First, default argument promotions 6.5.2.2 If the expression that denotes the called function has a type that does not include a prototype, the integer promotions are performed on each argument, and arguments that have type float are promoted to double. These are called the default argument promotions. Now for integer promotions: 6.3.1.1 The following may … Read more
This is due to integer promotions, both operands of the bit-wise shifts will be promoted to int in both cases. In the second case: x = (x << 1) >> 1; the result of x << 1 will be an int and therefore the shifted bit will be preserved and available to the next step … Read more
It doesn’t matter whether you put it on the right hand side or the left hand side; the == operator is completely symmetrical. If both operands to the == operator have arithmetic type, as in this case, then the “usual arithmetic conversions” are applied (C99 §6.5.9). In this case, the rule that applies is: If … Read more
The operands of % undergo the usual arithmetic conversions to bring them to a common type, before the division is performed. If the operands were int and size_t, then the int is converted to size_t. If size_t is 32-bit then -4 would become 4294967292 and then the result of the expression is 4294957292 % 7 … Read more
I purposefully did not compile and run then on my system, since this is the type of thing that could be compiler dependent. This is not compiler dependent. C++ clearly defines the order of these operations and how they are converted. How the conversion happens is dependent on the order of operations. double result1 = … Read more
[expr.unary.op] The operand of ~ shall have integral or unscoped enumeration type; the result is the one’s complement of its operand. Integral promotions are performed. [expr.shift] The shift operators << and >> group left-to-right. […] The operands shall be of integral or unscoped enumeration type and integral promotions are performed. What’s the integral promotion of uint8_t … Read more
The so-called usual arithmetic conversions apply to many binary operators, but not all of them. For example they do not apply to the bit shift operators, &&, ||, comma operator, and assignment operators. This is the rule for the bit shift operators: 6.5.7 … 3 Semantics … The integer promotions are performed on each of … Read more
What you’re seeing is the result of the so-called “usual arithmetic conversions” that occur during arithmetic expressions, particularly those that are binary in nature (take two arguments). This is described in §5/9: Many binary operators that expect operands of arithmetic or enumeration type cause conversions and yield result types in a similar way. The purpose … Read more
I think you are confusing two things. Promotion is the process by which values of integer type “smaller” that int/unsigned int are converted either to int or unsigned int. The rules are expressed somewhat strangely (mostly for the benefit of handling adequately char) but ensure that value and sign are conserved. Then there is the … Read more
This is because of the various implicit type conversion rules in C. There are two of them that a C programmer must know: the usual arithmetic conversions and the integer promotions (the latter are part of the former). In the char case you have the types (signed char) == (unsigned char). These are both small … Read more