3.2.5.2 simulate the C function scanf () function, 3.2.5.2scanf

Source: Internet
Author: User

3.2.5.2 simulate the C function scanf () function, 3.2.5.2scanf

In Python, there is no function directly equivalent to scanf (). Therefore, to format the input, you need to use the regular expression function. In addition, the function of the regular expression is better than that of scanf () more flexible and more powerful functions. The following describes some Equivalent Expressions:

Scanf () Format String

Regular Expression

% C

.

% 5c

. {5}

% D

[-+]? \ D +

% E, % E, % f, % g

[-+]? (\ D + (\. d *)? | \. \ D +) ([eE] [-+]? \ D + )?

% I

[-+]? (0 [xX] [\ dA-Fa-f] + | 0 [0-7] * | \ d +)

% O

[-+]? [0-7] +

% S

\ S +

% U

\ D +

% X, % X

[-+]? (0 [xX])? [\ DA-Fa-f] +

 

 

Example of a string:

/Usr/sbin/sendmail-0 errors, 4 warnings

For strings in the preceding format, if you use the C function scanf () for input, you must use the following format:

% S-% d errors, % d warnings

If we use a regular expression, it is represented as follows:

(\ S +)-(\ d +) errors, (\ d +) warnings

Example:

Print ('scanf ()')

Pattern = re. compile (r "(\ S +)-(\ d +) errors, (\ d +) warnings ")

Match = pattern. match ('/usr/sbin/sendmail-0 errors, 4 warnings ')

If match:

Print (match. groups ())

The output is as follows:

Scanf ()

('/Usr/sbin/sendmail', '0', '4 ')

 

Example of % c:

Print ('scanf () % C ')

Pattern = re. compile (r ".")

Match = pattern. match ('this is for test \ n ')

If match:

Print (match. group ())

The output is as follows:

Scanf () % c

T

 

Example of % 5c:

Print ('scanf () % 5c ')

Pattern = re. compile (r ". {5 }")

Match = pattern. match ('this is for test \ n ')

If match:

Print (match. group ())

The output is as follows:

Scanf () % 5c

This

 

Example of % e, % E, % f, % g:

Print ('scanf () % e, % E, % f, % G ')

Pattern = re. compile (r "[-+]? (\ D + (\. \ d *)? | \. \ D +) ([eE] [-+]? \ D + )? ")

Match = pattern. match ('+ 200.3721 \ n ')

If match:

Print (match. group ())

Match = pattern. match ('x9876 \ n ')

If match:

Print (match. group () # unmatched no output

The output is as follows:

Scanf () % e, % E, % f, % g

+ 200.3721

 

Example of % I:

Print ('scanf () % I ')

Pattern = re. compile (r "[-+]? (0 [xX] [\ dA-Fa-f] + | 0 [0-7] * | \ d + )")

Match = pattern. match ('0xaa55 \ n ')

If match:

Print (match. group ())

Match = pattern. match ('192. 56 \ n ')

If match:

Print (match. group ())

The output is as follows:

Scanf () % I

0xAA55

234

 

Example of % o of octal:

Print ('scanf () % o ')

Pattern = re. compile (r "[-+]? [0-7] + ")

Match = pattern. match ('2014 \ n ')

If match:

Print (match. group ())

Match = pattern. match ('2014 \ n ')

If match:

Print (match. group () # unmatched no output

The output is as follows:

Scanf () % o

0756

 

Example of string % s:

Print ('scanf () % s ')

Pattern = re. compile (r "\ S + ")

Match = pattern. match ('shenzhen is a fishing village \ n ')

If match:

Print (match. group ())

Match = pattern. match ('2014 \ n ')

If match:

Print (match. group ())

The output is as follows:

Scanf () % s

Shenzhen is a small fishing village

898

 

Example of % u:

Print ('scanf () % U ')

Pattern = re. compile (r "\ d + ")

Match = pattern. match ('2014 \ n ')

If match:

Print (match. group ())

Match = pattern. match ('-898 \ n ')

If match:

Print (match. group () # unmatched no output

The output is as follows:

Scanf () % u

756

 

Example of hexadecimal % x, % X:

Print ('scanf () % x % x ')

Pattern = re. compile (r "[-+]? (0 [xX]) [\ dA-Fa-f] + ")

Match = pattern. match ('0x756 \ n ')

If match:

Print (match. group ())

Match = pattern. match ('-898 \ n ')

If match:

Print (match. group () # unmatched no output

The output is as follows:

Scanf () % x % X

Zero x 756




Cai junsheng QQ: 9073204 Shenzhen

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