a + b和char类型的值

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.

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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?

Answer 1

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.

Answer 2

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 ]

Answer 3

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.