What is the # for when formatting using %s

I came across this example of an assertion and was wondering what the `#` is for:

#define ASSERT( x ) if ( !( x ) ) { \
int *p = NULL; \
DBGPRINTF("Assert failed: [%s]\r\n Halting.", #x); \
*p=1; \
}

`#x` is the stringification directive

`#define Stringify(x) #x`

means `Stringify(abc)` will be substituted with `"abc"`

as in

#define initVarWithItsOwnName(x) const char* p = #x

int main()
{
initVarWithItsOwnName(Var);
std::cout << Var; //will print Var
}

It is the "stringize" preprocessing operator.

It takes the tokens passed as the argument to the macro parameter `x` and turns them into a string literal.

#define ASSERT(x) #x

ASSERT(a b c d)
// is replaced by
"a b c d"

`#` is the preprocessor's ["stringizing" operator][1]. It turns macro parameters into string literals. If you called `ASSERT(foo >= 32)` the `#x` is expanded to `"foo >= 32"` during evaluation of the macro.

[1]:

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It's a preprocessor feature called [stringification][1]. It

> replaces [the macro parameter] with the literal text of the actual argument, converted to a string constant.


[1]:

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It's the stringizing operator.

[To see links please register here]

What you see is called *stringification*. It allows you to convert an argument of a macro into a string literal. You can read more about it here

[To see links please register here]

.

`#` is the stringizing operator defined in Section 6.10.3.2 (C99) and in Section 16.3.2. (C++03)

It converts macro parameters to string literals without expanding the parameter definition.

>If the replacement that results is not a valid character string literal, the
behavior is **undefined**. The order of evaluation of # operator is **unspecified**.

For instance, syntactically, occurrences of the backslash character in string literals are limited to escape sequences.

In the following example:

1 #define mkstr(x) # x
2 char *p = mkstr(a \ b);
/* "a \ b" violates the syntax of string literals */

the result of the `#` operator need not be `"a \ b"`.