What are the +: and -: Verilog/SystemVerilog operators? When and how do you use them? For example:
logic [15:0] down_vect;
logic [0:15] up_vect;
down_vect[lsb_base_expr +: width_expr]
up_vect [msb_base_expr +: width_expr]
down_vect[msb_base_expr -: width_expr]
up_vect [lsb_base_expr -: width_expr]
That particular syntax is called an indexed part select. It's very useful when you need to select a fixed number of bits from a variable offset within a multi-bit register.
Here's an example of the syntax:
reg [31:0] dword;
reg [7:0] byte0;
reg [7:0] byte1;
reg [7:0] byte2;
reg [7:0] byte3;
assign byte0 = dword[0 +: 8]; // Same as dword[7:0]
assign byte1 = dword[8 +: 8]; // Same as dword[15:8]
assign byte2 = dword[16 +: 8]; // Same as dword[23:16]
assign byte3 = dword[24 +: 8]; // Same as dword[31:24]
The biggest advantage with this syntax is that you can use a variable for the index. Normal part selects in Verilog require constants. So attempting the above with something like dword[i+7:i] is not allowed.
So if you want to select a particular byte using a variable select, you can use the indexed part select.
Example using variable:
reg [31:0] dword;
reg [7:0] byte;
reg [1:0] i;
// This is illegal due to the variable i, even though the width is always 8 bits
assign byte = dword[(i*8)+7 : i*8]; // ** Not allowed!
// Use the indexed part select
assign byte = dword[i*8 +: 8];
The purpose of this operator is when you need to access a slice of a bus, both MSB position and LSB positions are variables, but the width of the slice is a constant value, as in the example below:
bit[7:0] bus_in = 8'hAA;
int lsb = 3;
int msb = lsb+3; // Setting msb=6, for out bus of 4 bits
bit[3:0] bus_out_bad = bus_in[msb:lsb]; // ILLEGAL - both boundaries are variables
bit[3:0] bus_out_ok = bus_in[lsb+:3]; // Good - only one variable
5.2.1 Vector bit-select and part-select addressing
Bit-selects extract a particular bit from a vector net, vector reg, integer, or time variable, or parameter. The bit can be addressed using an expression. If the bit-select is out of the address bounds or the bit-select is x or z , then the value returned by the reference shall be x . A bit-select or part-select of a scalar, or of a variable orparameter of type real or realtime, shall be illegal.
Several contiguous bits in a vector net, vector reg, integer, or time variable, or parameter can be addressed and are known as part-selects. There are two types of part-selects, a constant part-select and an indexed part-select. A constant part-select of a vector reg or net is given with the following syntax:
vect[msb_expr:lsb_expr]
Both msb_expr and lsb_expr shall be constant integer expressions. The first expression has to address a more significant bit than the second expression.
An indexed part-select of a vector net, vector reg, integer, or time variable, or parameter is given with the following syntax:
reg [15:0] big_vect;
reg [0:15] little_vect;
big_vect[lsb_base_expr +: width_expr]
little_vect[msb_base_expr +: width_expr]
big_vect[msb_base_expr -: width_expr]
little_vect[lsb_base_expr -: width_expr]
The msb_base_expr and lsb_base_expr shall be integer expressions, and the width_expr shall be a positive constant integer expression. The lsb_base_expr and msb_base_expr can vary at run time. The first two examples select bits starting at the base and ascending the bit range. The number of bits selected is equal to the width expression. The second two examples select bits starting at the base and descending the bit range.
A part-select of any type that addresses a range of bits that are completely out of the address bounds of the net, reg, integer, time variable, or parameter or a part-select that is x or z shall yield the value x when read and shall have no effect on the data stored when written. Part-selects that are partially out of range shall, when read, return x for the bits that are out of range and shall, when written, only affect the bits that are in range.
