SystemVerilog constraint: constrain the read address has been written before - constraints

I want to write a constraint to make sure r_addr is only allowed when the same address has been used as w_addr before, but the following constraint doesn't work. Do you have any suggestion?
class try;
rand int w_addr;
rand int r_addr;
int ua[$];
int aa[int];
constraint unique_addr_c{
aa.size() == 0 || aa.exists(r_addr);
}
endclass
module test;
try a;
initial begin
a=new;
repeat(20) begin
if(a.randomize);
$display("add=%0d", a.w_addr);
$display("add=%0d", a.r_addr);
a.ua.push_back(a.w_addr);
a.aa[a.w_addr] = 1;
end
end
endmodule

You want to use the inside operator.
class try;
rand bit [31:0] w_addr;
rand bit [31:0] r_addr;
bit [31:0] ua[$];
constraint unique_addr_c{
ua.size() >0 -> r_addr inside {ua};
}
function void post_randomize();
ua.push_back(w_addr);
endfunction
endclass
module test;
try a;
initial begin
a=new;
repeat(50) begin
if (!a.randomize) $error("randomization failed");
$display("wadd=%0d", a.w_addr);
$display("radd=%0d", a.r_addr);
end
end
endmodule
Notes: you should used unsigned types for addresses. Your check for randomize should produce an error.

This uses a queue instead of an associated array.
class try;
rand int w_addr;
rand int r_addr;
int q[$];
function void post_randomize();
int sz;
int idx;
q.push_back(w_addr);
sz = q.size();
idx = $urandom_range(0, sz-1);
r_addr = q[idx];
endfunction
endclass
module test;
try a;
initial begin
a=new;
repeat(20) begin
a.randomize();
$display("sz=%0d w_addr=%x r_add=%x", a.q.size(), a.w_addr, a.r_addr);
end
end
endmodule

The following code works in mentor but not in synopsys
class bus;
rand bit [31:0] address;
rand bit rd_wr_en;
rand bit [1:0] interleaving;
endclass
class my_bus;
rand bus b1[20];
function new();
foreach(b1[n])
b1[n] = new();
endfunction
constraint c {
foreach (b1[n])
if (n>0)
if(b1[n-b1[n].interleaving].rd_wr_en ==1)
{
b1[n].address == b1[n-b1[n].interleaving].address;
b1[n].rd_wr_en == 0;
}
}
endclass
module tb;
my_bus b2;
initial begin
b2 = new();
repeat (10) begin
b2.randomize;
foreach(b2.b1[n])
$display ( "%p", b2.b1[n]);
end
end
endmodule

Related

Looking for Segmentation Fault in C script

Hi trying to learn C specifically how to use pointers.
I wrote this script to practice ideas I've learned, but it crashes with segmentation fault error.
Bit of research search suggests that I am trying to access something that I should not be accessing I think that is an uninitialized pointer but I can't find it.
#include <stdio.h>
struct IntItem {
struct IntItem* next;
int value;
};
struct IntList {
struct IntItem* head;
struct IntItem* tail;
};
void append_list(struct IntList* ls, int item){
struct IntItem* last = ls->tail;
struct IntItem addition = {NULL,item};
last->next = &addition;
ls->tail = &addition;
if (!ls->head) {
ls->head = &addition;
}
}
int sum(int x, int y){
return x + y;
}
int max(int x, int y){
return x*(x>y) + y*(y>x);
}
int reduce(struct IntList xs, int (*opy)(int, int)){
struct IntItem current = *xs.head;
int running = 0;
while (current.next) {
running = opy(running,current.value);
current = *current.next;
}
return running;
}
int main(void) {
struct IntList ls = {NULL, NULL};
printf("Start Script\n");
append_list(&ls, 1);
append_list(&ls, 2);
append_list(&ls, 3);
printf("List Complete\n");
printf("Sum: %i",reduce(ls,sum));
printf("Max: %i",reduce(ls,max));
return 0;
}
Hints:
When you call append_list(&ls, 1), then inside append_list, what is the value of last?
What does last->next = &addition do?
And for your next bug:
What happens to addition after append_list returns? What does that mean for pointers to it?

