How to find occurrences of a substring=“\r” in a string=“This is a \\rtest \n\r string” in java using Pattern and matchers

How to find occurrences of a substring="r" in a string="This is a \rtest nr string" in java using Pattern and matchers.

When I use

Pattern p4=Pattern.compile("\r",Pattern.LITERAL);


Even then this is not working

It worked well when I have string="This is a \rtest \n\r string".It have me correct count i.e..2

But for string="This is a \\rtest \n\r string".
It gave me incorrect count =1;

string="This is a \\rtest \n\r string"

function to be called:

static int countMatches(Pattern pattern, String string)


Matcher matcher=pattern.matcher(string);
int count=0;
int pos=0;
while(matcher.find(pos))


count++;
pos=matcher.start()+1;



return count;




2 Answers
2

The problem is that the String "This is a \rtest \n\r string" really does not contain two occurences of '\r'. the first occurence is interprated by the compiler as the two characters '\' (which is a single backsapce '') and 'r' (which is something unprintable)

"This is a \rtest \n\r string"

'\r'

'\'

''

'r'

This is also true for "This is a \\rtest \n\r string" or any other not-even number of backslashes

"This is a \\rtest \n\r string"

In string you can use \ if u want to escape character. So in your string

\

String string="This is a \rtest \n\r string";


In this string \rtest part is escaping the r part with double backslash \.

\rtest

r

\

This r has meaning in string library.
You can read this question : r diff n

r

So in your example r could not see in your variable string. If you change the string with this you can get result as 2 ;

String string="This is a \rtest \n\r string";


in your string second part as \n\r there is saving the n with and r with .

\n\r

n

r

In shortly , change the string to get result correctly. if you write double backspace before the anything , you will escape after char.

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