(unsigned int) problem

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Level 1

0 points

(unsigned int) problem

Hi there,

Not sure if I am in the correct place to ask these kind of questions. However in any case, some help will be greatly appreciated.

The problem is that I do not know what is wrong with this code:

#import <Foundation/Foundation.h>

int main (void)

{

@autoreleasepool {

unsigned int i, j; // the problem is solved if this line is interchanged by:

// int i, j;

unsigned int m=4, n=3;

double matrix[m][n];

for (i=0; i<m; i++)

for (j=0; j<n; j++)

matrix[i][j]=5*i-2*j;

printf("Matrix:\n");

for (i=0; i<m; i++)

for (j=0; j<n; j++)

printf(" matrix[%u][%u] = %.3f\n", i, j, matrix[i][j]);

}

return 0;

}

When running this code, I get these odd results:

matrix[0][1] = 4294967294.000

matrix[0][2] = 4294967292.000

which should be:

matrix[0][1] = -2.000

matrix[0][2] = -4.000

Any idea?

Thank you!

P.S. This problem is solved if both i and j are defined as numbers of type int. Why is this? I am using Xcode 4.2.1.

Xcode 4.2.1-OTHER, Mac OS X (10.7.3)

Posted on Mar 11, 2012 10:21 AM

Reply

Answer 1

etresoft

Level 9

56,686 points

Mar 11, 2012 10:37 AM in response to adarpodracir

That is how a negative number is represented when you stuff it into an unsigned type.

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Answer 2

Jongware

Level 2

265 points

Mar 11, 2012 10:52 AM in response to adarpodracir

What etre says.

This is the responsible line:

matrix[i][j]=5*i-2*j;

Since both i and j are unsigned ints at that point, the right halve is evaluated using unsigned ints. Unsigned numbers don't go lower than 0 (they are so typical in that behavior, they even got named for that). So the result is the next best thing, a wrap-around to the *maximum* integer value. Which is actually the way 'negative' numbers work (the maximum positive value is half the max you can get with unsigned numbers -- minus 1, because there also needs to be room for a 'zero' somewhere).

After this right halve is evaluated (using unsigned arithmetic) it is assigned to the double variable, but at that point it's "too late" -- unsigned numbers will be silently converted to these.

You cn solve this in one of two ways: make i and j signed ints after all, or force of the two to be a double *at evaluation time*. That, don't use "(double)(5*i-2*j)" because that will give the same result, but use "5*(double)i-2*j" or any variant thereof. Even "5.*i" is already enough.

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Answer 3

adarpodracir Author

Level 1

0 points

Mar 11, 2012 11:40 AM in response to Jongware

Dear Jongware/etresoft,

Thank you very much for reply.

I think I am beginning to understand. Since this expresion:

5*i-2*j

is in function of i and j, which are unsigned integers, and because their coeffiecients are also unsigned integers (assuming that the minus sign is used here for subtraction), all the expresion is then computed by the computer suggesting the result is an unsigned integer as well. The resulting number is then converted to double. Am I correct?

Thank you.

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Answer 4

Jongware

Level 2

265 points

Mar 11, 2012 12:51 PM in response to adarpodracir

That is correct. Even working around the minus sign only will help:

matrix[i][j]=5*i + (-2)*j;

but a better solution would be to force either one of the variables or one of the constants (5 and 2) to the target type "double". Note that using fractional constants would also "automatically" solve it:

matrix[i][j]=1.5*i-2.3*j;

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