Say we have a large block:
mod module {
pub const fiz: u32 = (1 << 0);
// etc...
}
flag = {
if (var & module::fiz) != 0 { module::fiz }
else if (var & module::foo) != 0 { module::foo }
else if (var & module::bar) != 0 { module::bar }
else if (var & module::baz) != 0 { module::baz }
// .. there could be many more similar checks
};
With simply replacement macro its possible to do:
#define TEST(f) ((var) & (f)) != 0 { f }
Allowing:
flag = {
if TEST(module::fiz)
else if TEST(module::foo)
else if TEST(module::bar)
else if TEST(module::baz)
}
It seems Rust doesn't allow a macro to declare part of an if statement.
I managed to avoid repetition using assignment, but its quite ugly.
flag = {
let f;
if (var & {f = module::fiz; f }) != 0 { f }
else if (var & {f = module::foo; f }) != 0 { f }
else if (var & {f = module::bar; f }) != 0 { f }
else if (var & {f = module::baz; f }) != 0 { f }
};
Does Rust provide some convenient/elegant way to allow repetition in this case?
I don't think flag checking is the important part of this question, the issue is that you may want to repeat content in the check again in the body of an if statement, e.g.:
if (foo && OTHER_EXPRESSION) { do(); something_with(OTHER_EXPRESSION) }
else if (foo && SOME_EXPRESSION) { do(); something_with(SOME_EXPRESSION) }
I think you have an X/Y problem here, so I am going to solve this without using if/else.
What you seem to be doing is checking for the presence of a bit pattern, and prioritise the order in which said patterns are checked for (unclear if it matters, but let's assume it does).
So, let's do this the functional way:
let constants = [fiz, foo, bar, baz];
let flag = constants.iter().filter(|v| var & *v == **v).next();
And it just works, no macro or repetitive stuff.
If you want to use macros, you can write it like this:
mod module {
pub const fiz: u32 = (1 << 0);
pub const foo: u32 = (1 << 1);
pub const bar: u32 = (1 << 2);
pub const baz: u32 = (1 << 3);
}
macro_rules! check_bits {
([$($Constant:expr),*]) => {
|var: u32| {
$(if ($Constant & var) != 0 {
return $Constant;
})*
return 0;
}
}
}
fn main() {
let var = 5;
let checker = check_bits!([module::bar, module::fiz, module::foo, module::baz]);
assert_eq!(checker(var), module::bar);
println!("All OK");
}
Related
can anybody help me to build the function mentioned above I am using dart in the flutter and want this function
make a function which calculate the number of same nearby character in flutter like aabcddaabb => 2abc2d2a2b
Same
void main() {
var input = 'aabcddaabb';
print(getret(input));
}
String getret(String input) {
var ret = '';
var cc = '';
var co = 0;
cut(){
var c = input[0];
input = input.substring(1);
return c;
}
write(){
if(co == 1) ret = '$ret$cc';
if(co > 1) ret = '$ret$co$cc';
}
while(input.isNotEmpty){
final c = cut();
if(c != cc){
write();
cc = c;
co = 1;
}else{
co ++;
}
}
write();
return ret; // 2abc2d2a2b
}
There's probably a smarter and shorter way to do it, but here's a possible solution:
String string = 'aaabcddaabb';
String result = '';
String lastMatch = '';
int count = 0;
while (string.isNotEmpty) {
if (string[0] != lastMatch) {
result += '${count > 1 ? count : ''}$lastMatch';
lastMatch = string[0];
count = 0;
}
count++;
string = string.substring(1);
}
result += '${count > 1 ? count : ''}$lastMatch';
print(result); //3abc2d2a2b
I also came up with this smarter solution. Even though it's nice that it's a single expression it's maybe not very readable:
String string = 'aaabcddaabb';
String result = string.split('').fold<String>(
'',
(r, e) => r.isNotEmpty && e == r[r.length - 1]
? r.length > 1 &&
int.tryParse(r.substring(r.length - 2, r.length - 1)) != null
? '${r.substring(0, r.length - 2)}${int.parse(r.substring(r.length - 2, r.length - 1)) + 1}$e'
: '${r.substring(0, r.length - 1)}2$e'
: '$r$e');
print(result); //3abc2d2a2b
I implemented RSA cryptosystem to create a simple digital signature system. It is for freeCodeCamp's upcoming Cryptography curriculum.
