Why does a variable of a function declared outside its function definition doesn't throw an error?

Question

Why is this Code with integer declaration in the middle of nowhere (in-between function definition), not throwing an error?

1) Why is it syntactically correct.
2) What is the use of doing so.?

#include <stdio.h>  
void func(int, int);
int main()
{
     int a, b;
     a = 10;
     b = 20;
     func(a, b);

     return 0;
}
void func(i, j)
int i,j;                  //why does this doesn't throw error.
{
     printf("a = i = %d\nb = j = %d\n", i, j);
}

As answered below, it's syntactically correct because it is valid -- it's the original way things were done. There isn't any use of using this syntax now though unless you happen to be using a K&R style compiler that does not support ANSI C. — mah, Feb 19 '16 at 17:35
And don't **EVER** provide a prototype for a function defined in K&R style. To find out why, change `int` to `char` in your function prototype and also in the definition. — Andrew Henle, Feb 19 '16 at 17:49

score 3 · Accepted Answer · edited May 23 '17 at 12:07

3

TL;DR - K&R style.

In your code

void func(i, j)
int i,j; {

does not throw an error, because it was a (the only) valid syntax, once upon a time.

Currently it is not invalid, but not used anymore.

You can read more about K&R style syntax here, if you want.

edited May 23 '17 at 12:07

Community

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answered Feb 19 '16 at 17:33

Sourav Ghosh

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score 1 · Answer 2 · edited May 23 '17 at 12:23

1

This

void func(i, j)
int i,j; //declare types of arguments
{
  //function body
}

is called the K&R syntax. It's obsolete but still used in some C projects like bash.

edited May 23 '17 at 12:23

Community

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answered Feb 19 '16 at 17:36

Petr Skocik

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Why does a variable of a function declared outside its function definition doesn't throw an error?

2 Answers2