Why String.replaceAll() in java requires 4 slashes "\\\\" in regex to actually replace "\"?

You need to esacpe twice, once for Java, once for the regex.

Java code is

"\\\\"

makes a regex string of

"\\" - two chars

but the regex needs an escape too so it turns into

\ - one symbol

@Peter Lawrey's answer describes the mechanics. The "problem" is that backslash is an escape character in both Java string literals, and in the mini-language of regexes. So when you use a string literal to represent a regex, there are two sets of escaping to consider ... depending on what you want the regex to mean.

But why is it like that?

It is a historical thing. Java originally didn't have regexes at all. The syntax rules for Java String literals were borrowed from C / C++, which also didn't have built-in regex support. Awkwardness of double escaping didn't become apparent in Java until they added regex support in the form of the Pattern class ... in Java 1.4.

So how do other languages manage to avoid this?

They do it by providing direct or indirect syntactic support for regexes in the programming language itself. For instance, in Perl, Ruby, Javascript and many other languages, there is a syntax for patterns / regexs (e.g. '/pattern/') where string literal escaping rules do not apply. In C# and Python, they provide an alternative "raw" string literal syntax in which backslashes are not escapes. (But note that if you use the normal C# / Python string syntax, you have the Java problem of double escaping.)

---

Why do text.replaceAll("\n","/"), text.replaceAll("\\n","/"), and text.replaceAll("\\\n","/") all give the same output?

The first case is a newline character at the String level. The Java regex language treats all non-special characters as matching themselves.

The second case is a backslash followed by an "n" at the String level. The Java regex language interprets a backslash followed by an "n" as a newline.

The final case is a backslash followed by a newline character at the String level. The Java regex language doesn't recognize this as a specific (regex) escape sequence. However in the regex language, a backslash followed by any non-alphabetic character means the latter character. So, a backslash followed by a newline character ... means the same thing as a newline.

1) Let's say you want to replace a single \ using Java's replaceAll method:

   \
   ˪--- 1) the final backslash

2) Java's replaceAll method takes a regex as first argument. In a regex literal, \ has a special meaning, e.g. in \d which is a shortcut for [0-9] (any digit). The way to escape a metachar in a regex literal is to precede it with a \, which leads to:

 \ \
 | ˪--- 1) the final backslash
 |
 ˪----- 2) the backslash needed to escape 1) in a regex literal

3) In Java, there is no regex literal: you write a regex in a string literal (unlike JavaScript for example, where you can write /\d+/). But in a string literal, \ also has a special meaning, e.g. in \n (a new line) or \t (a tab). The way to escape a metachar in a string literal is to precede it with a \, which leads to:

\\\\
|||˪--- 1) the final backslash
||˪---- 3) the backslash needed to escape 1) in a string literal
|˪----- 2) the backslash needed to escape 1) in a regex literal
˪------ 3) the backslash needed to escape 2) in a string literal