I have ran into a peculiar situation:
void f(signed int8 Var1) {
if ( Var1 < 0 )
Var1 = -Var1;
/// at this point I'm expecting Var1 to be >= 0
/// and it works, unless we call f(-128).
/// for -128 there is no corresponding positive number that
/// signed int8 can hold, and (-Var1) results in the same -128.
}
I ran into this while working on an embedded code, but I tested this and works the same funny way on Windows as well.
For now I added this to the code:
if (Var1 == -128)
Var1 = 127;
I'm wondering if there is better way of dealing with this.
the point is the result of |-128| cannot be expressed in an 8 bit signed integer. So what do you want to do in that case? One possibility is to not overwrite var1 but to have a working internal variable thats unsigned 8 bit. That can always hold the result of |u8|
I dont know if thats really a solution because you have not explained what happens next in f.
Here is my suggestion
#include <stdio.h>
#include <stdint.h>
void f(int8_t Val1){
uint8_t abs ;
if (Val1 < 0){
abs = -Val1;
} else{
abs = Val1;
}
printf("i=%d o=%u\n", Val1, abs);
}
int main()
{
f(1);
f(-4);
f(-128);
}
output
i=1 o=1
i=-4 o=4
i=-128 o=128
Related
I'm trying to parse a hex number byte by byte, and concatenate to a string the representation of each byte, in the order they're stored in memory. (for a little test on endianness, but that's not important I guess).
Here is the code (please ignore the glaring unit-test issues with it :D; also, some of the code might look weird since initially the display_bytes method took in a char* not an int8_t*, but I thought using an int8_t might make it more obvious to me, what the issue is)
TEST_CLASS(My001littlebigendian)
{
public:
TEST_METHOD(TestMethod1)
{
int i = 0x12345678;
display_bytes((int8_t*)&i, sizeof(i));
}
void display_bytes(int8_t* b, int length)
{
std::stringstream ss;
for (int i = 0; i < length; ++i)
{
int8_t signedCharRepresentation = *(b + i); //signed char has 1 byte
int8_t signed8ByteInt = (int8_t)signedCharRepresentation; //this is not ok
int32_t signed32ByteInt = (int32_t)signedCharRepresentation; //this is ok. why?
//ss << std::hex << signed8ByteInt; //this is not ok. why?
ss << std::hex << signed32ByteInt; //this is ok
}
std::string stringRepresentation = ss.str();
if (stringRepresentation.compare("78563412") == 0)
{
Assert::IsTrue(true, L"machine is little-endian");
}
else if(stringRepresentation.compare("01234567") == 0)
{
Assert::IsTrue(true, L"machine is big-endian");
}
else
{
Assert::IsTrue(true, L"machine is other-endian");
}
}
};
Now, what I don't understand (as hopefull the comments make clear) is why does this only work when I cast each byte to a 4 byte int, and not an 1 byte int. Since I am working with chunks of 1 byte. Intuitively it would make me think doing it like this should cause some sort of overflow? But it seems not.
I've not dug deeper into why this is the issue yet, since I was hoping to not need to. And maybe if someone with more knowledge in this area can give me nudge in the right direction, or maybe even an outright answer if I'm missing something very obvious. (which I do feel I might be, since I'm not used to working at this low level).
I have two rand arrays: pointer and value. Whatever values in the pointer should also come in value with same number of times. For eg: if pointer[i] == 2, then value should have a value 2 which occur two times and should be after 1.
Expected result is shown below.