XV6: pwd implementation

I was trying to implement pwd command in xv6 system. But i am getting a error in sysfile.c execution. The function is as follows showing the error:-
int sys_getcwd(void) {
char *p;
int n;
if(argint(1, &n) < 0 || argptr(0, &p, n) < 0)
return -1;
return name_for_inode(p, n, proc->cwd);
}
I get error as follows:
error: ‘proc’ undeclared (first use in this function)
return name_for_inode(p, n, proc->cwd);
But I also included proc.h in this file.
proc is not defined, and warning, it's a struct name.
You have to query the current process, you can do it with myproc() function
int sys_getcwd(void) {
char *p;
int n;
struct proc *curproc = myproc();
if(argint(1, &n) < 0 || argptr(0, &p, n) < 0)
return -1;
return name_for_inode(p, n, curproc->cwd);
}
Yes proc is not global variable so initialize it with current process on which cpu is working.
struct proc *proc = myproc();
return name_for_inode(p, n, proc->cwd);

timerfd mysteriously set int to 0 when read()

I am doing an timerfd hello world in ubuntu 14.04, but got a strange situation: the int count is reset after read timerfd but uint64_int not.
int main(int agrc, char **argv) {
unsigned int heartbeat_interval = 1;
struct itimerspec next_timer;
struct timespec now;
if (clock_gettime(CLOCK_REALTIME, &now) == -1)
err_sys((WHERE + std::string("timer error")).c_str());
next_timer.it_value.tv_sec = now.tv_sec;
next_timer.it_value.tv_nsec = 0;
next_timer.it_interval.tv_sec = heartbeat_interval;
next_timer.it_interval.tv_nsec = 0;
int timefd = timerfd_create(CLOCK_REALTIME, 0);
if (timerfd_settime(timefd, TFD_TIMER_ABSTIME, &next_timer, NULL) == -1) {
err_sys((WHERE).c_str());
}
uint64_t s;
int exp;
int count = 1;
uint64_t count1=0;
while (1) {
s = read(timefd, &exp, sizeof(uint64_t));
if (s != sizeof(uint64_t)) {
err_sys((WHERE).c_str());
}
}
}
int exp;
^^^
s = read(timefd, &exp, sizeof(uint64_t));
^^^ ^^^^^^^^
Unless your int and uint64_t types are the same size, this is a very bad idea. What's most likely happening is that the 64 bits you're reading are overwriting exp and whatever else happens to be next to it on the stack.
Actually, even if they are the same size, it's a bad idea. What you should have is something like:
s = read(timefd, &exp, sizeof(exp));
That way, you're guaranteed to never overwrite the data and your next line would catch the problem for you:
if (s != sizeof(uint64_t)) {
It won't solve the problem that an unsigned integral type and an integral type will be treated differently but you can fix that just by using the right type for exp.

Sizeof pointer of pointer in C [duplicate]

