Thread: confused by handling of data type

  1. 07-09-2002 #1
    The Gweech
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    confused by handling of data type

    I have a question regarding data types

    1. This code
    unsigned short num=65535;
    cout << num+2 << endl;

    outputs 65537 . I expected it to output 1 because of the value
    wrapping around itself if you get my meaning. Does the cout
    object have an ability to change to a more suitable data type?
    Can anybody shed some light?

    Thanks!
  2. 07-09-2002 #2
    SilentStrike
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    It gets promoted to an integer to do the addition. If you force it back into an unsigned short, you will get the wrapping effect.

    Code:
    #include <iostream>
using namespace std;

int main() {
	unsigned short num=65535;
	cout << num+2 << endl;

	cout << (unsigned short) (num+ 2) << endl;
	return 0;
}
    Prove you can code in C++ or C# at TopCoder, referrer rrenaud
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