

typedef struct { 
    char north; 
    uint16_t lat; 
    uint16_t lat_frac; 
    char east; 
    uint16_t lon; 
    uint16_t lon_frac; 
} gpscoords; 

如何打包和解包的gpscoords到緩衝區指向的緩衝區變量？我想在包裝之前爲整數到網絡順序轉換，但我這樣做如下錯誤：

void pack_gpscoords(gpscoords *pos, char *buffer) { 
    pack_gpscoords(pos, &buffer); 
    memcpy(buffer, pos, sizeof(pos)+1); 
} 

void unpack_gpscoords(gpscoords *pos, char *buffer) { 
    unpack_gpscoords(pos, &buffer); 
    memcpy(pos, buffer, sizeof(buffer)+1); 
} 

Answer 1

你應該把字節緩衝區unsigned char代替char。

這裏是你如何能做到這一點：

void pack_gpscoords(gpscoords *pos, unsigned char *buffer) { 
    buffer[0] = pos->north; 
    buffer[1] = pos->lat >> 8; 
    buffer[2] = pos->lat & 255; 
    buffer[3] = pps->lat_frac >> 8; 
    buffer[4] = pps->lat_frac & 255; 
    buffer[5] = pos->east; 
    buffer[6] = pos->lon >> 8; 
    buffer[7] = pos->lon & 255; 
    buffer[8] = pps->lon_frac >> 8; 
    buffer[9] = pps->lon_frac & 255; 
} 

void unpack_gpscoords(gpscoords *pos, const unsigned char *buffer) { 
    pos->north = buffer[0]; 
    pos->lat = (buffer[1] << 8) | buffer[2]; 
    pos->lat_frac = (buffer[3] << 8) | buffer[4]; 
    pos->east = buffer[5]; 
    pos->lon = (buffer[6] << 8) | buffer[7]; 
    pos->lon_frac = (buffer[8] << 8) | buffer[9]; 
} 

注：

• 的面向字節的打包表示假定字段的順序相同。
• 網絡順序表示big-endian：第一個字節包含高位8位，第二個字節包含低位8位。