First off, here is some code:
int main()
{
int days[] = {1,2,3,4,5};
int *ptr = days;
printf("%u\n", sizeof(days));
printf("%u\n", sizeof(ptr));
return 0;
}
Is there a way to find out the size of the array that ptr is pointing to (instead of just giving its size, which is four bytes on a 32-bit system)?
No, you can't. The compiler doesn't know what the pointer is pointing to. There are tricks, like ending the array with a known out-of-band value and then counting the size up until that value, but that's not using sizeof().
Another trick is the one mentioned by Zan, which is to stash the size somewhere. For example, if you're dynamically allocating the array, allocate a block one int bigger than the one you need, stash the size in the first int, and return ptr+1 as the pointer to the array. When you need the size, decrement the pointer and peek at the stashed value. Just remember to free the whole block starting from the beginning, and not just the array.
The answer is, "No."
What C programmers do is store the size of the array somewhere. It can be part of a structure, or the programmer can cheat a bit and malloc() more memory than requested in order to store a length value before the start of the array.
For dynamic arrays (malloc or C++ new) you need to store the size of the array as mentioned by others or perhaps build an array manager structure which handles add, remove, count, etc. Unfortunately C doesn't do this nearly as well as C++ since you basically have to build it for each different array type you are storing which is cumbersome if you have multiple types of arrays that you need to manage.
For static arrays, such as the one in your example, there is a common macro used to get the size, but it is not recommended as it does not check if the parameter is really a static array. The macro is used in real code though, e.g. in the Linux kernel headers although it may be slightly different than the one below:
#if !defined(ARRAY_SIZE)
#define ARRAY_SIZE(x) (sizeof((x)) / sizeof((x)[0]))
#endif
int main()
{
int days[] = {1,2,3,4,5};
int *ptr = days;
printf("%u\n", ARRAY_SIZE(days));
printf("%u\n", sizeof(ptr));
return 0;
}
You can google for reasons to be wary of macros like this. Be careful.
If possible, the C++ stdlib such as vector which is much safer and easier to use.
There is a clean solution with C++ templates, without using sizeof(). The following getSize() function returns the size of any static array:
#include <cstddef>
template<typename T, size_t SIZE>
size_t getSize(T (&)[SIZE]) {
return SIZE;
}
Here is an example with a foo_t structure:
#include <cstddef>
template<typename T, size_t SIZE>
size_t getSize(T (&)[SIZE]) {
return SIZE;
}
struct foo_t {
int ball;
};
int main()
{
foo_t foos3[] = {{1},{2},{3}};
foo_t foos5[] = {{1},{2},{3},{4},{5}};
printf("%u\n", getSize(foos3));
printf("%u\n", getSize(foos5));
return 0;
}
Output:
3
5
As all the correct answers have stated, you cannot get this information from the decayed pointer value of the array alone. If the decayed pointer is the argument received by the function, then the size of the originating array has to be provided in some other way for the function to come to know that size.
Here's a suggestion different from what has been provided thus far,that will work: Pass a pointer to the array instead. This suggestion is similar to the C++ style suggestions, except that C does not support templates or references:
#define ARRAY_SZ 10
void foo (int (*arr)[ARRAY_SZ]) {
printf("%u\n", (unsigned)sizeof(*arr)/sizeof(**arr));
}
But, this suggestion is kind of silly for your problem, since the function is defined to know exactly the size of the array that is passed in (hence, there is little need to use sizeof at all on the array). What it does do, though, is offer some type safety. It will prohibit you from passing in an array of an unwanted size.
int x[20];
int y[10];
foo(&x); /* error */
foo(&y); /* ok */
If the function is supposed to be able to operate on any size of array, then you will have to provide the size to the function as additional information.
For this specific example, yes, there is, IF you use typedefs (see below). Of course, if you do it this way, you're just as well off to use SIZEOF_DAYS, since you know what the pointer is pointing to.
If you have a (void *) pointer, as is returned by malloc() or the like, then, no, there is no way to determine what data structure the pointer is pointing to and thus, no way to determine its size.