The below program logs "Failure" when I use some bigger prime numbers like p = 11, q = 17. I am hashing my message and the result is smaller than N. I don't know why this is happening. Please guide me in the right direction.
This is for teaching millions of people out there. If you have any suggestions for improvement on the current implementation then please feel free to comment :)
Git Repo: https://github.com/vkweb/digital-signature-system
Thanks! Happy coding <3
const firstPrime =11;
const secondPrime = 19;
const N = firstPrime * secondPrime;
const phiOfN = (firstPrime - 1) * (secondPrime - 1);
let publicKey = 0;
function hashTheMessage(message) {
let hashValue = 0;
for (let i = 0, msgLength = message.length; i < msgLength; ++i) {
hashValue += message.charCodeAt(i);
}
return hashValue % N === 0 ? 1 : hashValue % N;
}
function isCoPrime(smallerNum, largerNum) {
for (let i = 2; i <= smallerNum; ++i) {
if (smallerNum % i === 0 && largerNum % i === 0) {
return false;
}
}
return true;
}
function generatePrivateKey() {
for (let i = 2; i < phiOfN; ++i) {
if (isCoPrime(i, N) && isCoPrime(i, phiOfN)) {
return i;
}
}
console.log("\nPrivate key can't be generated.");
return 0;
}
function generatePublicKey(privateKey) {
if (!privateKey) {
console.log("\nPublic key can't be generated.");
} else {
publicKey = 1;
while (privateKey) {
if ((publicKey * privateKey) % phiOfN === 1 && privateKey !== publicKey) {
return;
}
publicKey += 1;
}
}
}
function generateSignature(hashValue, privateKey) {
return Math.pow(hashValue, privateKey) % N;
}
function decryptSignature(digitalSignature) {
return Math.pow(digitalSignature, publicKey) % N;
}
let hashValue = hashTheMessage("Hello world!");
let privateKey = generatePrivateKey();
generatePublicKey(privateKey);
let signature = generateSignature(hashValue, privateKey);
let decryptedSignature = decryptSignature(signature);
if(decryptedSignature === hashValue)
{
console.log("Success!");
}
else
{
console.log("Failure!");
}
```
I don't know how convert java "continue" to scala.
I can use marker from bool + break, but its bad idea
Google did not help :(
It's my first program in scala... yep.. it's horrible
sort java
def sort(in: Array[Int], a:Int, b:Int)
{
var i,j,mode;
double sr=0;
if (a>=b) return; // size =0
for (i=a; i<=b; i++) sr+=in[i];
sr=sr/(b-a+1);
for (i=a, j=b; i <= j;)
{
if (in[i]< sr) { i++; continue; } // left > continue
if (in[j]>=sr) { j--; continue; } // right, continue
int c = in[i]; in[i] = in[j]; in[j]=c;
i++,j--; // swap and continue
}
if (i==a) return;
sort(in,a,j); sort(in,i,b);
}
sort scala...
def SortMerger(in:List[Int], a:Int, b:Int):Unit = {
var i = 0;
var j = 0;
var mode = 0;
var sr = 0.0;
if(a>=b) return;
i=a
while(i<=b)
{
sr+=in.ElementOf(i);
i += 1
}
sr=sr/(b-a+1)
i=a
j=b
while(i<=j)
{
if(in.ElementOf(i)<sr) {
i += 1;
// where continue??? ><
}
}
return
}
Scala has no continue statement, but what you are trying to do can be done with a simple if/else structure.
while(i<=j)
{
if(in(i) < sr) {
i += 1
} else if (in(j) >= sr) {
j -= 1
} else {
int c = in(i)
in(i) = in(j)
in(j) = c
i += 1
j -= 1
}
}
Note that the type of in here should be Array, not List
I am practicing my array form of data structure with swift.
I made a class "student"
and there are functions like display() and delete()
However, the application is not working.
