C ++支持 scanf 函数。没有简单的选择，尤其是如果您要使用所有怪癖来复制 scanf（）的确切语义。

C++ supports the scanf function. There is no simple alternative, especially if you want to replicate the exact semantics of scanf() with all the quirks.

但是请注意，您的代码有几个问题：

Note however that your code has several issues:

• 您没有将最大字符数传递给 ps1 和 ps2 。任何足够的输入序列都会导致缓冲区溢出，并带来严重后果。

• You do not pass the maximum number of characters to read into ps1 and ps2. Any sufficiently input sequence will cause a buffer overflow with dire consequences.

您可以简化第一种格式％* [\t＼ ＼n] ，格式字符串中只有一个空格。这也将允许不存在空格字符的情况。如当前所写， scanf（）将失败，并且如果在 0 。 c> 。

You could simplify the first format %*[ \t\n] with just a space in the format string. This would also allow for the case where no whitespace characters are present. As currently written, scanf() would fail and return 0 if no whitspace characters are present before the ".

类似地，如果在第二个之前没有非字母或没有其他字符 ， scanf 会返回简短的 0 或 1 并将一个或两个目标数组保留为不确定状态。

Similarly, if no non letters or if no other characters follow before the second ", scanf would return a short count of 0 or 1 and leave one or both destination array in an indeterminate state.

对于所有这些原因，在C语言中，首先使用 fgets（）读取一行输入并使用 sscanf（）$ c会更安全和可预测

For all these reasons, it would be much safer and predictable in C to first read a line of input with fgets() and use sscanf() or parse the line by hand.

在C ++中，您肯定要使用 std :: regex 在® regex.h> 中定义的软件包。

In C++, you definitely want to use the std::regex package defined in <regex.h>.