how I can set binary data in a char[]

Question

I have const binary data that I need insert to buffer for example

 char buf[] = "1232\0x1";

but how can do it when binary data is at first like below

 char buf[] = "\0x11232";

compiler see it like a big hex number but my perpose is

 char buf[] = {0x1,'1','2','3','2'};

`\0x1` isn't hex for 1 but a NULL byte followed by the characters 'x' and '1' — Seth Carnegie, May 15 '12 at 07:22
You can't use either of your first two snippets. Just do it long hand like your third snippet. — Corbin, May 15 '12 at 07:23

ThiefMaster · Accepted Answer · 2012-05-15T07:31:25.933

4

You can use compile-time string concatenation:

char buf[] = "\x01" "1232";

~~However, with a 2-digit number after \x it also works without:~~

char buf[] = "\x011232";

edited May 15 '12 at 07:31

answered May 15 '12 at 07:25

ThiefMaster

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This is incorrect. The escape sequence will consume more than two characters in the second example. See http://stackoverflow.com/q/5784969/163956 – Greg Inozemtsev May 15 '12 at 07:28

score 3 · Answer 2 · answered May 15 '12 at 07:31

You can create a single string literal by composing it of adjacent strings - the compiler will concatenate them:

char buf[] = "\x1" "1232";

is equivalent to:

char buf[] = {0x1,'1','2','3','2', 0};  // note the terminating null, which may or may not be important to you

Felice Pollano · Answer 3 · 2012-05-15T07:29:16.520

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You have to write it in two byte or four byte format:

\xhh = ASCII character in hexadecimal notation

\xhhhh = Unicode character in hexadecimal notation if this escape sequence is used in a wide-character constant or a Unicode string literal.

so in your case you have to write "\x0112345"

edited May 15 '12 at 07:29

answered May 15 '12 at 07:26

Felice Pollano

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how I can set binary data in a char[]

3 Answers3