"Type Mismatch in redeclaration"

can you post your entire code?

int main()
{
int i=2,j=3,k;
k=sum(i,j);
printf("%d\n",k);

return 0;
}

int sum(int p,int q)

{
int m;
m=p+q;
return m;
}


-chirag

oh no, i'm too late, Alf already massacre the question =)

Tongue in cheek
Alf, how many lines would it take you to answer the question: 1 + 1?
Tongue out of cheek

/me pretending to be Alf

well, 1+1 is an interesting question,  if you go back a bit to the dinosaur age and perform the calculation using dinosaur bone....

....
....< a year later>
.....

and that's how we arrive at the modern age answer that 1+1=2

=)

in C, without a function declaration, you will get this because the function definition will assume type

void sum(void);

and you will get your error.  I imagine that you want sum to return an int.  simply return m from your current and make the following declaration before you ever USE sum:

int sum(int,int);

int main(...){...}

int sum(int i, int j){...}

Then you won't get this error.

sarithara, post your code

I will send the code. Please give time

I wouldn't go back to dinosaurs but I could show the formal answer assuming Peano's postulates ;)

1 + 1 = 1 + s(0) = s(1 + 0) = s(1) = 2

There ;) It was only one line! ;) Of course, if we have to explicitely state the postulates and definitions required it will be more than one single line:

D1.   s(0)       === 1  (=== means "equal by definition").
D2.   s(1)       === 2
D3.   x + 0     === x
D4.   x + s(y) === s(x + y)

this gets:
1 + 1 = 1 + s(0)  // since s(0) = 1 by definition D1.
1 + s(0) = s(1 + 0) // since x + s(y) = s(x + y) by definition D4.
s(1 + 0) = s(1) // since 1 + 0 = 1 by definition D3.
s(1) = 2 by definition D2.

so 1 + 1 = 2 QED.

Now, try using those same definitions (actually only D3 and D4 are needed) and show that x + y = y + x for all y and all x. It is possible but it's far from trivial. You can also show that x + (y + z) = (x + y) + z using only D3 and D4.

Good luck ;)

Alf