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Addition of two chars produces int
Given the following C++ code:
unsigned char a = 200;
unsigned char b = 100;
unsigned char c = (a + b) / 2;
The output is 150 as logically expected, however shouldn't there be an integer overflow in the expression (a + b)?
Obviously there must be an integer promotion to deal with the overflow here, or something else is happening that I cannot see. I was wondering if someone could enlighten me, so I can know what it is I can and shouldn't rely on in terms of integer promotion and overflow.
297
The rules for detecting overflow in a two's complement sum are simple: If the sum of two positive numbers yields a negative result, the sum has overflowed. If the sum of two negative numbers yields a positive result, the sum has overflowed. Otherwise, the sum has not overflowed.
Write a “C” function, int addOvf(int* result, int a, int b) If there is no overflow, the function places the resultant = sum a+b in “result” and returns 0. Otherwise it returns -1. The solution of casting to long and adding to find detecting the overflow is not allowed.
An integer overflow occurs when you attempt to store inside an integer variable a value that is larger than the maximum value the variable can hold. The C standard defines this situation as undefined behavior (meaning that anything might happen).
Neither C++ not C perform arithmetical computations withing "smaller" integer types like, char and short. These types almost always get promoted to int before any further computations begin. So, your expression is really evaluated as
unsigned char c = ((int) a + (int) b) / 2;
P.S. On some exotic platform where the range of int does not cover the range of unsigned char, the type unsigned int will be used as target type for promotion.
177
No, this is not an error.
The compiler always calculates at minimum of integer precision, the result will be converted back to unsigned char on assignment only.
This is in the standard.
39
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