Is it possible to write something like this:
printf(#"
-
-
-
-
");
I can do it in C#, but can't in C. It gives me an error in CodeBlocks. Am I allowed to do such ?
Error message: error: stray '#' in program.
No. That syntax doesn't exist in C.
If you want a multiple-line string, write it as multiple double-quoted strings with no other tokens in between them. They will be combined.
printf(
"some string"
"more of the string"
"even more of the string"
);
(You will, of course, need to add a \n at the end of each line if that's what you want.)
No that's not a syntax that C understands, C doesn't have raw literals.
You can use \ as the last character to continue on the next line:
const char *str = "hello\n\
world";
Also, consecutive string literals will be concatenated. So you can do e.g.
const char *str = "Hello\n"
"world\n";
C#'s verbatim strings are not available in C. If you have some characters to escape, like " or \, escape them with '\', there is no there option in this language.
If you want to embed multiple lines in a string literal, you can either insert \n at the appropriate location in your string, or escape the return character as well:
printf("Here's\
a multiline\
string litteral");
Line continuation with \ at the end of the line.
printf("\
\
-\
-\
-\
-\
");
String literals in C may not contain newlines. You have two workarounds:
Use implicit string concatenation (done by the compiler).
printf("The quick brown"
" fox jumps over"
" the sleazy dog.");
Escape the newline by placing a backslash in front of it.
printf("The quick brown\
fox jumps over\
the sleazy dog.");
Personally, I prefer the first form since the second looks ugly (my opinion) and forces you to ruin your code indentation.
In either case, the string will simply not contain the newlines. So if you really meant for them to be there, you'll have to add them via \n.
Related
(I am very new to C.)
Visual newlines seem to be unimportant in C.
For instance:
int i; int j;
is same as
int i;
int j;
and
int k = 0 ;
is same as
int
k
=
0
;
so why is
"hello
hello"
not the same as
"hello hello"
It is because a line that contains a starting quote character and not an ending quote character was more likely a typing mistake or other error than an attempt to write a string across multiple lines, so the decision was made that string literals would not span source lines, unless deliberately indicated with \ at the end of a line.
Further, when such an error occurs, the compile would be faced with reading possibly thousands of lines of code before determining there was no closing quote character (end of file was reached) or finding what was intended as an opening quote character for some other string literal and then attempting to parse the contents of that string literal as C code. In addition to burdening early compilers with limited compute resources, this could result in confusing error messages in a part of the source code far removed from the missing quote character.
This choice is effected in C 2018 6.4.5 1, which says that a string literal is " s-char-sequenceopt ", where s-char-sequence is any member of the character set except the quote character, a backslash, or a new-line character (and a string literal may also have an encoding prefix, a u8, u, U, or L before the first ").
Strings can be continued over newlines by putting a backslash immediately before the newline:
"hello \
hello"
Or (better), using string concatenation:
"hello "
"hello"
Note that the space has been carefully preserved so that these are equivalent to "hello hello" except for the line numbering in the file after the appearance.
The backslash-newline line elimination is done very early in the translation process — in phase 2 of the conceptual translation phases.
Note that there is no stripping leading blanks or anything. If you write:
printf("Some long string with maybe an integer %d in it\
and some more data on the next line\n", i);
Then the string has a sequence of (at least) 8 blanks in it between in it and and some. The count of 8 assumes that the printf() statement is aligned in the left margin; if it is indented, you'd need to add the extra white space corresponding to the indentation.
1- using double quotes for each string :
char *str = "hello "
"hello" ;
** One problem with this approach is that we need to escape specially characters such as quotation mark " itself.
2- Using - \ :
char *str = "hello \
hello" ;
** This form is a lot easier to write, we don't need to write quotation marks for each line.
We can think of a C program as a series of tokens: groups of characters that can't be split up without changing their meaning. Identifiers and keywords are tokens. So are operators like + and -, punctuation marks such as the comma
and semicolon, and string literals.
For example, the line
int i; int j;
consists of 6 tokens: int, i, ;, int, j and ;. Most of the time, and particularly in this case, the amount of space (space, tab and newline characters) is not critical. That's why the compiler will treat
int i
;int
j;
The same.
Writing
"Hello
Hello"
Is like writing
un signed
and hope that the compiler treat it as
unsigned
Just like space is not allowed between a keyword, newline character is not allowed in a string literal token. But it can be included using the newline escape '\n' when needed.
To write strings across lines use string concatenation method
"Hello"
"Hello"
Although the above method is recommended, you can also use a backslash
"Hello \
Hello"
With the backslash method, beware of the beginning space in a new line. The string will include everything in that line until it finds a closing quote or another backslash.
Basically, I can't figure this out, I want my C program to store the entire plaintext of a batch program then insert in a file and then run.
I finished my program, but holding the contents is my problem. How do I insert the code in a string and make it ignore ALL special characters like %s \ etc?
You have to escape special characters with a \, you can escape backslash itself with another backslash (i.e. \\).
As Ian previously mentioned, you can escape characters that aren't allowed in normal C strings with \; for instance, newline becomes \n, double-quote becomes \", and backslash becomes \\.
If you're unable or unwilling to do this for whatever reason, then you may be out of luck if you're solution must be in C. However, if you're willing to switch to C++, then you can use raw strings:
const char* s1 = R"foo(
Hello
World
)foo";
This is equivalent to
const char* s2 = "\nHello\nWorld\n";
A raw string must begin with R" followed by an arbitrary delimiter (made of any source character but parentheses, backslash and spaces; can be empty; and at most 16 characters long), then (, and must end with ) followed by the delimiter and ". The delimiter must be chosen such that the termination substring (), delimiter, ") does not appear within the string.
