rondino - 2 months ago 6

C Question

Reading about the integer promotion and integer conversion rank I found this link

- 1.If both operands have the same type, then no further conversion is needed.

- 2.Otherwise, if both operands have signed integer types or both have unsigned integer types, the operand with the type of lesser integer

conversion rank is converted to the type of the operand with greater

rank.

- 3.Otherwise, if the operand that has unsigned integer type has rank greater or equal to the rank of the type of the other operand, then

the operand with signed integer type is converted to the type of the

operand with unsigned integer type.

- 4.Otherwise, if the type of the operand with signed integer type can represent all of the values of the type of the operand with unsigned

integer type, then the operand with unsigned integer type is converted

to the type of the operand with signed integer type.

- 5.Otherwise, both operands are converted to the unsigned integer type corresponding to the type of the operand with signed integer type.

The points 1 2 3 are totally clear but I still not come up with example for the case 4 and 5. Can someone provide please an example concerning any implementation ?

As I know the integer conversion rank is:

_Bool < char < short < int < long < long long int

Whatever the size of bytes related to the types are equal or higher. Right?

Concerning the promotion or conversion from one type to the other. Are the bits added to the lowest type zero or 1 or the left extreme bit has effect on that ?

I want to know how is the process in the

For the integer promotion it can always conserve the value and the sign without a doubt.

Answer

Case 4 applies if you have an unsigned type that is smaller in rank than the signed type it is operating with and they have **different sizes**. Case 5 then if the two are the **same size**.

For example, on my system `int`

is 32-bit, `long`

is 64-bit, and `long long`

is 64-bit. If you then have the following:

```
unsigned int a; // range: 0 to 4294967295
long b; // range: -9223372036854775808 to 9223372036854775807
unsigned long c; // range: 0 to 18446744073709551615
long long d; // range: -9223372036854775808 to 9223372036854775807
```

For an expression involving `a`

and `b`

, which are `unsigned int`

and `long`

, any valid `unsigned int`

can fit in a `long`

. So `a`

is converted to `long`

in that situation.

Conversely, For an expression involving `c`

and `d`

, which are `unsigned long`

and `long long`

, a `long long`

cannot hold all values of an `unsigned long`

. So both operands are converted to `unsigned long long`

.

Regarding what happens during a promotion / conversion on the bit level, let's first assume that the lower rank type is smaller than the higher rank type, and that signed types use 2's complement representation.

For a conversion from a 32 bit `int`

to a 64 bit `long`

, if the value is positive, 4 bytes containing all 0 bits are added on the left. If the value is negative, 4 bytes containing all 1 bits are added on the left. For example, the representation of value `5`

changes from `0x00000005`

to `0x0000000000000005`

. For the value `-5`

, the representation changes from `0xfffffffb`

to `0xfffffffffffffffb`

.

Source (Stackoverflow)

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