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It will give a syntax error as "x undeclared" .. The preprocessor replaces all the statements followed by the #define statements within the scope..

This will work without any error-
void main()
{
#define x 1
funct();
}

funct()
{
printf("%d",x);
}

whereas this will give an error-
void main()
{
funct();
printf("%d",x);
}

funct()
{
#define x 1
}

They are defined in global var or static var area and have global scope.

#define is a preprocessor. it doesn't take any area inside the program memory. Before compilation it replace with the actual value inside the code.

i dont agree with tejas if variables defined by #define r placed in global area then it does not matter where they r declared they should be visible

Yes deestiny. my explanation for the code is wrong .. however, output is correct..
The correct reason is.. preprocessor scans the code from beginning and replaces all the statements followed by a #define statement with the corresponding value..
g()
{
#define b 6
}
int main()
{
g();
f();
printf("%d",b);
return 0;
}
f()
{
#define b 4
}
Now.. the b value in main() is 6..
If you remove the call to method g(), then also output will be 6, as preprocessor doesnt care the execution sequence.. It just scans the files executing the preprocessor statements..

For the above snippet,it would give a compilation error stating that x was undeclared.However if you reorder the code.

void funct()
{
#define x 1
}

void main()
{
funct();
printf("%d",x);
}

this would print 1.

The subquestion to be answered is where does b is stored?
--> I think it is not stored in Data Segment or BSS. All the instances of b in code segment are replaced by 6. And hence b is not actually stored.

Answer is compilation error.
so how does it work and where is it stored. pre-processor parses and does textual replacement of the value. So the value is nothing but some constant, it is stored in global/static area and the type used will be based on the value.

to some of the people ho had doubt abut why it prints 4 at one place and 6 at other place bcoz pre-processor would have substituted so and when it comes to compilation stage the values are already substituted and these values will be then stored in global/static area.

Answer is 1.
I compiled and ran the program.

I executed it too.
Throws an error if f() called after printf..
prints 1 if f() called and defined before.. tried in both c and c++