如何将BYTE字符串写为char * [英] How to write a BYTE string as a char*
问题描述
你好
我有这样的字节加载：
unsigned char mybyte [] = {0xa1,0x14,0x02， 0x01,0x01,0x02,0x01,0x0a，
0x30,0x0c，0x80，
0x03,0x32,0x30,0x31,0x62,0x05,0x80， 0x03,0x32,0x30,0x32};
但是我想把它们写成一个字符串。所以我试过这个：
char szSend [] =
" \ x0a1 \ x014 \ x002 \ x001 \ x001 \ x002 \ x00100 \ x00a \ x030 \ x00 \\ nc \ x080 \ x003201 \x062 \ x005 \ x
080 \ x003202" ;;
但是这给了我一个编译错误
错误C2022：''12801''：对于角色来说太大了
错误C2022：''12802 ''：对于角色来说太大了
我可以不这样做吗？对于我来说，将这些非打印的b / b $ b字符放入字符串中最简单的方法是什么？或者我需要使用像sprintf这样的东西吗？
这样做吗？
Hello
I have a load of bytes like this:
unsigned char mybyte[] = {0xa1, 0x14, 0x02, 0x01, 0x01, 0x02, 0x01, 0x0a,
0x30, 0x0c, 0x80,
0x03, 0x32, 0x30, 0x31, 0x62, 0x05, 0x80, 0x03, 0x32, 0x30, 0x32};
But I want to write them as a string. so I tried this:
char szSend[] =
"\x0a1\x014\x002\x001\x001\x002\x001\x00a\x030\x00 c\x080\x003201\x062\x005\x
080\x003202";
But that gave me a compile error
error C2022: ''12801'' : too big for character
error C2022: ''12802'' : too big for character
Can I not do this? What is easiest way for me to put these non printing
characters into a string? Or will I need to use something like sprintf to
do this?
推荐答案
在文章< eo ******************* @ news.demon.co.uk> ;,
Angus< no **** @ gmail.comwrote：
In article <eo*******************@news.demon.co.uk>,
Angus <no****@gmail.comwrote:
> unsigned char mybyte [] = {0xa1,0x14,0x02,0x01,0x01,0x02,0x01,0x0a，
0x30,0x0c，0x80，
0x03,0x32,0x30,0x31,0x62,0x05,0x80,0x03,0x32,0x30,0x32};
>unsigned char mybyte[] = {0xa1, 0x14, 0x02, 0x01, 0x01, 0x02, 0x01, 0x0a,
0x30, 0x0c, 0x80,
0x03, 0x32, 0x30, 0x31, 0x62, 0x05, 0x80, 0x03, 0x32, 0x30, 0x32};
>但我想把它们写成一个字符串。所以我试过这个：
char szSend [] =
" \ x0a1 \ x014 \ x002 \ x001 \ x001 \ x002 \ x001 \ x00a \ x030 \ x0 0c \x080 \ x003201 \x062 \ x005 \ x
080 \ x003202" ;;
>But I want to write them as a string. so I tried this:
char szSend[] =
"\x0a1\x014\x002\x001\x001\x002\x001\x00a\x030\x0 0c\x080\x003201\x062\x005\x
080\x003202";
这两个应该匹配吗？在第一个你有03 32 30 31和
在第二个003201。
如果你在一个字符串上添加一个零，那么第一个字符串非常好结束。
- 理查德
-
考虑需要多达32个字符
在某些字母表中" - X3.4,1963。
Are those two supposed to match? In the first you have 03 32 30 31 and
in the second 003201.
The first is perfectly good as a string if you add a zero on the end.
-- Richard
--
"Consideration shall be given to the need for as many as 32 characters
in some alphabets" - X3.4, 1963.
>
我可以不这样做吗？对于我来说，将这些非打印的b / b $ b字符放入字符串中最简单的方法是什么？或者我需要使用像sprintf这样的东西吗？
这样做吗？
>
Can I not do this? What is easiest way for me to put these non printing
characters into a string? Or will I need to use something like sprintf to
do this?
您认为哪些字符是不可打印的？
Which characters do you think are non-printable?
Angus写道：
Angus wrote:
你好
我有这样的字节加载：
unsigned char mybyte [] = {0xa1,0x14,0x02,0x01,0x01,0x02,0x01,0x0a，
0x30,0x0c，0x80，
0x03,0x32,0x30,0x31,0x62,0x05,0x80,0x03,0x32,0x30,0x32};
但我想把它们写成字符串。所以我试过这个：
char szSend [] =
" \ x0a1 \ x014 \ x002 \ x001 \ x001 \ x002 \ x00100 \ x00a \ x030 \ x00 \\ nc \ x080 \ x003201 \x062 \ x005 \ x
080 \ x003202" ;;
但是这给了我一个编译错误
错误C2022：''12801''：对于角色来说太大了
错误C2022：''12802 ''：对于字符来说太大了
Hello
I have a load of bytes like this:
unsigned char mybyte[] = {0xa1, 0x14, 0x02, 0x01, 0x01, 0x02, 0x01, 0x0a,
0x30, 0x0c, 0x80,
0x03, 0x32, 0x30, 0x31, 0x62, 0x05, 0x80, 0x03, 0x32, 0x30, 0x32};
But I want to write them as a string. so I tried this:
char szSend[] =
"\x0a1\x014\x002\x001\x001\x002\x001\x00a\x030\x00 c\x080\x003201\x062\x005\x
080\x003202";
But that gave me a compile error
error C2022: ''12801'' : too big for character
error C2022: ''12802'' : too big for character
你的初始化字符串包含两个十六进制常量，看起来好像是
太大而不能成为字符。 $>
\ x003201 == 12801
\ x003202 == 12802
更正这些常量，使它们各自代表可以
存储在char中，编译器将停止抱怨。
HTH
-
Lew
Your initialization string contains two hex constants that appear to be
too big to be characters.
\x003201 == 12801
\x003202 == 12802
Correct these constants so that they each represent something that can
be stored in a char, and your compiler will stop complaining.
HTH
--
Lew
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