RE: Some queries related to expression type

From: Kausik Datta (kausikd@cal.interrasystems.com)
Date: Tue Jun 28 2005 - 21:46:25 PDT

  • Next message: Steven Sharp: "RE: Some queries related to expression type"

    Hi Steven,
    Thanks for your reply. For question 4 here is the section (4.1.13) of
    LRM.(See the last line)

    The evaluation of a conditional operator shall begin with the evaluation of
    expression1. If expression1 evaluates
    to false (0), then expression3 shall be evaluated and used as the result of
    the conditional expression. If
    expression1 evaluates to true (known value other than 0), then expression2
    is evaluated and used as the
    result. If expression1 evaluates to ambiguous value (x or z), then both
    expression2 and expression3 shall be
    evaluated and their results shall be combined, bit by bit, using Table 28 to
    calculate the final result unless
    expression2 or expression3 is real, in which case the result shall be 0. If
    the lengths of expression2 and
    expression3 are different, the shorter operand shall be lengthened to match
    the longer and zero-filled from
    the left (the high-order end).

    Thanks
    Kausik

    -----Original Message-----
    From: owner-etf@boyd.com [mailto:owner-etf@boyd.com] On Behalf Of Steven
    Sharp
    Sent: Wednesday, June 29, 2005 5:42 AM
    To: etf@boyd.com; kausikd@cal.interrasystems.com
    Subject: Re: Some queries related to expression type

    > 1. What will be the sign type of expression : (0 || 1.234) at line 2.
    > Is the logical or operator (||) treated as comparison operator?

    The LRM fails to specify this. The result of the logical AND and OR
    operators (|| and &&) is supposed to be the same as the result of a
    comparison operator: 1 bit unsigned.

    These operators are not actually comparison operators, however. Their
    operands are fully self-determined, and do not affect each other's size.
    The table of bit lengths for self-determined expressions has been fixed for
    the 2005 revision, to put these in a different table entry from the
    comparisons.

    > 2. What will be the sign type of expression (3.33333 == 3.33333000) ? (0
    ||
    > 1.234) : (4 && (1 - 1)) at line 2?
    > How does it depend on (3.33333 == 3.33333000) , (0 || 1.234) or (4 &&
    > (1 - 1))

    It will be unsigned. The (3.33333 == 3.33333000) does not affect the result
    type at all, since it is a self-determined operand of the ?:
    operator. The other two operands do affect the result type. Both are
    unsigned (from question 1 above), so the result is unsigned.

    > 3. What will be the decimal value of 1'sb1

    It has a value of -1. A 1-bit signed value can only represent two numerical
    values: 0 and -1. There is no way to represent 1. This is a natural
    consequence of how twos-complement is defined, applied to the degenerate
    case of one bit.

    > 4. Will be the value of r and r1 same? According to LRM they will be
          different.
          r will be 0011 and r1 will be 1111. Is it a bug in LRM?

    Where does the LRM say this? Both will be 4'b1111.

    In the expression "1==1 ? 2'sb11 : 1", the operand "1==1" is self-
    determined, so the fact that its result is unsigned has no effect on the
    result type of the conditional operator. The other two operands are
    context-determined. Both are signed, so the result of the conditional
    operator is signed (and 32 bits wide). So the result of 2'sb11 is coerced
    to signed and extended to 32 bits.
    Then the assignment truncates this back down to 4 bits, or 1111.

    Steven Sharp
    sharp@cadence.com

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