Sample code:
class ABC;
rand int unsigned pointer[$];
rand int unsigned value[20];
int count;
constraint c_mode {
pointer.size() == count;
solve pointer before value;
//======== Pointer constraints =========//
// To avoid duplicates
unique {pointer};
foreach(pointer[i]) {
// Make sure pointer is inside 1 to 4
pointer[i] inside {[1:4]};
// Make sure in increasing order
if (i>0)
pointer[i] > pointer[i-1];
}
//======== Value constraints =========//
//Make sure Pointer = 2 has to come two times in value, but this is not working as expected
foreach(pointer[i]) {
value.sum with (int'(item == pointer[i])) == pointer[i];
}
// Ensure it will be in increasing order but not making sure that pointers are not grouping together
// For eg: if pointer = 2, then 2 has to come two times together and after 1 in the array order. This is not met with the below constraint
foreach(value[i]) {
foreach(value[j]) {
((i>j) && (value[i] inside pointer) && (value[j] inside pointer)) -> value[i] >= value[j];
}
}
}
function new(int num);
count = num;
endfunction
endclass
module tb;
initial begin
int unsigned index;
ABC abc = new(4);
abc.randomize();
$display("-----------------");
$display("Pointer = %p", abc.pointer);
$display("Value = %p", abc.value);
$display("-----------------");
end
endmodule
I would implement this using a couple of helper arrays:
class pointers_and_values;
rand int unsigned pointers[];
rand int unsigned values[];
local rand int unsigned values_dictated_by_pointers[][];
local rand int unsigned filler_values[][];
// ...
endclass
The values_dictated_by_pointers array will contain the groups of values that your pointers mandate. The other array will contain the dummy values that come between these groups. So, the values array will contain filler_values[0], values_dictated_by_pointers[0], filler_values[1], values_dictated_by_pointers[1], etc.
Computing the values mandated by the pointers is easy:
constraint compute_values_dicated_by_pointers {
values_dictated_by_pointers.size() == pointers.size();
foreach (pointers[i]) {
values_dictated_by_pointers[i].size() == pointers[i];
foreach (values_dictated_by_pointers[i,j])
values_dictated_by_pointers[i][j] == pointers[i];
}
}
You need as many groups as you need pointers. In each group you have as many elements as the pointer value for that group. Also, each element of a group has the same value as the group's pointer value.
For the filler values you didn't mention what they should look like. I interpreted your problem description to say that the values in the pointers array should only come in the patters described above. This means that they are not allowed as filler values. Depending on whether you want to allow filler values before the first value, you will need either as many filler groups as you have pointers or one extra. In the following code I allowed filler values before the "real" values:
constraint compute_filler_values {
filler_values.size() == pointers.size() + 1;
foreach (filler_values[i, j])
!(filler_values[i][j] inside { pointers });
}
You'll also need to constrain the size of each of the filler value groups, otherwise the solver will leave them as 0. Here you can change the constraints to match your requirements. I chose to always insert filler values and to never insert more than 3 filler values.
constraint max_number_of_filler_values {
foreach (filler_values[i]) {
filler_values[i].size() > 0;
filler_values[i].size() <= 3;
}
}
For the real values array, you can compute its value in post_randomize() by interleaving the other two arrays:
function void post_randomize();
values = filler_values[0];
foreach (pointers[i])
values = { values, values_dictated_by_pointers[i], filler_values[i] };
endfunction
If you need to be able to constrain values as well, then you'll have to implement this interleaving operation using constraints. I'm not going to show this, as this is probably pretty complicated in itself and warrants an own question.
Be aware that the code above might not work on all EDA tools, because of spotty support for random multi-dimensional arrays. I only got this to work on Aldec Riviera Pro on EDA Playground.
I was given an assignment to create a procedure that scans a float, called getfloat.
for some reason, I am getting random values. If I enter "1" it prints 49.Why does this happen? And also, when i input values, I can't see them on the screen? when I use scanf for example i see what i hit, on the little black screen. but now the screen is just blank, and when i click enter it shows a bad output:
Example - input: -1. Output: 499.00000
Here is my code:
#include <stdio.h>
#include <conio.h>
#include <math.h>
#include <ctype.h>
void getfloat(float* num);
void main()
{
float num=0;
printf("Enter the float\n");
getfloat(&num);
printf("\nThe number is %lf\n",num);
getch();
}
void getfloat(float* num)
{
float c,sign=1,exponent=10;
c=getch();
if((!isdigit(c))&&(c!='+')&&(c!='-')) //if it doesnt start with a number a + or a -, its not a valid input
{
printf("Not a number\n");
return;
}
if(c=='-') //if it starts with a minus, make sign negative one, later multiply our number by sign
sign=-1;
for(*num=0;isdigit(c);c=getch())
*num=(*num*10)+c; //scan the whole part of the number
if(c!='.') //if after scanning whole part, c isnt a dot, we finished
return;
do //if it is a dot, scan fraction part
{
c=getch();
if(isdigit(c))
{
*num+=c/exponent;
exponent*=10;
}
}while(isdigit(c));
*num*=sign;
}
There are a number of issues.
