我有以下代码块(不是由我编写)用于进行ASCII字符到EBCDIC的映射和重新编码。
// Variables.
CodeHeader* tchpLoc = {};
...
memset(tchpLoc->m_ucpEBCDCMap, 0xff, 256);
for (i = 0; i < 256; i++) {
if (tchpLoc->m_ucpASCIIMap[i] != 0xff) {
ucTmp2 = i;
asc2ebn(&ucTmp1, &ucTmp2, 1);
tchpLoc->m_ucpEBCDCMap[ucTmp1] = tchpLoc->m_ucpASCIIMap[i];
}
}
< p > CodeHeader
的定义如下:
typedef struct {
...
UCHAR* m_ucpASCIIMap;
UCHAR* m_ucpEBCDCMap;
} CodeHeader;
我遇到问题的方法是
void asc2ebn(char* szTo, char* szFrom, int nChrs)
{
while (nChrs--)
*szTo++ = ucpAtoe[(*szFrom++) & 0xff];
}
[注意,unsigned char
数组ucpAtoe[256]
的内容已在问题末尾附上以供参考]。
现在,我有一个旧的C应用程序和我的C++11转换并行运行,这两个代码都会写入一个巨大的.bin文件,但存在微小的差异,我已经追踪到以上代码。对于两种代码,发生的情况是:
...
if (tchpLoc->m_ucpASCIIMap[i] != 0xff) {
ucTmp2 = i;
asc2ebn(&ucTmp1, &ucTmp2, 1);
tchpLoc->m_ucpEBCDCMap[ucTmp1] = tchpLoc->m_ucpASCIIMap[i];
}
i = 32
被输入,asc2ebn
方法返回ucTmp1
为64
或'@'
,对于C和C++变体都是如此。下一个输入是i = 48
,对于这个值,asc2ebn
方法将ucTmp1
返回为240
或'ð'
,而C++代码将ucTmp1
返回为-16
或'ð'
。我的问题是为什么这个查找/转换会为完全相同的输入和查找数组(以下是复制的)产生不同的结果?在这种情况下,旧的C代码被认为是正确的,因此我希望C++能够为这个查找/转换产生相同的结果。谢谢你的时间。static UCHAR ucpAtoe[256] = {
'\x00','\x01','\x02','\x03','\x37','\x2d','\x2e','\x2f',/*00-07*/
'\x16','\x05','\x25','\x0b','\x0c','\x0d','\x0e','\x0f',/*08-0f*/
'\x10','\x11','\x12','\xff','\x3c','\x3d','\x32','\xff',/*10-17*/
'\x18','\x19','\x3f','\x27','\x22','\x1d','\x35','\x1f',/*18-1f*/
'\x40','\x5a','\x7f','\x7b','\x5b','\x6c','\x50','\xca',/*20-27*/
'\x4d','\x5d','\x5c','\x4e','\x6b','\x60','\x4b','\x61',/*28-2f*/
'\xf0','\xf1','\xf2','\xf3','\xf4','\xf5','\xf6','\xf7',/*30-37*/
'\xf8','\xf9','\x7a','\x5e','\x4c','\x7e','\x6e','\x6f',/*38-3f*/
'\x7c','\xc1','\xc2','\xc3','\xc4','\xc5','\xc6','\xc7',/*40-47*/
'\xc8','\xc9','\xd1','\xd2','\xd3','\xd4','\xd5','\xd6',/*48-4f*/
'\xd7','\xd8','\xd9','\xe2','\xe3','\xe4','\xe5','\xe6',/*50-57*/
'\xe7','\xe8','\xe9','\xad','\xe0','\xbd','\xff','\x6d',/*58-5f*/
'\x79','\x81','\x82','\x83','\x84','\x85','\x86','\x87',/*60-67*/
'\x88','\x89','\x91','\x92','\x93','\x94','\x95','\x96',/*68-6f*/
'\x97','\x98','\x99','\xa2','\xa3','\xa4','\xa5','\xa6',/*70-77*/
'\xa7','\xa8','\xa9','\xc0','\x6a','\xd0','\xa1','\xff',/*78-7f*/
'\xff','\xff','\xff','\xff','\xff','\xff','\xff','\xff',/*80-87*/
'\xff','\xff','\xff','\xff','\xff','\xff','\xff','\xff',/*88-8f*/
'\xff','\xff','\xff','\xff','\xff','\xff','\xff','\xff',/*90-97*/
'\xff','\xff','\xff','\x4a','\xff','\xff','\xff','\xff',/*98-9f*/
'\xff','\xff','\xff','\xff','\xff','\xff','\xff','\xff',/*a0-a7*/
'\xff','\xff','\xff','\xff','\xff','\xff','\xff','\xff',/*a8-af*/
'\xff','\xff','\xff','\x4f','\xff','\xff','\xff','\xff',/*b0-b7*/
'\xff','\xff','\xff','\xff','\xff','\xff','\xff','\xff',/*b8-bf*/
'\xff','\xff','\xff','\xff','\xff','\x8f','\xff','\xff',/*c0-c7*/
'\xff','\xff','\xff','\xff','\xff','\xff','\xff','\xff',/*c8-cf*/
'\xff','\xff','\xff','\xff','\xff','\xff','\xff','\xff',/*d0-d7*/
'\xff','\xff','\xff','\xff','\xff','\xff','\xff','\xff',/*d8-df*/
'\xff','\xff','\xff','\xff','\xff','\xff','\xff','\xff',/*e0-e7*/
'\xff','\xff','\xff','\xff','\xff','\xff','\xff','\xff',/*e8-ef*/
'\xff','\xff','\xff','\x8c','\xff','\xff','\xff','\xff',/*f0-f7*/
'\xff','\xff','\xff','\xff','\xff','\xff','\xff','\xff' };
char
类型来说,240
和-16
是相同的值,对吗? - Joker_vDunsigned char
而不是char
吗?因为char
可以是无符号的也可以是有符号的。 - Shafik YaghmourucTmp2
用于“from”索引,而unTmp1
用于“to”索引。 - MoonKnightucpAtoe[48]
产生的位模式是0xf0
。如果将其解释为有符号字符/整数,它的值为-16,但如果将其解释为无符号字符/整数,该模式的值为240。这正好是你所看到的两个值,所以似乎你有一个情况将结果视为有符号,另一个情况将其视为无符号。 - twalberg