Trying to Write a Perfect Number function....and getting syntax errors...?

Question


#define n 9999
bool IsPerfect(int n);


main()
{
int j;

for(j = 1; j <=n; j++)
{
if(IsPerfect(n) == TRUE)
{
printf("%d", IsPerfect(n));
}
}
}
bool IsPerfect(int n)
{
int total, i;
total = 0;

for(i=1; i <= n; i++)
{

if(n % i == 0)
{
total += i;


}
if(total == i)
{
return(TRUE);
}
else
{


return(FALSE);
}
}
}


Ok so I get a syntax error on the two lines that have "bool IsPerfect(int n)" and I have no idea why.....help?

Ok so I made it "bool IsPerfect()" but now all i get is 0....probably 9999 of them....whats wrong with my math in the IsPerfect function??? too many ifs?

Answer 1

It's this: You have a #define n 9999. What this means is that the C/C++ parser (after the preprocessor has done its thing) sees the two bool IsPerfect( int n ) lines as bool IsPerfect( int 9999). You can't have a constant as a formal parameter.

Hope that helps.

Answer 2

You seem to have an extra right brace below the "printf("%d", IsPerfect(n)" expression, or above the "bool IsPerfect(int n)" expression. I'm a little rusty on C++, but I seem to recall that you won't encounter errors specifically saying "extra parenthesis" or "extra brace", but they will cause syntax errors. With the extra brace above it makes the compiler think you are trying to close out main() at that step.

Answer 3

You didn't provide enough detail. Is this C or C++? What *exactly* is the syntax error you're getting. Also, I'm pretty sure that in both C and C++, the main() function must always return an integer.

You can email me if you need help, but you need to give me more info.

Answer 4

i don't be attentive to why it truly is happening even though it truly is happeneing to me to so i anticipate it truly is something incorrect with the full internet site! I logged on and had a couplequestions to respond to and have been given an identical message. supply it an hour and see if its back to widely used!

Answer 5

instead of using #define use

const int n=99999