For example:
reg [31: 0] big_vect;
reg [0 :31] little_vect;
reg [63: 0] dword;
integer sel;
big_vect[ 0 +: 8] // == big_vect[ 7 : 0]
big_vect[15 -: 8] // == big_vect[15 : 8]
little_vect[ 0 +: 8] // == little_vect[0 : 7]
little_vect[15 -: 8] // == little_vect[8 :15]
dword[8sel +: 8] // variable part-select with fixed width*
Example 1—The following example specifies the single bit of acc vector that is addressed by the operand
index :
acc[index]
The actual bit that is accessed by an address is, in part, determined by the declaration of acc . For instance, each of the declarations of acc shown in the next example causes a particular value of index to access a different bit:
reg [15:0] acc;
reg [2:17] acc
Example 2—The next example and the bullet items that follow it illustrate the principles of bit addressing. The code declares an 8-bit reg called vect and initializes it to a value of 4. The list describes how the separate bits of that vector can be addressed.
reg [7:0] vect;
vect = 4; // fills vect with the pattern 00000100
// msb is bit 7, lsb is bit 0
— If the value of addr is 2, then vect[addr] returns 1.
— If the value of addr is out of bounds, then vect[addr] returns x.
— If addr is 0, 1, or 3 through 7, vect[addr] returns 0.
— vect[3:0] returns the bits 0100.
— vect[5:1] returns the bits 00010.
— vect[ expression that returns x ] returns x.
— vect[ expression that returns z ] returns x.
— If any bit of addr is x or z , then the value of addr is x.
NOTE 1—Part-select indices that evaluate to x or z may be flagged as a compile time error.
NOTE 2—Bit-select or part-select indices that are outside of the declared range may be flagged as a compile time error.
Related
I'm trying to implement the 'Sport Scheduling Problem' (with a Round-Robin approach to break symmetries). The actual problem is of no importance. I simply want to declare the value at x[1,1] to be the set {1,2} and base the sets in the same column upon the first set. This is modelled as in the code below. The output is included in a screenshot below it. The problem is that the first set is not printed as a set but rather some sort of range while the values at x[2,1] and x[3,1] are indeed printed as sets and x[4,1] again as a range. Why is this? I assume that in the declaration of x that set of 1..n is treated as an integer but if it is not, how to declare it as integers?
EDIT: ONLY the first column of the output is of importance.
int: n = 8;
int: nw = n-1;
int: np = n div 2;
array[1..np, 1..nw] of var set of 1..n: x;
% BEGIN FIX FIRST WEEK $
constraint(
x[1,1] = {1, 2}
);
constraint(
forall(t in 2..np) (x[t,1] = {t+1, n+2-t} )
);
solve satisfy;
output[
"\(x[p,w])" ++ if w == nw then "\n" else "\t" endif | p in 1..np, w in 1..nw
]
Backend solver: Gecode
(Here's a summarize of my comments above.)
The range syntax is simply a shorthand for contiguous values in a set: 1..8 is a shorthand of the set {1,2,3,4,5,6,7,8}, and 5..6 is a shorthand for the set {5,6}.
The reason for this shorthand is probably since it's often - and arguably - easier to read the shorthand version than the full list, especially if it's a long list of integers, e.g. 1..1024. It also save space in the output of solutions.
For the two set versions, e.g. {1,2}, this explicit enumeration might be clearer to read than 1..2, though I tend to prefer the shorthand version in all cases.
I have trouble understanding what happens when calling &*pointer
int j=8;
int* p = &j;
When I print in my compiler I get the following
j = 8 , &j = 00EBFEAC p = 00EBFEAC , *p = 8 , &p = 00EBFEA0
&*p= 00EBFEAC
cout << &*p gives &*p = 00EBFEAC which is p itself
& and * have same operator precedence.I thought &*p would translate to &(*p)--> &(8) and expected compiler error.
How does compiler deduce this result?
You are stumbling over something interesting: Variables, strictly spoken, are not values, but refer to values. 8 is an integer value. After int i=8, i refers to an integer value. The difference is that it could refer to a different value.
