Value of a+b and char type

Question

I am working in C++ and I had (as an exercise) to write on paper 2 answers. The first question: if we have the following declarations and initialisations of variables:

unsigned char x=250, z=x+7, a='8';

What is the value of the expression?

z|(a-'0') // (here | is bitwise disjunction)

We have unsigned char, so the number z=x+7 is reduced mod 256, thus, after writing the numbers in binary, the answer is 9.

The next question: a and b are int variables, a=1 and b=32767.

The range of int is [-32768, 32767]. We don't have an unsigned type here. My question is: what is the value of a+b? How does this work with signed data types if the value of a certain variable is greater than the range of that data type?

Actually, there is only one question. I solved the first and explained the way I solved it. The first question is only to explain my struggle. — shibuya, Apr 27 '16 at 11:56
What does the first question have to do with the second question? I'm really confused. — Lightness Races in Orbit, Apr 27 '16 at 12:03
Because unsigned types allow mod 2^n arithmetic and I was wondering if this also works for signed data types without to be considered undefined behavior. — shibuya, Apr 27 '16 at 12:04

score 3 · Accepted Answer · edited Jun 20 '20 at 09:12

3

The next question: a and b are int variables, a=1 and b=32767.

[...]My question is: what is the value of a+b?

Its undefined behavior. We cant tell you what it will be. We could make a reasonable guess but as far as C++ is concerned signed integer overflow is undefined behavior.

edited Jun 20 '20 at 09:12

Community

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answered Apr 27 '16 at 11:45

NathanOliver

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undefined behavior or implementation defined? I mean the difference described here: http://stackoverflow.com/questions/2397984/undefined-unspecified-and-implementation-defined-behavior – PiotrNycz Apr 27 '16 at 11:47
@user6261180 No problem. It is a little short but I wanted didn't want to explain what is happening for you since it may not happen to everyone. – NathanOliver Apr 27 '16 at 11:58

Maxim Egorushkin · Answer 2 · 2016-04-27T11:55:25.430

There is no operator+(unsigned char, unsigned char) in C++, it first promotes these unsigned char arguments to int and only then does the addition, so that the type of the expression is int.

And then that int whose value is too big to fit in unsigned char gets converted to unsigned char.

The standard says:

A prvalue of an integer type can be converted to a prvalue of another integer type. A prvalue of an unscoped enumeration type can be converted to a prvalue of an integer type. If the destination type is unsigned, the resulting value is the least unsigned integer congruent to the source integer (modulo 2**n where n is the number of bits used to represent the unsigned type). [ Note: In a two’s complement representation, this conversion is conceptual and there is no change in the bit pattern (if there is no truncation). — end note ]

score 0 · Answer 3 · 2016-04-27T11:52:55.937

0

For the second question, the answer is undetermined.

You can verify it yourself like this :

#include <iostream>

using namespace std;

int main()
{
    int a = 1;
    int b = 32767;
    int c = a+b;
    cout << c << endl;
}

The result will depend on your machine.

edited Apr 27 '16 at 11:52

answered Apr 27 '16 at 11:46

An `int` can have the range [-32768, 32767]. – Pete Becker Apr 27 '16 at 11:50
*For the second question, the answer is indeed -32768* it may be on your machine and the OP's machine but in C++ it is UB. You cannot say it will always be -32768 – NathanOliver Apr 27 '16 at 11:51

Value of a+b and char type

3 Answers3