Problem with loop in C - c

Question

i'm trying to compute "2^0 + 2^1 + 2^2 + ... + 2^14", using the following program(i'm a newbie and can only compute a exponent by multiply itself a certain times). The result should be 32767, but i ran it and got 270566475, i thought for long but can't figure out why...
#include <stdio.h>
int main(void)
{
int i, e, exponent, sum;
e = 1;
exponent = 1;
sum = 1;
for (i = 1; i <=14; i++)
{
for (e = 1; e <= i; e++)
{
exponent *= 2;
}
sum += exponent;
}
printf("%d\n", sum);
return 0;
}
So what's wrong with this??? Thanks!!!

Answer 1

You don't need the inner loop. Just execute exponent *= 2 once, directly inside the outer loop. BTW, I think you have to do it after the sum += ....
Also, you could start with sum = 0 and i = 0, which is closer to the math you described.

Answer 2

Look at your inner loop by itself. It's trying to calculate, for one specific value of i, 2^i.
But exponent does not start at 1 every time. So you go into that loop with exponent already some very large value.
for (i = 1; i <=14; i++)
{
exponent = 1;
for (e = 1; e <= i; e++)
{
exponent *= 2;
}
sum += exponent;
}
Now you've reset exponent (which, to be clear, isn't the exponent at all but the calculated result) for each new power of 2.

Answer 3

If you have right to create a function it better to do it like this with a recursive function :
#include <stdio.h>
int power(int x, int exp) {
if (exp == 0)
return 1;
else
return x * power(x, exp-1);
}
int main (int argc, const char * argv[])
{
int i;
int sum = 0;
for (i = 0; i <= 14; i++) {
sum += power(2, i);
}
printf("%d",sum);
return 0;
}
I hope it helps.
You just need one loop because each you already have the result of n-1 value. I had correct your code it works.
#include <stdio.h>
int main (int argc, const char * argv[])
{
int i, e, exponent, sum;
e = 1;
exponent = 1;
sum = 1;
for (i = 1; i <= 14; i++)
{
exponent *= 2;
sum += exponent;
}
printf("%d\n", sum);
return 0;
}
Both codes work