In order to obtain the value, i must be dereferenced, i.e. the value stored in the memory location which i stands for must be obtained. This dereferencing is performed implicitly in C whenever a value of the type which the variable references is requested: i=8; printf("%d", i) results in the same output as printf("%d", 8). That is funny because variables are essentially aliases for addresses, while numeric literals are aliases for immediate values. In C these very different things are syntactically treated identically. A variable can stand in for a literal in an expression and will be automatically dereferenced. The resulting machine code makes that very clear. Consider the two functions below. Both have the same return type, int. But f has a variable in the return statement which must be dereferenced so that its value can be returned (in this case, it is returned in a register):
int i = 1;
int g(){ return 1; } // literal
int f(){ return i; } // variable
If we ignore the housekeeping code, the functions each translate into a sigle machine instruction. The corresponding assembler (from icc) is for g:
movl $1, %eax #5.17
That's pretty starightforward: Put 1 in the register eax.
By contrast, f translates to
movl i(%rip), %eax #4.17
This puts the value at the address in register rip plus offset i in the register eax. It's refreshing to see how a variable name is just an address (offset) alias to the compiler.
The necessary dereferencing should now be obvious. It would be more logical to write return *i in order to return 1, and write return i only for functions which return references — or pointers.
In your example it is indeed illogical to a degree that
int j=8;
int* p = &j;
printf("%d\n", *p);
prints 8 (i.e, p is actually dereferenced twice); but that &(*p) yields the address of the object pointed to by p (which is the address value stored in p), and is not interpreted as &(8). The reason is that in the context of the address operator a variable (or, in this case, the L-value obtained by dereferencing p) is not implicitly dereferenced the way it is in other contexts.
When the attempt was made to create a logical, orthogonal language — Algol68 —, int i=8 indeed declared an alias for 8. In order to declare a variable the long form would have been refint m = loc int := 3. Consequently what we call a pointer or reference would have had the type ref ref int because actually two dereferences are needed to obtain an integer value.
j is an int with value 8 and is stored in memory at address 00EBFEAC.
&j gives the memory address of variable j (00EBFEAC).
int* p = &j Here you define a variable p which you define being of type int *, namely a value of an address in memory where it can find an int. You assign it &j, namely an address of an int -> which makes sense.
*p gives you the value associated with the address stored in p.
The address stored in p points to an int, so *p gives you the value of that int, namely 8.
& p is the address of where the variable p itself is stored
&*p gives you the address of the value the memory address stored in p points to, which is indeed p again. &(*p) -> &(j) -> 00EBFEAC
Think about &j itself (or even &(j)). According to your logic, shouldn't j evaluate to 8 and result in &8, as well? Dereferencing a pointer or evaluating a variable results in an lvalue, which is a value that you can assign to or take the address of.
The L in "lvalue" refers to the left in "left hand side of the assignment", such as j = 10 or *p = 12. There are also rvalues, such as j + 10, or 8, which obviously cannot be assigned to.
That's just a basic explanation. In C++ there's a lot more to it, with various classes of values (but that thread might be too advanced for your current needs).
I found the following code snippet in verilog code for AES.
function [7:0] xtime;
input [7:0] b; xtime={b[6:0],1'b0}^(8'h1b&{8{b[7]}});
endfunction
Please explain what does this do. The more elaborate explanation, the better.
b is a 8 bit input.
b[6:0],1'b0 last 7 bits left shifted, and padded with 0
^ xor
8'h1b 8 bits hex 1b anded with the sign bit.
Explained in one line: If msb is set xor with 0x1b otherwise just *2
A quick search of xtime and AES leads me to this c implementation’s comment:
// xtime is a macro that finds the product of {02} and the argument to
// xtime modulo {1b}
#define xtime(x) ((x<<1) ^ (((x>>7) & 1) * 0x11b))
looks like it maybe is doing about the same.