#include <stdio.h>
#define NUM_DAYS 5
typedef int days_t[ NUM_DAYS ];
#define SIZEOF_DAYS ( sizeof( days_t ) )
int main() {
days_t days;
days_t *ptr = &days;
printf( "SIZEOF_DAYS: %u\n", SIZEOF_DAYS );
printf( "sizeof(days): %u\n", sizeof(days) );
printf( "sizeof(*ptr): %u\n", sizeof(*ptr) );
printf( "sizeof(ptr): %u\n", sizeof(ptr) );
return 0;
}
Output:
SIZEOF_DAYS: 20
sizeof(days): 20
sizeof(*ptr): 20
sizeof(ptr): 4
There is no magic solution. C is not a reflective language. Objects don't automatically know what they are.
But you have many choices:
Obviously, add a parameter
Wrap the call in a macro and automatically add a parameter
Use a more complex object. Define a structure which contains the dynamic array and also the size of the array. Then, pass the address of the structure.
You can do something like this:
int days[] = { /*length:*/5, /*values:*/ 1,2,3,4,5 };
int *ptr = days + 1;
printf("array length: %u\n", ptr[-1]);
return 0;
My solution to this problem is to save the length of the array into a struct Array as a meta-information about the array.
#include <stdio.h>
#include <stdlib.h>
struct Array
{
int length;
double *array;
};
typedef struct Array Array;
Array* NewArray(int length)
{
/* Allocate the memory for the struct Array */
Array *newArray = (Array*) malloc(sizeof(Array));
/* Insert only non-negative length's*/
newArray->length = (length > 0) ? length : 0;
newArray->array = (double*) malloc(length*sizeof(double));
return newArray;
}
void SetArray(Array *structure,int length,double* array)
{
structure->length = length;
structure->array = array;
}
void PrintArray(Array *structure)
{
if(structure->length > 0)
{
int i;
printf("length: %d\n", structure->length);
for (i = 0; i < structure->length; i++)
printf("%g\n", structure->array[i]);
}
else
printf("Empty Array. Length 0\n");
}
int main()
{
int i;
Array *negativeTest, *days = NewArray(5);
double moreDays[] = {1,2,3,4,5,6,7,8,9,10};
for (i = 0; i < days->length; i++)
days->array[i] = i+1;
PrintArray(days);
SetArray(days,10,moreDays);
PrintArray(days);
negativeTest = NewArray(-5);
PrintArray(negativeTest);
return 0;
}
But you have to care about set the right length of the array you want to store, because the is no way to check this length, like our friends massively explained.
This is how I personally do it in my code. I like to keep it as simple as possible while still able to get values that I need.
typedef struct intArr {
int size;
int* arr;
} intArr_t;
int main() {
intArr_t arr;
arr.size = 6;
arr.arr = (int*)malloc(sizeof(int) * arr.size);
for (size_t i = 0; i < arr.size; i++) {
arr.arr[i] = i * 10;
}
return 0;
}
No, you can't use sizeof(ptr) to find the size of array ptr is pointing to.
Though allocating extra memory(more than the size of array) will be helpful if you want to store the length in extra space.
int main()
{
int days[] = {1,2,3,4,5};
int *ptr = days;
printf("%u\n", sizeof(days));
printf("%u\n", sizeof(ptr));
return 0;
}
Size of days[] is 20 which is no of elements * size of it's data type.
While the size of pointer is 4 no matter what it is pointing to.
Because a pointer points to other element by storing it's address.
In strings there is a '\0' character at the end so the length of the string can be gotten using functions like strlen. The problem with an integer array, for example, is that you can't use any value as an end value so one possible solution is to address the array and use as an end value the NULL pointer.
#include <stdio.h>
/* the following function will produce the warning:
* ‘sizeof’ on array function parameter ‘a’ will
* return size of ‘int *’ [-Wsizeof-array-argument]
*/
void foo( int a[] )
{
printf( "%lu\n", sizeof a );
}
/* so we have to implement something else one possible
* idea is to use the NULL pointer as a control value
* the same way '\0' is used in strings but this way
* the pointer passed to a function should address pointers
* so the actual implementation of an array type will
* be a pointer to pointer
*/
typedef char * type_t; /* line 18 */
typedef type_t ** array_t;
int main( void )
{
array_t initialize( int, ... );
/* initialize an array with four values "foo", "bar", "baz", "foobar"
* if one wants to use integers rather than strings than in the typedef
* declaration at line 18 the char * type should be changed with int
* and in the format used for printing the array values
* at line 45 and 51 "%s" should be changed with "%i"
*/
array_t array = initialize( 4, "foo", "bar", "baz", "foobar" );
