测试文章-二进制和ASCII码转换代码

//二进制转字符串(outbuff存放转换结果，必须拥有inbuff的两倍空间以上)//成功返回true 暂时不会失败bool BinToA(const void *inbuff, int inlen, void *outbuff){//将一个字节拆成前后四位，并分别加上64转换成两个可见字符 assert(inbuff != NULL && inlen > 0 && outbuff != NULL);memcpy((char *) outbuff + inlen, (char *) inbuff, inlen);char *pInPos = (char *) outbuff + inlen;char *pOutPos = (char *)outbuff;for (int i = 0; i < inlen; ++i){*pOutPos++ = ( (*pInPos >> 4) & 0x0f ) | 0x40; //前四字节加64(| 0x40)*pOutPos++ = ( (*pInPos++) & 0x0f ) | 0x40; }return true;}//字符串转二进制(outbuff存放转换结果，必须拥有inbuff的一半空间以上) inlen必须是2的位数(因为2个字节合成1个)bool BinFromA(const void *inbuff, int inlen, void *outbuff){//取两个字节的低四位和低四位，合并成一个字节(先取出来的字节低四位做为合并后字节高四位assert(inbuff != NULL && (inlen > 0 && (inlen % 2) == 0) && outbuff != NULL);char *pInPos = (char *)inbuff;char *pOutPos = (char *)outbuff;char ch;for (int i = 0; i < inlen; i += 2){//注意，不能合并为 *pOutPos++ = (*pInPos++ & 0x0f) | (*pInPos++ & 0x0f); 因为没有顺序点ch = (*pInPos++ & 0x0f) << 4;*pOutPos++ = ( ch ) | (*pInPos++ & 0x0f);}return true;}