Regular expressions and printf

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Regular expressions can be used to pick apart a printf string for further processing. Here is an example in C. For more examples, see Printf strings .

Here is the C code.

#include <stdio.h>

int main() {
	int x1 = 2;
	int y1 = 3;
	int z1 = x1+y1;
	printf("The sum of %d and %d is %d.\n",x1,y1,z1);
	return 0;
	}

Here is the output of the C code.

The sum of 2 and 3 is 5.

The text string being used is the following.

"The sum of %d and %d is %d."

The goal is to use regular expressions to pick apart the text and format string parts into lists.

Here are the characters in the string. Regular expressions match characters. Blanks are important so any blanks in a pattern need to be matched in the string being matched.

Here is the JavaScript code.

var s1 = "The sum of %d and %d is %d.";
var n1 = s1.length;
for (i1=0; i1 <= n1-1; i1++) {
	var ch1 = s1.substring(i1,i1+1);
	console.log(i1+". "+ch1+"\n");
	}

Here is the output of the JavaScript code.

0. T

1. h

2. e

3.  

4. s

5. u

6. m

7.  

8. o

9. f

10.  

11. %

12. d

13.  

14. a

15. n

16. d

17.  

18. %

19. d

20.  

21. i

22. s

23.  

24. %

25. d

26. .

A common way to find multiple patterns in a string using a regular expression, where further processing is desired, is to match a prefix, the pattern, a suffix, and then do further matching on the suffix, saving intermediate results as needed.

The pattern ".*" matches zero or more of any character.
The pattern ".+" matches one or more of any character.
The pattern "x?" matches zero or more "x" characters (i.e., optional)
The vertical bar "|" within parentheses separates alternative matches.

Here is the JavaScript code.

var data1 = "The sum of %d and %d is %d.";
var pat1 = "^(.*)(%d)(.*)$";

var reg1 = new RegExp(pat1);
var match1 = reg1.exec(data1);
if (match1) {
	var prefix1 = match1[1];
	var middle1 = match1[2];
	var suffix1 = match1[3];
	console.log("1=["+prefix1+"] 2=["+middle1+"] 3=["+suffix1+"]\n");
	}

Here is the output of the JavaScript code.

1=[The sum of %d and %d is ] 2=[%d] 3=[.]

The problem here is that the last match is found, not the first match.

A greedy (or eager) match attempts to match as much of the text as possible (i.e., to the last match).

A lazy match stop as soon as possible (i.e., at the first match).

In the above case, a greedy match found the last match but ignored the others. In this case, a lazy match is needed.

The ".*" pattern matches in a greedy (eager) way.

The ".*?" pattern matches in a lazy way. That is, with the question mark "?" following the asterisk "*".

Here is the JavaScript code.

var data1 = "The sum of %d and %d is %d.";
var pat1 = "^(.*?)(%d)(.*)$";

var reg1 = new RegExp(pat1);
var match1 = reg1.exec(data1);
if (match1) {
	var prefix1 = match1[1];
	var middle1 = match1[2];
	var suffix1 = match1[3];
	console.log("1=["+prefix1+"] 2=["+middle1+"] 3=["+suffix1+"]\n");
	}

Here is the output of the JavaScript code.

1=[The sum of ] 2=[%d] 3=[ and %d is %d.]

Some languages have ways to help find multiple matches. But the following way works in any language (that has the regular expression pattern matching).

... to be added , as done in class ...