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<HTML><HEAD> <META NAME="GENERATOR" CONTENT="Adobe PageMill 2.0 Mac"> <LINK REL="STYLESHEET" HREF="ch04.css"> <TITLE> 4.4 Expression bit lengths </TITLE></HEAD><BODY BGCOLOR="#ffffff"><P><A NAME="pgfId=749"></A><HR ALIGN=LEFT></P><P><A HREF="ch04.htm">Chapter start</A> <A HREF="ch04.3.htm">Previous page</A></P><H1>4.4 Expression bit lengths</H1><P><P CLASS="Body"><A NAME="pgfId=750"></A>Controlling the number of bitsthat are used in expression evaluations is important if consistent resultsare to be achieved. Some situations have a simple solution, for example,if a bit-wise and operation is specified on two 16-bit registers, then theresult is a 16-bit value. However, in some situations it is not obvioushow many bits are used to evaluate an expression, or what size the resultshould be.</P><P><P CLASS="Body"><A NAME="pgfId=508"></A>For example, should an arithmeticadd of two 16-bit values perform the evaluation using 16 bits, or shouldthe evaluation use 17 bits in order to allow for a possible carry overflow?The answer depends on the type of device being modeled, and whether thatdevice handles carry overflow. The Verilog HDL uses the bit length of theoperands to determine how many bits to use while evaluating an expression.The bit length rules are shown below. In the case of the addition operator,the bit length of the largest operand, including the left-hand side of anassignment, shall be used.</P><PRE><B>reg</B> [15:0] a, b; // 16 bit registers<B>reg</B> [15:0] sumA; // 16 bit register<B>reg</B> [16:0] sumB; // 17 bit registersumA = a + b; // expression evaluates using 16 bitssumB = a + b; // expression evaluates using 17 bits</PRE><P><P CLASS="SubSection"><A NAME="pgfId=765"></A>Rules for expression bitlengths</P><P><P CLASS="Body"><A NAME="pgfId=766"></A>The rules governing the expressionbit lengths have been formulated so that most practical situations havea natural solution.</P><P><P CLASS="Body"><A NAME="pgfId=767"></A>The number of bits of an expression(known as the size of the expression) shall be determined by the operandsinvolved in the expression and the context in which the expression is given.</P><P><P CLASS="Body"><A NAME="pgfId=768"></A>A <I>self-determined expression</I>is one where the bit length of the expression is solely determined by theexpression itself--for example, an expression representing a delay value.</P><P><P CLASS="Body"><A NAME="pgfId=769"></A>A <I>context-determined expression</I>is one where the bit length of the expression is determined by the bit lengthof the expression and by the fact that it is part of another expression.For example, the bit size of the right-hand side expression of an assignmentdepends on itself and the size of the left-hand side.</P><P><P CLASS="Body"><A NAME="pgfId=770"></A>Table 4-23 shows how theform of an expression shall determine the bit lengths of the results ofthe expression. In Table 4-23, <CODE>i</CODE> , <CODE>j</CODE> , and<CODE>k</CODE> represent expressions of an operand, and<CODE> L(i)</CODE>represents the bit length of the operand represented by <CODE>i</CODE> .</P><P><TABLE BORDER="1" CELLSPACING="2" CELLPADDING="0"><CAPTION ALIGN="TOP"><P CLASS="TableTitle"><A NAME="pgfId=878"></A>Table 4-25: Bit lengthsresulting from expressions</CAPTION><TR><TH><P CLASS="CellHeading"><A NAME="pgfId=464"></A>Expression</TH><TH><P CLASS="CellHeading"><A NAME="pgfId=486"></A>Bit length</TH><TH><P CLASS="CellHeading"><A NAME="pgfId=753"></A>Comments</TH></TR><TR><TD><P CLASS="CellBody"><A NAME="pgfId=771"></A>unsized constant number</TD><TD><P CLASS="CellBody"><A NAME="pgfId=772"></A>same as integer</TD><TD><P CLASS="CellBody"><A NAME="pgfId=776"></A> </TD></TR><TR><TD><P CLASS="CellBody"><A NAME="pgfId=777"></A>sized constant number</TD><TD><P CLASS="CellBody"><A