int size( array_t );
/* print array size */
printf( "size %i:\n", size( array ));
void aprint( char *, array_t );
/* print array values */
aprint( "%s\n", array ); /* line 45 */
type_t getval( array_t, int );
/* print an indexed value */
int i = 2;
type_t val = getval( array, i );
printf( "%i: %s\n", i, val ); /* line 51 */
void delete( array_t );
/* free some space */
delete( array );
return 0;
}
/* the output of the program should be:
* size 4:
* foo
* bar
* baz
* foobar
* 2: baz
*/
#include <stdarg.h>
#include <stdlib.h>
array_t initialize( int n, ... )
{
/* here we store the array values */
type_t *v = (type_t *) malloc( sizeof( type_t ) * n );
va_list ap;
va_start( ap, n );
int j;
for ( j = 0; j < n; j++ )
v[j] = va_arg( ap, type_t );
va_end( ap );
/* the actual array will hold the addresses of those
* values plus a NULL pointer
*/
array_t a = (array_t) malloc( sizeof( type_t *) * ( n + 1 ));
a[n] = NULL;
for ( j = 0; j < n; j++ )
a[j] = v + j;
return a;
}
int size( array_t a )
{
int n = 0;
while ( *a++ != NULL )
n++;
return n;
}
void aprint( char *fmt, array_t a )
{
while ( *a != NULL )
printf( fmt, **a++ );
}
type_t getval( array_t a, int i )
{
return *a[i];
}
void delete( array_t a )
{
free( *a );
free( a );
}
#include <stdio.h>
#include <string.h>
#include <stddef.h>
#include <stdlib.h>
#define array(type) struct { size_t size; type elem[0]; }
void *array_new(int esize, int ecnt)
{
size_t *a = (size_t *)malloc(esize*ecnt+sizeof(size_t));
if (a) *a = ecnt;
return a;
}
#define array_new(type, count) array_new(sizeof(type),count)
#define array_delete free
#define array_foreach(type, e, arr) \
for (type *e = (arr)->elem; e < (arr)->size + (arr)->elem; ++e)
int main(int argc, char const *argv[])
{
array(int) *iarr = array_new(int, 10);
array(float) *farr = array_new(float, 10);
array(double) *darr = array_new(double, 10);
array(char) *carr = array_new(char, 11);
for (int i = 0; i < iarr->size; ++i) {
iarr->elem[i] = i;
farr->elem[i] = i*1.0f;
darr->elem[i] = i*1.0;
carr->elem[i] = i+'0';
}
array_foreach(int, e, iarr) {
printf("%d ", *e);
}
array_foreach(float, e, farr) {
printf("%.0f ", *e);
}
array_foreach(double, e, darr) {
printf("%.0lf ", *e);
}
carr->elem[carr->size-1] = '\0';
printf("%s\n", carr->elem);
return 0;
}
#define array_size 10
struct {
int16 size;
int16 array[array_size];
int16 property1[(array_size/16)+1]
int16 property2[(array_size/16)+1]
} array1 = {array_size, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
#undef array_size
array_size is passing to the size variable:
#define array_size 30
struct {
int16 size;
int16 array[array_size];
int16 property1[(array_size/16)+1]
int16 property2[(array_size/16)+1]
} array2 = {array_size};
#undef array_size
Usage is:
void main() {
int16 size = array1.size;
for (int i=0; i!=size; i++) {
array1.array[i] *= 2;
}
}
Most implementations will have a function that tells you the reserved size for objects allocated with malloc() or calloc(), for example GNU has malloc_usable_size()
However, this will return the size of the reversed block, which can be larger than the value given to malloc()/realloc().
There is a popular macro, which you can define for finding number of elements in the array (Microsoft CRT even provides it OOB with name _countof):
#define countof(x) (sizeof(x)/sizeof((x)[0]))
Then you can write:
int my_array[] = { ... some elements ... };
printf("%zu", countof(my_array)); // 'z' is correct type specifier for size_t

Pointer back and next for a node

I'm new to C++. I'm now trying to create a class with back and forth pointer. My code is listed below:
#include<iostream>
using namespace std;
class Node
{
public:
Node(int d, Node*k = NULL, Node*q = NULL) :data(d), back(k), next(q){};
int data;
Node*next; // point to next value on the list
Node*back; // point to back value on the list
};
int main()
{
int n;
Node*p = NULL;
Node*k = NULL; //k is back
while (cin >> n)
{
p = new Node(n,k);
p->back->next = p;
k = p;
}
for (; p; p = p->back)
cout << p->data << "->";
cout << "*\n";
system("pause");
}
However, I always have this error: "Access violation writing location"
I wonder if anybody have a solution ? Thanks
In the first iteration of the loop p->back is NULL. You get the access violation because you dereference it. Write this instead:
while (cin >> n)
{
p = new Node(n,k);
if (p->back != NULL) // p->back == NULL in the first iteration
p->back->next = p;
k = p;
}

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