How can print raw escape characters as \t and \n in cout?

问题

how can print escape characters without further processing and as \t or \n or ... in std::cout? I dont want to process text manually before sending it to output? Is there any switch to std::cout for this purpose?

回答1:

Basically a raw string literal is a string in which the escape characters (like \n \t or \" ) of C++ are not processed. A raw string literal starts with R"( and ends in )", let's see it in an example the difference between a normal string and a raw string in C++:

string raw_str=R"(First line.\nSecond line.\nEnd of message.\n)";
cout<<raw_str<<endl;

result:

~$ ./a.out
First line.\nSecond line.\nEnd of message.\n

回答2:

If you add one extra slash there as \\t you can see \t in the output of std::cout

For example: cout<<"\\t hello" will print \t hello.

I hope this helps

来源：https://stackoverflow.com/questions/38843417/how-can-print-raw-escape-characters-as-t-and-n-in-cout