CODEWITHSUNDEEP
ctype-conversionbitwise-operatorsparentheses
talkaboutquality
talkaboutquality

Reputation: 1362

Can parentheses in C change the result type of operands of a bitwise operation?

I have fed the following code through a static analysis tool:

u1 = (u1 ^ u2); // OK

u1 = (u1 ^ u2) & u3;  // NOT OK

u1 = (u1 ^ u2) & 10; // NOT OK

u1 = (u1 ^ u2) & 10U; // NOT OK

u1 = (unsigned char)(u1 ^ u2) & 10U; // OK

u1 = (unsigned char)(u1 ^ u2) & u3;  // OK

"OK" means the static analysis tool did not complain. "NOT OK" means the static analysis tool did complain -- claiming that some operand of a bitwise operation is not an unsigned integer.

The results from the last 2 lines show that the parentheses are causing either

a. an actual type conversion to signed

b. something that the static analysis tool thinks is a type conversion to signed

I will ask the static analysis tool developer about (b).

But before I do, I would like to know if perhaps the C language is known to do (a)?

Upvotes: 5

Views: 403

Answers (2)

pmg
pmg

Reputation: 108938

Nothing in C is done below int: eg when adding two unsigned chars, even before the addition, the operands are converted to int according to the default promotions.

unsigned char u1, u2, u3;
u1 = 0;
u2 = 42;
u3 = u1 + u2;

In the last line, first u1 and u2 are converted to int, then the + operator is applied to obtain a int value and then that value is converted back to unsigned char (of course the compiler can use shortcuts!)

Upvotes: 6

Lindydancer
Lindydancer

Reputation: 26094

This is because, in C, the resulting type of an operation on two unsigned char:s is int. The static analysis tool correctly (though not very intuitive) reports that the & is applied to an int.

Upvotes: 3

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