给定一个计数的字符串(一个字符数组或一个包装器,如std :: string),是否有一个“正确”的方法来在C或C中转义和/或解除它,这样“特殊”的字符(如null人物)成为C风格逃脱,“正常”人物保持原样？
或者我必须用手做吗？
解决方法
@H_301_7@ 这是处理单个字符的功能：/*
** Does not generate hex character constants.
** Always generates triple-digit octal constants.
** Always generates escapes in preference to octal.
** Escape question mark to ensure no trigraphs are generated by repetitive use.
** Handling of 0x80..0xFF is locale-dependent (might be octal,might be literal).
*/
void chr_cstrlit(unsigned char u,char *buffer,size_t buflen)
{
if (buflen < 2)
*buffer = '\0';
else if (isprint(u) && u != '\'' && u != '\"' && u != '\\' && u != '\?')
sprintf(buffer,"%c",u);
else if (buflen < 3)
*buffer = '\0';
else
{
switch (u)
{
case '\a': strcpy(buffer,"\\a"); break;
case '\b': strcpy(buffer,"\\b"); break;
case '\f': strcpy(buffer,"\\f"); break;
case '\n': strcpy(buffer,"\\n"); break;
case '\r': strcpy(buffer,"\\r"); break;
case '\t': strcpy(buffer,"\\t"); break;
case '\v': strcpy(buffer,"\\v"); break;
case '\\': strcpy(buffer,"\\\\"); break;
case '\'': strcpy(buffer,"\\'"); break;
case '\"': strcpy(buffer,"\\\""); break;
case '\?': strcpy(buffer,"\\\?"); break;
default:
if (buflen < 5)
*buffer = '\0';
else
sprintf(buffer,"\\%03o",u);
break;
}
}
}
这是处理一个以null结尾的字符串(使用上面的函数)的代码：
void str_cstrlit(const char *str,size_t buflen)
{
unsigned char u;
size_t len;
while ((u = (unsigned char)*str++) != '\0')
{
chr_cstrlit(u,buffer,buflen);
if ((len = strlen(buffer)) == 0)
return;
buffer += len;
buflen -= len;
}
}
