Include double-quote (") in C-string

Question

I would like a to define a variable string in C that contains the following set of characters: a-zA-Z0-9'-_”.

Therefore I would do it like this:

char str[64] = "abcdefghijklmnopqrstuwxyzABCDEFGHIJKLMNOPQRSTUWXYZ0123456789'-_""

As you can see the problem is at the end with " character.

Question 1: How can I work around that?

Question 2: Is there a better way than my way to define a string like that?

PS: I didn't really know how to title my question, so if you got a better one, please edit it.

Escape it with the backslash: `"\""`. Note that with `str[64]` you have enough space for all your 64 characters, but not for a null-terminator. Depending on how you plan to use the string this might be a problem. With this initialisation though, specifying the length is optional so `char str[] = "...";` is valid and will automatically make the buffer big enough to contain the string + null-terminator. — Kninnug, Dec 08 '13 at 20:15
Unless you've omitted some letters that I didn't notice that's 66 characters. — Hot Licks, Dec 08 '13 at 20:22
@HotLicks I copied & pasted it into Notepad++ which counted 64 characters: there are no v's in it (upper- nor lower-case). — Kninnug, Dec 08 '13 at 20:24
Second question: it depends. What are you using this string for? Can the same thing be done using code? Can the same thing be done using its inverse -- a list of printable ASCII characters that are *not* in this list (about 30, with or without "v")? Can you use a bitmap array instead of literal characters? Can you use `ctype` instead? — Jongware, Dec 08 '13 at 21:09

Glenn Teitelbaum · Accepted Answer · 2013-12-08T20:36:11.270

17

use the backslash: "\"" is a string containing "

like this:

char str[67] = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789'-_\"";

added one for the implicit '\0' at the end (and put in the missing vV) - this could also be:

char str[] = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789'-_\"";

and let the compiler count for you - then you can get the count with sizeof(str);

How does it add up to 67?

a-z 26
A-Z 26
0-9 10
'-_" 4
'\0' 1
    ---
    67

edited Dec 08 '13 at 20:36

answered Dec 08 '13 at 20:16

Glenn Teitelbaum

Ok, thanks. But is there a better way to define this string? – Dragos Rizescu Dec 08 '13 at 20:19
If you want it as a string - that is what you need to do -- if you you want a regex (regular expression) then you can express it according to the regex engine you use -- in C++11 there is `std::regex` – Glenn Teitelbaum Dec 08 '13 at 20:20
Should be `char str[66] = ...`, shouldn't it? – Hot Licks Dec 08 '13 at 20:22
Or you could just omit the array-length, as it's optional with such an initialization. Also: there are no v's in it, so 65 is enough for the string + null-terminator. – Kninnug Dec 08 '13 at 20:25
@HotLicks - 67 editted to show math – Glenn Teitelbaum Dec 08 '13 at 20:31
@Kninnug I think the missing Vs are a typo since the question asked about A-Z and a-z – Glenn Teitelbaum Dec 08 '13 at 20:34

score 6 · Answer 2 · answered Dec 08 '13 at 20:16

6

Use "\"" (backslash") for putting " in a string

answered Dec 08 '13 at 20:16

2 Answers2