3.2.5.2 analog C function scanf () function

In Python , there is no function function that is directly equivalent to scanf (), so you need to format the input, you need to use the function of regular expression, and the function of regular expression is more than scanf() more flexible and more powerful, here are some of the equivalent expressions:

scanf () format string	Regular expressions
%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]+

Enter an example of a string:

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

For strings in the format above, if you use the C function scanf() to enter, you need to use the following format to implement:

%s-%d errors,%d warnings

If we use a regular expression to represent it 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 resulting output is as follows:

scanf ()

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

Example of%c:

Print (' scanf ()%c ')

Pattern = Re.compile (r ".")

Match = Pattern.match (' This was for test\n ')

If match:

Print (Match.group ())

The resulting output is as follows:

scanf ()%c

T

examples of%5c:

Print (' scanf ()%5c ')

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

Match = Pattern.match (' This was for test\n ')

If match:

Print (Match.group ())

The resulting output is as follows:

scanf ()%5c

This

%e,%e,%f,%g Example:

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 ()) # mismatch no output

The resulting output is as follows:

scanf ()%e,%e,%f,%g

+200.3721

examples 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 (' 234.56\n ')

If match:

Print (Match.group ())

The resulting output is as follows:

scanf ()%i

0xaa55

234

An example of an octal %o :

Print (' scanf ()%o ')

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

Match = Pattern.match (' 0756\n ')

If match:

Print (Match.group ())

Match = Pattern.match (' 898\n ')

If match:

Print (Match.group ()) # mismatch no output

The resulting 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 small fishing village \ n ')

If match:

Print (Match.group ())

Match = Pattern.match (' 898\n ')

If match:

Print (Match.group ())

The resulting 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 (' 756\n ')

If match:

Print (Match.group ())

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

If match:

Print (Match.group ()) # mismatch no output

The resulting output is as follows:

scanf ()%u

756

example of hex %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 ()) # mismatch no output

The resulting output is as follows:

scanf ()%x

0x756

Cai Junsheng qq:9073204 Shenzhen

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3.2.5.2 analog C function scanf () function