在C中转换转义序列 [英] Converting escape sequence in C
问题描述
我从某些用户输入中获得了一个名为ArrayA的char数组，其中可能包含转义序列字符.我想逐个字符地将ArrayA复制到ArrayB中.这样说，我将如何将ArrayA中的转义字符表示为ArrayB?如果我要正确打印ArrayB，仅复制字符实际上不会将\ t转换为选项卡?它将\ t放置为字符而不是正确的吗?我可以只输入0x09到\ t转义序列的ArrayB中，以便在打印ArrayB时实际上在该数组的特定位置打印一个选项卡吗?
I have a char array called ArrayA from some user input that may contain escape sequence characters in it. I want to copy ArrayA into ArrayB character by character. With that said how would i then represent the escape characters from ArrayA into ArrayB? Simply copying the characters won't actually convert \t to a tab if i were to print ArrayB correct? It would just place \t as a character instead correct? Can i just input 0x09 into ArrayB for the \t escape sequence so that it will actually print a tab at that specific place in the array when printing ArrayB?
推荐答案
函数cstrlit_chr()
如何?
Function cstrlit_chr()
How about this:
/* Convert C Character Literal in (str..end] (excluding surrounding quotes) */
/* to character, returning converted char or -1 if string is invalid. */
/* Convert string containing C character literal to character value */
/* Returns -1 if character literal is invalid, otherwise 0x00..0xFF */
/* Does not support extension \E for ESC \033. */
/* Does not support any extension for DEL \177. */
/* Does not support control-char notation ^A for CTRL-A \001. */
/* Accepts \z as valid z when z is not otherwise special. */
/* Accepts \038 as valid CTRL-C \003; next character starts with the 8. */
/* Accepts \x3Z as valid CTRL-C \003; next character starts with the Z. */
/* Treats invalid octal escape \8 or \9 as 8 or 9 */
int cstrlit_chr(const char *str, const char *end, char const ** const eptr)
{
unsigned char u;
int rv;
if (str >= end)
rv = -1; /* String contains no data */
else if ((u = *str++) != '\\')
rv = u;
else if (str == end)
rv = -1; /* Just a backslash - invalid */
else if ((u = *str++) == 'x')
{
/**
** Hex character constant - \xHH or \xH, where H is a hex digit.
** Technically, can be \xHHH too, if CHAR_BIT > 8; this nicety
** is being studiously ignored.
*/
int x1;
int x2;
if (str == end)
rv = -1;
else if ((x1 = basedigit(*str++, 16)) < 0)
{
rv = -1; /* Invalid hex constant */
str--;
}
else if (str == end)
rv = x1; /* Single digit hex constant */
else if ((x2 = basedigit(*str++, 16)) < 0)
{
rv = x1; /* Single-digit hex constant */
str--;
}
else
rv = (x1 << 4) | x2; /* Double-digit hex constant */
}
else if (isdigit(u))
{
/**
** Octal character constant - \O or \OO or \OOO, where O is an
** octal digit. Technically, the constant extends for an
** indefinite number of octal digits; this nicety is being
** studiously ignored. Treat \8 as 8 and \9 as 9.
*/
int o1;
int o2;
int o3;
if ((o1 = basedigit(u, 8)) < 0)
rv = u; /* Invalid octal constant (\8 or \9) */
else if (str == end)
rv = o1; /* Single-digit octal constant */
else if ((o2 = basedigit(*str++, 8)) < 0)
{
rv = o1; /* Single-digit octal constant */
str--;
}
else if (str == end)
rv = (o1 << 3) | o2; /* Double-digit octal constant */
else if ((o3 = basedigit(*str++, 8)) < 0)
{
rv = (o1 << 3) | o2; /* Double-digit octal constant */
str--;
}
else if (o1 >= 4)
rv = -1; /* Out of range 0x00..0xFF (\000..\377) */
else
rv = (((o1 << 3) | o2) << 3) | o3;
}
else
{
/* Presumably \a, \b, \f, \n, \r, \t, \v, \', \", \? or \\ - or an error */
switch (u)
{
case 'a':
rv = '\a';
break;
case 'b':
rv = '\b';
break;
case 'f':
rv = '\f';
break;
case 'n':
rv = '\n';
break;
case 'r':
rv = '\r';
break;
case 't':
rv = '\t';
break;
case 'v':
rv = '\v';
break;
case '\"':
rv = '\"';
break;
case '\'':
rv = '\'';
break;
case '\?':
rv = '\?';
break;
case '\\':
rv = '\\';
break;
case '\0': /* Malformed: solitary backslash followed by NUL */
rv = -1;
break;
default:
rv = u; /* Nominally invalid: \X but X not special; return X. */
break;
}
}
if (eptr != 0)
*eptr = str;
return(rv);
}
它处理C89字符序列；它不处理Unicode(通用)字符(\uXXXX或\U00XXXXXX).
It handles C89 character sequences; it does not handle Unicode (universal) characters (\uXXXX or \U00XXXXXX).
/*
** Convert character to digit in given base,
** returning -1 for invalid bases and characters.
*/
int basedigit(char c, int base)
{
int i;
#if (('z' - 'a') != 25 || ('Z' - 'A') != 25)
#error Faulty Assumption
This code assumes the code set is ASCII, ISO 646, ISO 8859, or something similar.
#endif /* Alphabet test */
if (base < 2 || base > 36)
i = -1;
else if (c >= '0' && c <= '9')
i = c - '0';
else if (c >= 'A' && c <= 'Z')
i = c - 'A' + 10;
else if (c >= 'a' && c <= 'z')
i = c - 'a' + 10;
else
i = -1;
return((i < base) ? i : -1);
}
样品用量
/* Sample usage */
#include <stdio.h>
#include <string.h>
int main(void)
{
const char data[] = "ab\\xFF\\03\\7\\377\\t\\?\\'\\\\yz";
const char *end = data + strlen(data);
const char *start = data;
const char *next;
int c;
while ((c = cstrlit_chr(start, end, &next)) != -1)
{
char buffer[20];
snprintf(buffer, sizeof(buffer), "[[%.*s]]", (int)(next-start), start);
printf("%3d (0x%.2X) %-10s - [[%s]]\n", c, c & 0xFF, buffer, next);
start = next;
}
return 0;
}
请注意，扫描范围是由指向起始字符的指针标识的，而结尾是由指向范围结束后的字符的指针标识的(在示例中，字符串末尾的'\0'，但是该函数可以处理任意数据，并且不需要以null终止.输入字符串在源代码中具有加倍的反斜杠，因此实际的字符串仅包含单个反斜杠.
Note that the range scanned is identified by a pointer to the start character and the end is identified by a pointer to the character after the end of the range (in the example, the '\0' at the end of the string, but the function works on arbitrary data and does not require that it is null terminated. The input string has backslashes doubled up in the source code so that the actual string contains single backslashes.
如果为c == -1，则转换失败.否则，c包含字符，并且end是指向转换完成位置的指针.
If c == -1, conversion failed. Otherwise, c contains the character, and end is a pointer to where the conversion finished.
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