Lets clean up the code and breakdown some of the assignments:
function [7:0] xtime;
input [7:0] b;
reg [7:0] temp1;
reg [7:0] temp2;
begin
temp1 = {b[6:0],1'b0};
temp2 = ( 8'h1b & {8{b[7]}} );
xtime = temp1 ^ temp2;
end
endfunction
Function called xtime outputs the variable xtime 8 bits wide. Has 8 bit wide input called b.
temp1 left shifts input b, padding LSB (Least Significant Bit) with 0 and throwing away MSB (Most Significant Bit).
temp2 bitwise ANDs 8'h1B (8'b0001_1011) with the MSB of input b.
b[7] selects bit 7 of b.
{8{value}} is the replication operator.
{8{1'b1}} => 8'b1111_1111
{4{2'b10}} => 8'b1010_1010
xtime performs the bitwise XOR of temp1 and temp2.
I'm trying to synthesize an Altera circuit using as few logic elements as possible. Also, embedded multipliers do not count against logic elements, so I should be using them. So far the circuit looks correct in terms of functionality. However, the following module uses a large amount of logic elements. It uses 24 logic elements and I'm not sure why since it should be using 8 + a couple of combinational gates for the case block.
I suspect the adder but I'm not 100% sure. If my suspicion is correct however, is it possible to use multipliers as a simple adder?
module alu #(parameter N = 8)
(
output logic [N-1:0] alu_res,
input [N-1:0] a,
input [N-1:0] b,
input [1:0] op,
input clk
);
wire [7:0] dataa, datab;
wire [15:0] result;
// instantiate embedded 8-bit signed multiplier
mult mult8bit (.*);
// assign multiplier operands
assign dataa = a;
assign datab = b;
always_comb
unique case (op)
// LW
2'b00: alu_res = 8'b0;
// ADD
2'b01: alu_res = a + b;
// MUL
2'b10: alu_res = result[2*N-2:N-1]; // a is a fraction
// MOV
2'b11: alu_res = a;
endcase
endmodule
Your case statement will generate a 4 input mux with op as the select which uses a minimum of 2 logic cells. However since your assigning an 8-bit variable in the case block you will require 2 logic elements for each bit of the output. Therefore total logic elements is 8*2 for the large mux and 8 for the adder giving you 24 as the total.
I'm doing this project too so I won't give too much away about how to optimise this. However what I will tell you is that both the mux's and the adder can be implemented using multipliers, 8 at most. With that said I don't think this architecture is optimal for a multiplier implementation.
I wrote a code for concatenation as below:
module p2;
int n[1:2][1:3] = {2{{3{1}}}};
initial
begin
$display("val:%d",n[2][1]);
end
endmodule
It is showing errors.
Please explain?
Unpacked arrays require a '{} format. See IEEE Std 1800-2012 § 5.11 (or search for '{ in the LRM for many examples).
Therefore update your assignment to:
int n[1:2][1:3] = '{2{'{3{1}}}};
int n[1:2][1:3] = {2{{3{1}}}};
Just looking at {3{1}} this is a 96 bit number 3 integers concatenated together.
It is likely that {3{1'b1}} was intended.
The main issue looks to be the the left hand side is an unpacked array, and the left hand side is a packed array.
{ 2 { {3{1'b1}} } } => 6'b111_111
What is required is [[3'b111],[3'b111]],
From IEEE std 1800-2009 the array assignments section will be of interest here
10.9.1 Array assignment patterns
Concatenation braces are used to construct and deconstruct simple bit vectors.
A similar syntax is used to support the construction and deconstruction of arrays. The expressions shall match element for element, and the braces shall match the array dimensions. Each expression item shall be evaluated in the context of an
assignment to the type of the corresponding element in the array. In other words, the following examples are not required to cause size warnings:
bit unpackedbits [1:0] = '{1,1}; // no size warning required as
// bit can be set to 1
int unpackedints [1:0] = '{1'b1, 1'b1}; // no size warning required as
// int can be set to 1’b1
A syntax resembling replications (see 11.4.12.1) can be used in array assignment patterns as well. Each replication shall represent an entire single dimension.
unpackedbits = '{2 {y}} ; // same as '{y, y}
int n[1:2][1:3] = '{2{'{3{y}}}}; // same as '{'{y,y,y},'{y,y,y}}