delphi - What is the behaviour of shl and shr for non register sized operands?

Question

This question is inspired by my attempts to answer another question: Converting decimal/integer to binary - how and why it works the way it does?

The only documentation for the bitwise shift operators that I can find says:

The operations x shl y and x shr y shift the value of x to the left or right by y bits, which (if x is an unsigned integer) is equivalent to multiplying or dividing x by 2^y; the result is of the same type as x. For example, if N stores the value 01101 (decimal 13), then N shl 1 returns 11010 (decimal 26). Note that the value of y is interpreted modulo the size of the type of x. Thus for example, if x is an integer, x shl 40 is interpreted as x shl 8 because an integer is 32 bits and 40 mod 32 is 8.

Consider this program:

{$APPTYPE CONSOLE}
program BitwiseShift;
var
  u8: Byte;
  u16: Word;
  u32: LongWord;
  u64: UInt64;
begin
  u8 := $ff;
  Writeln((u8 shl 7) shr 7);
  // expects: 1 actual: 255

  u16 := $ffff;
  Writeln((u16 shl 15) shr 15);
  // expects: 1 actual: 65535

  u32 := $ffffffff;
  Writeln((u32 shl 31) shr 31);
  // expects: 1 actual: 1

  u64 := $ffffffffffffffff;
  Writeln((u64 shl 63) shr 63);
  // expects: 1 actual: 1
end.

I have run this with both XE3 and XE5, for both 32 and 64 bit Windows compilers, and the outupts are consistent, as commented in the code above.

I expected that (u8 shl 7) shr 7 would be evaluated entirely in the context of an 8 bit type. So when bits are shifted beyond the end of that 8 bit type, those bits are lost.

My question is why the program behaves as it does.

---

Interestingly I translated the program to C++ and on my 64 bit mingw 4.6.3 obtained the same output.

#include <cstdint>
#include <iostream>

int main()
{
    uint8_t u8 = 0xff;
    std::cout << ((u8 << 7) >> 7) << std::endl;

    uint16_t u16 = 0xffff;
    std::cout << ((u16 << 15) >> 15) << std::endl;

    uint32_t u32 = 0xffffffff;
    std::cout << ((u32 << 31) >> 31) << std::endl;

    uint64_t u64 = 0xffffffffffffffff;
    std::cout << ((u64 << 63) >> 63) << std::endl;
}

Answer 1

The reason is type promotion:

One special case of implicit type conversion is type promotion, where the compiler automatically expands the binary representation of objects of integer or floating-point types. Promotions are commonly used with types smaller than the native type of the target platform's ALU prior to arithmetic and logical operations in order to make such operations possible, or more efficient if the ALU can work with more than one type. C and C++ perform such promotion for objects of boolean, character, wide character, enumeration, and short integer types which are promoted to int, and for objects of type float, which are promoted to double. Unlike some other type conversions, promotions never lose precision or modify the value stored in the object.

So in the following code

var
  u8: Byte;

begin
  u8 := $ff;
  Writeln((u8 shl 7) shr 7);
..

the u8 value is promoted to 32-value before shl; to fix the result you need explicit type conversion:

  Writeln(Byte(u8 shl 7) shr 7);

---

C++ Standard, Section 4.5 Integral promotions:

An rvalue of type char, signed char, unsigned char, short int, or unsigned short int can be converted to an rvalue of type int if int can represent all the values of the source type; otherwise, the source rvalue can be converted to an rvalue of type unsigned int.

---

To check if Delphi follows the same convention in type promotion I've written the following application:

var
  u8: Byte;
  u16: Word;
  u32: LongWord;

procedure Test(Value: Integer); overload;
begin
  Writeln('Integer');
end;

procedure Test(Value: Cardinal); overload;
begin
  Writeln('Cardinal');
end;

begin
  u8 := $ff;
  Test(u8);     // 'Integer'
  u16 := $ffff;
  Test(u16);    // 'Integer'
  u32 := $ffffffff;
  Test(u32);    // 'Cardinal'
  Readln;
end.

So I believe there should be no difference between Delphi and C++ here.