Difference between revisions of "CSC231 Mystery C Program with signed numbers"
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=Solution= | =Solution= | ||
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| + | ;Answer 1 | ||
* First we need to find the unsigned version of -32756, because that is the value we expected, but this value is too large to be represented in 2's complement. | * First we need to find the unsigned version of -32756, because that is the value we expected, but this value is too large to be represented in 2's complement. | ||
* The difference between a negative 16-bit number in 2's complement, and its unsigned equivalent is 2<sup>16</sup>, of 65536. | * The difference between a negative 16-bit number in 2's complement, and its unsigned equivalent is 2<sup>16</sup>, of 65536. | ||
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* Therefore x = 32700 | * Therefore x = 32700 | ||
* Plug these numbers in the program and verify that we get, indeed, the two numbers listed in the comment section of the program. | * Plug these numbers in the program and verify that we get, indeed, the two numbers listed in the comment section of the program. | ||
| + | <br /> | ||
| + | ;Answer 2 | ||
| + | * We could switch from shorts to ints | ||
| + | * We could use unsigned shorts | ||
</onlydft> | </onlydft> | ||
Revision as of 08:19, 13 November 2014
Mystery Program
- The program below is incomplete.
- The two values that were used to initialize x and y have been removed.
- We do have the output of the program, though, corresponding to the original x and y values.
- Question 1
- Figure out the value of x and y in the original program.
- Question 2
- Find a way to fix the program so that it outputs the correct information
You may use this converter to help you out...
/* mystery.cpp
// D. T.
// mystery program
// To compile and run this program:
//
// g++ mystery.cpp
// a.out
//
// The output of the program is the following
//
// 32740
// -32756
//
// With what positive values were x and y initialized
// at the beginning of the program. Explain why.
// (a short int contains 16 bits)
*/
#include <stdio.h>
int main() {
short int x = ???? ;
short int y = ???? ;
x = x+y;
printf( "%d\n", x );
x = x+y;
printf( "%d\n", x );
return 0;
}
For reference, 215 = 32768.
Solution!
</onlydft>
