[SOLVED] C math question

errigour · 09-29-2012, 07:29 AM

How come maxi equals -15?

Code:

#include <stdio.h>
#include <limits.h>
int main()
{
     int imin = INT_MIN; // minimum value
     int imax = INT_MAX; // maximum value
     int maxi, j;
     int pow(int a, int b)
     {
	  int sum, j = 0;
	  for ( j = 0; j < b; j++)
	       sum *= a;
	  return sum;
     }
     maxi += pow(2, 4) + 1;
     printf("%d:%d\r\n", sizeof (int), imin);
     printf("%d:%d\r\n", sizeof (int), imax);
     printf("%d:%d\r\n", sizeof (int), maxi);
}

johnsfine · 09-29-2012, 07:48 AM

Quote:

Originally Posted by errigour

How come maxi equals -15?

Look at every line that mentions maxi

Quote:

Code:

     int maxi, j;
...
     maxi += pow(2, 4) + 1;
...
     printf("%d:%d\r\n", sizeof (int), maxi);
}

The first line creates a local variable named maxi, uninitialized meaning the value could be anything.

The second line adds something to that unpredictable starting value yielding a still unpredictable value.

Since you make a similar error with sum inside the nested function, the result would still be unpredictable even if you fixed the first error.

dwhitney67 · 09-29-2012, 07:49 AM

Consider initializing 'maxi' to something (e.g. zero). Also, sizeof() returns an unsigned long value; fix your printf() statements to use the formatter "%lu" for the sizeof() values.

Also, just so you know, your program does not conform to ISO C standards. You may want to consider placing your pow() function outside the scope of main(). Recompile your program using -pedantic as a compiler option to see the warning.

EDIT:

Also initialize 'sum' to some sane value in pow().

errigour · 09-29-2012, 07:55 AM

How come maxi here yields -64

Code:

#include <stdio.h>
#include <limits.h>
int power(int a, int b)
{
     int sum, j = 0;
     for ( j = 0; j < b; j++)
     sum *= a;
     return sum;
}
int main()
{
     int imin = INT_MIN; // minimum value
     int imax = INT_MAX; // maximum value
     int maxi = 0;

     maxi = power(2, 6);
     printf("%lu:%d\r\n", sizeof (int), imin);
     printf("%lu:%d\r\n", sizeof (int), imax);
     printf("%lu:%d\r\n", sizeof (int), maxi);
}

dwhitney67 · 09-29-2012, 07:56 AM

Code:

...
int sum, j = 0;
...

Initialize 'sum' to zero.

And it should be:

Code:

...
sum += a;
...

Don't multiply, add.

EDIT: 'sum' should be multiplied, not added. Scratch my thoughts above.

errigour · 09-29-2012, 07:57 AM

I thought int sum, j =0; initialized it but even after sum = 0 maxi ends up to be 0.

errigour · 09-29-2012, 07:58 AM

why add I want to know what 2 * 2 * 2 * 2 * 2 * 2 is

dwhitney67 · 09-29-2012, 07:59 AM

Quote:

Originally Posted by errigour

I thought int sum, j =0; initialized it but even after sum = 0 maxi ends up to be 0.

You can initialize like this:

Code:

int sum = a, j = 0;

---------- Post added 09-29-12 at 08:59 ----------

Quote:

Originally Posted by errigour

why add I want to know what 2 * 2 * 2 * 2 * 2 * 2 is

Dooh! You are right... I meant to initialize 'sum' to the value of 'a' (your first parameter).

firstfire · 09-29-2012, 08:01 AM

Hi.

You should initialize sum to 1 (or to a if iterate to j<b-1), because sum is actually a product: sum = 1*a*a*a... You also do not need to initialize j, because it is initialized in the loop.

errigour · 09-29-2012, 08:01 AM

Wow I completly missed that

---------- Post added 09-29-12 at 09:02 AM ----------

sorry and thanx

johnsfine · 09-29-2012, 08:17 AM

Quote:

Originally Posted by dwhitney67

Initialize 'sum'

Yes.

Quote:

to zero.

No.

Edit: Oops, I missed several posts despite refreshing the thread, so I gave an answer already given in more detail above.