1) Your posted code does not match your example "input: -1. Output: 499.00000", I get 0 due the lack of a getch() after finding a '-'. See #6.
1) 'c' is a character. When you enter '1', c took on a code for the letter 1, which in your case being ASCII coding, is 49. To convert a digit from its ASCII value to a number value, subtract 48 (the ASCII code for the letter '0', often done as c - '0'
*num=(*num*10)+c;
*num+=c/exponent;
becomes
*num = (*num*10) + (c-'0');
*num += (c-'0')/exponent;
2) Although you declare c as a float, recommend you declare it as an int. int is the return type from getch().
3) Function getch() is "used to get a character from console but does not echo to the screen". That is why you do not see them. Consider getchar() instead.
4) [Edit: delete Avoid =-. Thank-you #Daniel Fischer]
5) Your exponential calculation needs rework. Note: your exponent could receive a sign character.
6) When you test if(c=='-'), you do not then fetch another c. You also might want to test for else if(c=='+') and consume that c.
Good luck in your C journey.
49 is the Ascii code for the number 1. So when (0'<=c && c <='9') you need to subtract '0' to get the number itself.
A small hint: 49 is the ASCII for the character 1. You are using getch(), which gives you the return value char.
i'm trying to write a program that gets from the user a value x and prints sinx using taylor series. but my output is bad. the output i get is not even a number, its -1.#IND00 regardless of what i input.
here's my code
#include <stdio.h>
#include <conio.h>
void main()
{
int i;
double x,sum,last;
sum=(double)0;
scanf("%f",&x);
last=x;
sum=last;
for(i=1;i<10;i++)
{
last*=(double)(-x*x)/((2*i)*(2*i+1));
sum+=last;
}
printf("%f",sum);
getch();
}
I can see one problem:
scanf("%f",&x);
x is a double, so you need the l, i.e. "%lf".
[true but irrelevant point about how this isn't the right formula for sinh, even though sinh is nowhere mentioned in the question, redacted..]
The function gets an integer and a digit, and should return true
if the digit appears an even number of times in the integer, or false if not.
For example:
If digit=1 and num=1125
the function should return true.
If digit=1 and num=1234
the function should return false.
bool isEven(int num, int dig)
{
bool even;
if (num < 10)
even = false;
else
{
even = isEven(num/10,dig);
This is what I've got so far, and I'm stuck...
This is homework so please don't write the answer but hint me and help me get to it by myself.
To set up recursion, you need to figure out two things:
The base case. What is are the easy cases that you can handle outright? For example, can you handle single-digit numbers easily?
The rule(s) that reduce all other cases towards the base case. For example, can you chop off the last digit and somehow transform the solution for the remaning partial number into the solution for the full number?
I can see from your code that you've made some progress on both of these points. However, both are incomplete. For one thing, you are never using the target digit in your code.
The expression num%10 will give you the last digit of a number, which should help.
Your base case is incorrect because a single digit can have an even number of matches (zero is an even number). Your recursive case also needs work because you need to invert the answer for each match.
This funtion isEven() takes a single integer and returns the true if the number of occurence of numberToCheck is even.
You can change the base as well as the numberToCheck which are defined globally.
#include <iostream>
using std::cout;
using std::endl;
// using 10 due to decimal [change it to respective base]
const int base = 10;
const int numberToCheck = 5;
//Checks if the number of occurence of "numberToCheck" are even or odd
bool isEven(int n)
{
if (n == 0)
return 1;
bool hasNumber = false;
int currentDigit = n % base;
n /= base;
if (currentDigit == numberToCheck)
hasNumber = true;
bool flag = isEven(n);
// XOR GATE
return ((!hasNumber) && (flag) || (hasNumber) && (!flag));
};
int main(void)
{
// This is the input to the funtion IsEven()
int n = 51515;
if (isEven(n))
cout << "Even";
else
cout << "Odd";
return 0;
}
Using XOR Logic to integrate all returns
// XOR GATE
return ((!hasNumber) && (flag) || (hasNumber) && (!flag));