I am reading the Kernigan and Ritchie manual on C.
The following example arises:
printf("hello, world
");
The book states that the C compiler will produce an error message. How exactly does the compiler detect this, and why is this an issue? Shouldn't it just read the newline (presumably at the end of world) as no space?
You are not allowed to have a newline in a string literal, we can see this by looking at the grammar from the C99 draft standard section 6.4.5 String literals:
string-literal:
" s-char-sequenceopt "
L" s-char-sequenceopt "
s-char-sequence:
s-char
s-char-sequence s-char
s-char:
any member of the source character set except
the double-quote ", backslash \, or new-line character
We can see that s-char allows any character except ", \ and new-line.
The compiler detects the error because it finds a \n newline (ie, the one right after the d in world) before it has found a " (ie, to match the opening ").
It does this because, as other commenters have said, that is what the specification requires that it do.
There's "\n" in the argument. Never hit enter while writing arguments for printf() :)
In C it is not normally possible to use ' for printf of a string. However, I have text which are full of double quote ", and I need to escape all of them as
printf("This is \"test\" for another \"text\"");
Is it possible to printf in a way without escaping ". I mean using another character for wrapping the string.
Not recommended, but you can use a macro:
#include <stdio.h>
#define S(x) #x
int main() {
printf(S(This "is" a string (with nesting)!\n));
}
This prints
This "is" a string (with nesting)!
Now the delimiters are balanced () characters. However, to escape single ), ", or ' characters, you have to write something like S(right paren: ) ")" S(!\n), which is quite ugly. This technique is not recommended for writing maintainable code.
No, that is not possible in the C language. There is only one syntax for string literals, and that is that they are delimited by double quotes.
The only way to write unescaped quotation marks is as character literals inside character arrays, which is uglier and more difficult to write, so there's very little reason to do so in a case like this:
char array[] = {'T', 'h', 'i', 's', ' ', 'i', 's', ' ', '"'}; // etc.
printf("%s", array);
No there is not other way, the draft C99 standard in section 6.4.5 String literals has the following grammar:
string-literal:
" s-char-sequenceopt "
L" s-char-sequenceopt "
No, it's not possible in standard C.
C11 6.4.5 String literals
The same considerations apply to each element of the sequence in a string literal as if it
were in an integer character constant (for a character or UTF−8 string literal) or a wide
character constant (for a wide string literal), except that the single-quote ' is representable either by itself or by the escape sequence \', but the double-quote " shall be represented by the escape sequence \".
First of all, separate a program's requirements from the solutions to meet those requirements. Given the minimum amount of info. in this question, the requirement is to print, using C, a string that has double quotes. There are several ways to do this in C.
For example, the following code fragment:
char string[] = "This string \" has one double quote.";
printf("This string %cprints%c with %cdouble%c quotes", '"', '"', '"', '"');
printf("%s", string);
produces:
This string "prints" with "double" quotes.
This string " has one double quote.
Your application might have more requirements that you have not mentioned, but it should be possible to achieve what you want, just NOT the way you initially believe it should be done (welcome to the world of "Needs Analysis").
//R "delimiter( raw_characters )delimiter"
printf(R"SOME/\STRING");
Raw string will terminate after the first )" it sees.
Therefore, if )" is in the string, you have to add delimiter ("a" is used below)
/* Print dog without any escape. */
printf(R"a(|\_/|
|q p| /}
( 0 )"""\
|"^"` |
||_/=\\__|)a");
}
It's C++11 feature and you can find more information in
document
Simliar question have been answered escape R"()" in a raw string in C++
I'm a perl programmer and surprised to find c language has no a convenient way to store a paragraph, like:
my $a = <<'dd';
hello wolrd..
1
2
3
dd
So how do I the smiliary operation in C?
It's done like this:
char a[] =
"hello world\n"
" 1\n"
" 2\n"
" 3\n";
You've two different answers to your question which provide many of the same functionalities.
Both
char *str="Line 1\nLine 2\nLine 3";
and
char str[]="Line 1\nLine 2\nLine 3";
Allow you to print a paragraph as such:
printf("%s",str);
However, the first declaration (char *str) creates a string in what is generally read-only memory, whereas the second allows the string to be edited during run-time. This delineation is important, but not always clear. See this question for a few more details.
The character \n is the line feed character, and you should check to ensure that it behaves the way you expect it to on your target platform. For instance, on DOS you may need to use `"\r\n", which is carriage return + line feed. Wiki has an article about this.
Another difference in these forms, as one commenter pointed out, is that *str works as a pointer whereas str[] does not. They often, but not always, have the same behavior; this question has more information regarding this.
As some commenters have pointed out, there is a limit on the length of string literals in some compilers. MSVC has a limit of 2048 characters (see here) whereas GCC has no limit, by some accounts. A length of at least 509 one-byte characters is guaranteed by C90; this was increased to 4095 in C99.
Regardless, if you want to avoid this length limit or you want to organize the text in a prettier way, you can use this format (note that newlines and quotes must be used explicitly, the compiler treats the adjacent strings as being concatenated):
char *str =
"Line 1\n"
"Line 2\n"
"Line 3\n";
or this (the backslashes at the end of the line escape the newline you've inserted for the formatting, if you indent your code here, that will become part of the string):
char *str =
"Line 1 \
Line 2 \
Line 3";
you may try
char *str = "hello world\n 1\n 2\n 3\n";