如何表示空字符？

You can use c[i]= '\0' or simply c[i] = (char) 0.

The null/empty char is simply a value of zero, but can also be represented as a character with an escaped zero.

The empty space char would be ' '. If you're looking for null that would be '\0'.

Yes, c[i]='' is not a valid code. We parenthesis character constant between ' ', e.g. c[i] = 'A'; char A. but you don't write any char in between ''.

Empty space is nothing but suppose if you wants to assigned space then do:

c[i] = ' ';
//      ^  space

if wants to assigned nul char then do:

c[i] = '\0';
//       ^ null symbol

Example: Suppose if c[] a string (nul \0 terminated char array) if you having a string. for example:

char c[10] = {'a', '2', 'c', '\0'};

And you replace second char with space:

c[1] = ' ';

and if you print it using printf as follows:

printf("\n c: %s", c);

then output would be:

  c:  a  c
//      ^ space printed

And you replace second char with '\0':

c[1] = '\0';

then output would be:

  c:  a

because string terminated with \0.

You can't store "no character" in a character - it doesn't make sense.

As an alternative you could store a character that has a special meaning to you - e.g. null char '\0' - and treat this specially.

To represent the fact that the value is not present you have two choices:

1) If the whole char range is meaningful and you cannot reserve any value, then use char* instead of char:

char** c = new char*[N];
c[0] = NULL; // no character
*c[1] = ' '; // ordinary character
*c[2] = 'a'; // ordinary character
*c[3] = '\0' // zero-code character

Then you'll have c[i] == NULL for when character is not present and otherwise *c[i] for ordinary characters.

2) If you don't need some values representable in char then reserve one for indicating that value is not present, for example the '\0' character.

char* c = new char[N];
c[0] = '\0'; // no character
c[1] = ' '; // ordinary character
c[2] = 'a'; // ordinary character

Then you'll have c[i] == '\0' for when character is not present and ordinary characters otherwise.

There is no such thing as the "empty character" ''.

If you need a space character, that can be represented as a space: c[i] = ' ' or as its ASCII octal equivalent: c[i] = '\040'. If you need a NUL character that's c[i] = '\0'.

It might be useful to assign a null in a string rather than explicitly making some index the null char '\0'. I've used this for testing functions that handle strings ensuring they stay within their appropriate bounds.

With:

char test_src[] = "fuu\0foo";

This creates an array of size 8 with values:

{'f', 'u', 'u', '\0', 'f', 'o', 'o', '\0'}

There are two ways to do the same instruction, that is, an empty string. The first way is to allocate an empty string on static memory:

char* my_variable = "";

or, if you want to be explicit:

char my_variable = '\0';

The way posted above is only for a character. And, the second way:

#include <string.h>
char* my_variable = strdup("");

Don't forget to use free() with this one because strdup() use malloc inside.

String before = EMPTY_SPACE+TAB+"word"+TAB+EMPTY_SPACE;

This is a case of single quotes and double quotes having different meanings.

"" is translated to (const char[1])"" by the compiler. This lets you use it in initializations of character arrays.

'' is not, and would be an unterminated empty string. Because you can't tell if a string is empty without terminating it, this is not valid code. Hence the error.

You almost certainly wanted to do

c[i] = '\0';

If c was indeed a text string, this sets the string's length to i by terminating it on that character.

If c was not actually intended as a text string, that's still the value you are suppose to use to mean that there's nothing there, because it's false, and any other character is true.

If you actually meant to put a space there, then you wanted

c[i] = ' ';