Checks if the given character is a punctuation character in the current C locale. The default C locale classifies the characters !"#$%&'()*+,-./:;<=>?@[\]^_`{|}~
as punctuation.
The behavior is undefined if the value of ch is not representable as unsigned char and is not equal to EOF.
[edit] Parameters ch - character to classify [edit] Return valueNon-zero value if the character is a punctuation character, zero otherwise.
[edit] Example#include <ctype.h> #include <locale.h> #include <stdio.h> int main(void) { unsigned char c = '\xd7'; // the character à (multiplication sign) in ISO-8859-1 printf("In the default C locale, \\xd7 is %spunctuation\n", ispunct(c) ? "" : "not " ); setlocale(LC_ALL, "en_GB.iso88591"); printf("In ISO-8859-1 locale, \\xd7 is %spunctuation\n", ispunct(c) ? "" : "not " ); }
Possible output:
In the default C locale, \xd7 is not punctuation In ISO-8859-1 locale, \xd7 is punctuation[edit] References
ispunct
iswpunct
\x0
â\x8
\0
â\10
control codes (NUL
, etc.) â 0
0
0
0
0
0
0
0
0
0
0
0
9 \x9
\11
tab (\t
) â 0
0
â 0
â 0
0
0
0
0
0
0
0
0
10â13 \xA
â\xD
\12
â\15
whitespaces (\n
, \v
, \f
, \r
) â 0
0
â 0
0
0
0
0
0
0
0
0
0
14â31 \xE
â\x1F
\16
â\37
control codes â 0
0
0
0
0
0
0
0
0
0
0
0
32 \x20
\40
space 0
â 0
â 0
â 0
0
0
0
0
0
0
0
0
33â47 \x21
â\x2F
\41
â\57
!"#$%&'()*+,-./
0
â 0
0
0
â 0
â 0
0
0
0
0
0
0
48â57 \x30
â\x39
\60
â\71
0123456789
0
â 0
0
0
â 0
0
â 0
0
0
0
â 0
â 0
58â64 \x3A
â\x40
\72
â\100
:;<=>?@
0
â 0
0
0
â 0
â 0
0
0
0
0
0
0
65â70 \x41
â\x46
\101
â\106
ABCDEF
0
â 0
0
0
â 0
0
â 0
â 0
â 0
0
0
â 0
71â90 \x47
â\x5A
\107
â\132
GHIJKLMNOP
QRSTUVWXYZ
0
â 0
0
0
â 0
0
â 0
â 0
â 0
0
0
0
91â96 \x5B
â\x60
\133
â\140
[\]^_`
0
â 0
0
0
â 0
â 0
0
0
0
0
0
0
97â102 \x61
â\x66
\141
â\146
abcdef
0
â 0
0
0
â 0
0
â 0
â 0
0
â 0
0
â 0
103â122 \x67
â\x7A
\147
â\172
ghijklmnop
qrstuvwxyz
0
â 0
0
0
â 0
0
â 0
â 0
0
â 0
0
0
123â126 \x7B
â\x7E
\173
â\176
{|}~
0
â 0
0
0
â 0
â 0
0
0
0
0
0
0
127 \x7F
\177
backspace character (DEL
) â 0
0
0
0
0
0
0
0
0
0
0
0
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