NAME="pgfId=778"></A>as given</TD><TD><P CLASS="CellBody"><A NAME="pgfId=779"></A> </TD></TR><TR><TD><P CLASS="CellBody"><A NAME="pgfId=780"></A>i op j, where op is:<BR>+ - * / % & | ^ ^~ ~^</TD><TD><P CLASS="CellBody"><A NAME="pgfId=782"></A>max(L(i),L(j))</TD><TD><P CLASS="CellBody"><A NAME="pgfId=783"></A> </TD></TR><TR><TD><P CLASS="CellBody"><A NAME="pgfId=784"></A>op i, where op is:<BR>+, -, ~</TD><TD><P CLASS="CellBody"><A NAME="pgfId=785"></A>L(i)</TD><TD><P CLASS="CellBody"><A NAME="pgfId=786"></A> </TD></TR><TR><TD><P CLASS="CellBody"><A NAME="pgfId=790"></A>i op j, where op is:<BR>=== !== == != && || > >= < <=</TD><TD><P CLASS="CellBody"><A NAME="pgfId=935"></A>1 bit</TD><TD><P CLASS="CellBody"><A NAME="pgfId=936"></A>All operands are self-determined</TD></TR><TR><TD><P CLASS="CellBody"><A NAME="pgfId=938"></A>op i, where op is:<BR>& ~& | ~| ^ ~^ ^~</TD><TD><P CLASS="CellBody"><A NAME="pgfId=940"></A>1 bit</TD><TD><P CLASS="CellBody"><A NAME="pgfId=941"></A>All operands are self-determined</TD></TR><TR><TD><P CLASS="CellBody"><A NAME="pgfId=943"></A>i op j, where op is:<BR>>>, <<</TD><TD><P CLASS="CellBody"><A NAME="pgfId=944"></A>L(i)</TD><TD><P CLASS="CellBody"><A NAME="pgfId=945"></A>j is self-determined</TD></TR><TR><TD><P CLASS="CellBody"><A NAME="pgfId=946"></A>i ? j : k</TD><TD><P CLASS="CellBody"><A NAME="pgfId=947"></A>max(L(j),L(k))</TD><TD><P CLASS="CellBody"><A NAME="pgfId=948"></A>i is self-determined</TD></TR><TR><TD><P CLASS="CellBody"><A NAME="pgfId=949"></A>{i,...,j}</TD><TD><P CLASS="CellBody"><A NAME="pgfId=950"></A>L(i)+..+L(j)</TD><TD><P CLASS="CellBody"><A NAME="pgfId=951"></A>All operands are self-determined</TD></TR><TR><TD><P CLASS="CellBody"><A NAME="pgfId=953"></A>{i{j,..,k}}</TD><TD><P CLASS="CellBody"><A NAME="pgfId=954"></A>i*(L(j)+..+L(k))</TD><TD><P CLASS="CellBody"><A NAME="pgfId=955"></A>All operands are self-determined</TD></TR></TABLE></P><P><P CLASS="SubSection"><A NAME="pgfId=754"></A>An example of an expressionbit length problem</P><P><P CLASS="Body"><A NAME="pgfId=755"></A>During the evaluation of an expression,interim results shall take the size of the largest operand (in case of anassignment, this also includes the left-hand side). Care must be taken toprevent loss of a significant bit during expression evaluation. The examplebelow describes how the bit lengths of the operands could result in theloss of a significant bit.</P><P><P CLASS="Body"><A NAME="pgfId=862"></A>Given the following declarations</P><PRE><B>reg</B> [15:0] a, b, answer; // 16-bit registers</PRE><P><P CLASS="Body"><A NAME="pgfId=865"></A>The intent is to evaluate theexpression:</P><PRE>answer = (a + b) >> 1; // won't work properly</PRE><P><P CLASS="Body"><A NAME="pgfId=1163"></A>Where <CODE>a</CODE> and <CODE>b</CODE>are to be added, which may result in an overflow, and then shifted rightby one bit to preserve the carry bit in the 16-bit <CODE>answer</CODE> .</P><P><P CLASS="Body"><A NAME="pgfId=1164"></A>A problem arises, however, becauseall operands in the expression are of a 16-bit width. Therefore, the expression(<CODE> a + b</CODE> ) produces an interim result that is only 16 bits wide,thus losing the carry bit before the evaluation performs the 1-bit rightshift operation.</P><P><P CLASS="Body"><A NAME="pgfId=1165"></A>The solution is to force theexpression (<CODE> a + b</CODE> ) to evaluate using at least 17 bits. Forexample, adding an integer value of 0 to the expression will cause the evaluationto be performed using the bit size of integers. The following example willproduce the intended result:</P><PRE>answer = (a + b + 0) >> 1; // will work correctly</PRE><P><HR ALIGN=LEFT></P><P><A HREF="ch04.htm">Chapter start</A> <A HREF="ch04.3.htm">Previous page</A></BODY></HTML>⌨️ 快捷键说明
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