CODEWITHSUNDEEP
cmacros
angrysumit
angrysumit

Reputation: 221

Variable undeclared error when used inside a function

Variable 'a' doesn't resolve by pre processor macro when function fun() is defined above main() but working fine when defined below main().

#include <stdio.h>

void fun()
{
    printf("%d",a);
}

int main()
{
    #define a 5
    fun();
    return 0 ;
}

What is the reason behind it? And how can we make it work so that I don't have to move the function fun() definition.

Upvotes: 1

Views: 1600

Answers (4)

P.P
P.P

Reputation: 121357

The C preprocessor runs top-down and replaces the #defines ("text replacement"). So, when #define a 5 in main(), it can't "go back" and replace a in func(); it can only replace the a's, if any, below its #define.

You just need to define it above func().

From C11 draft, 6.10.3 Macro replacement:

A preprocessing directive of the form

# define identifier replacement-list new-line

defines an object-like macro that causes each subsequent instance of the macro name 171) to be replaced by the replacement list of preprocessing tokens that constitute the remainder of the directive. The replacement list is then rescanned for more macro names as specified below.

This has been the same since pre standard C. From Dennis Ritchie's C Reference Manual:

12.1 Token replacement A compiler-control line of the form

# define identifier token-string

(note: no trailing semicolon) causes the preprocessor to replace subsequent instances of the identifier with the given string of tokens (except within compiler control lines). The replacement token-string has comments removed from it, and it is surrounded with blanks. No rescanning of the replacement string is attempted.

(emphasis mine).

Upvotes: 5

unwind
unwind

Reputation: 399753

Of course you can't reference a preprocessor macro before defining it, that's simply not how it works.

There's no "variable" here though, your usage of the word is very confusing. The preprocessor is simply doing text replacement, the compiler will never see the a symbol, it will see the 5 instead (if the macro was defined).

I don't know a way to "make it work" since the goal is very strange. If you want to access a global variable, use one:

#include <stdio.h>

int fun(int x)
{
    extern int a;
    return a + x;
}

int a = 12;

int main(void) {
    printf("a=%d, fun() returned %d\n", a, fun(11));
    return 0;
}

Here extern makes us get away with defining a after fun(), but it can't be inside main().

Upvotes: 4

AlexP
AlexP

Reputation: 4430

The preprocessing step is done before the compiling step. Moreover, the preprocessor does not know C; for the preprocessor the source file is just text.

In your example, the symbol a used in function fun() is not declared before use, which is an error. It does not matter that somewhere further down there is a preprocessor definition for a. Since the definition is further down it does not apply in the body of fun(): always remember that for the proeprocessor the source file is just text, it has no knowledge of C. Preprocessing directives apply at the point where they appear in text, never before.

Upvotes: 3

Colin
Colin

Reputation: 3524

The preprocessor will replace any instances of a with 5 after the point of macro definition. You could place the #define line above the function fun.

As a side note, a is not a variable, the preprocessor will just do a text replacement before it is compiled. It would change fun to

void fun()
{
    printf("%d", 5);
}

Upvotes: 3

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