如何（un）转义C / C ++中的字符串？

给定一个计数字符串（一个字符数组，或者像std::string这样的包装器），是否存在一种“正确”的方法来转义和/或在C或C ++中取消它，这样“特殊”字符（如null）角色）成为C风格的逃脱和“正常”的角色保持原样？

或者我必须手动完成？

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; } } 

Answer 2

而不是分配一个新的缓冲区来包含转义的字符串，我喜欢在将它写入流时转义我的字符串。

以下函数可实现可读且简洁的代码。

 struct Escaped { const char* str; friend inline std::ostream& operator<<(std::ostream& os, const Escaped& e) { for (const char* char_p = e.str; *char_p != '\0'; char_p++) { switch (*char_p) { case '\a': os << "\\a"; break; case '\b': os << "\\b"; break; case '\f': os << "\\f"; break; case '\n': os << "\\n"; break; case '\r': os << "\\r"; break; case '\t': os << "\\t"; break; case '\v': os << "\\v"; break; case '\\': os << "\\\\"; break; case '\'': os << "\\'"; break; case '\"': os << "\\\""; break; case '\?': os << "\\\?"; break; default: os << *char_p; } } return os; } }; int main() { std::cout << Escaped{ "foo\n\tbar" } << std::endl; } 

产生

 foo\n bar