c++ - 如何在 C/C++ 中(取消)转义字符串？

Question

给定一个counted 字符串(可以是字符数组，也可以是像std::string 这样的包装器)，是否有一种“正确”的方式来转义和/或在 C 或 C++ 中对其进行转义，以便“特殊”字符(如空字符)变为 C 风格转义，而“普通”字符保持原样？

还是我必须手动完成？

Answer 1

这是一个处理单个字符的函数:

/*
** 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;
        }
    }
}

这是处理以 null 结尾的字符串的代码(使用上面的函数):

void str_cstrlit(const char *str, char *buffer, 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;
    }
    *buffer = '\0';
}