There is an error message that
EXC_BAD_INSTRUCTION (code=EXC_I386_INVOP, sub code=0x0).
I think this error is about "optional" problem.
Here is my code.
class student
{
var studentArray = [[String]?]()
var numberOfStudents : Int = 10;
func display()
{
for (var i = 0; i < numberOfStudents ; i++)
{
print("{");
for (var j = 0; j < 2; j++)
{
print(studentArray[i]![j] + " ");
}
print("}");
}
}
func delete( value : String)
{
var i = 0
for ( i = 0; i < numberOfStudents ; i++)
{
if (value == studentArray[i]![1])
{
break;
}
}
if (i == numberOfStudents - 1 )
{
print("not found");
}
else
{
for (var k = i; k < numberOfStudents - 1 ; k++)
{
studentArray[k]![1] = studentArray[k+1]![1];
studentArray[k]![0] = studentArray[k+1]![0];
}
numberOfStudents--;
}
}
}
var hello = student()
hello.studentArray = [["0","0ee"],["9","9ee", ]]
hello.display() // I have a error at this point
hello.studentArray
Could anyone explain what is about it for me?
There are several mistakes in your code. The actual error is caused by your numberOfStudents variable, which is hard coded to 10, even though the array only contains 2 elements. Use studentArray.count in your for loop, not 10. Then read the Swift manual. You should not be using optionals nor C-style for loops in this example.
Here's how I would do it...
class Student { // Capitalise your classes
// Unnecessary whitespace removed
var studentArray: [[String]] = [] // No need for optionals here
/*
var numberOfStudents : Int = 10; // var is useless & wrong, also no need for semi-colon
*/
func display() {
/* A Swift-ier way to do this is
for student in studentArray {
print("{")
for field in student {
print(field + " ")
}
print("}")
}
However, using indexing:
*/
for i in 0 ..< studentArray.count {
print("{")
for j in 0 ..< studentArray[i].count { // Don't *know* this will be 2
print(studentArray[i][j] + " ") // Don't need semi-colons unless you want to put multiple statements on the same line
}
print("}")
}
}
/* func delete() not used in question, so removed from answer */
}
var hello = Student()
hello.studentArray = [["0","0ee"], ["9","9ee", ]] // Note spurious (but not wrong) comma
hello.display()
hello.studentArray
I believe this can be achieved by counting the instances for each character in that string. Even if a single character in that string is repeated at least twice, we can declare that string as a palindrome.
For example: bbcccc can be rewritten as bccccb or ccbbcc.
edified can be rewritten as deified.
Some book mentioned we should be using hash table. I think we can just use a list and check for the character count.
Do you think the logic is correct?
Yes, the main idea is to count the times of each char existing in the string. And it will be true if the string has at most one char occurs odd times and all others even times.
For example:
aabbcc => acbbca
aabcc => acbca
aabbb => abbba
No. You don't have to use a hash map (as some of the other answers suggest). But the efficiency of the solution will be determined by the algorithm you use.
Here is a solution that only tracks odd characters. If we get 2 odds, we know it can't be a scrambled palindrome. I use an array to track the odd count. I reuse the array index 0 over and over until I find an odd. Then I use array index 1. If I find 2 odds, return false!
Solution without a hash map in javascript:
function isScrambledPalindrome(input) {
// TODO: Add error handling code.
var a = input.split("").sort();
var char, nextChar = "";
var charCount = [ 0 ];
var charIdx = 0;
for ( var i = 0; i < a.length; ++i) {
char = a[i];
nextChar = a[i + 1] || "";
charCount[charIdx]++;
if (char !== nextChar) {
if (charCount[charIdx] % 2 === 1) {
if (charCount.length > 1) {
// A scrambled palindrome can only have 1 odd char count.
return false;
}
charIdx = 1;
charCount.push(0);
} else if (charCount[charIdx] % 2 === 0) {
charCount[charIdx] = 0;
}
}
}
return true;
}
console.log("abc: " + isScrambledPalindrome("abc")); // false
console.log("aabbcd: " + isScrambledPalindrome("aabbcd")); // false
console.log("aabbb: " + isScrambledPalindrome("aabbb")); // true
console.log("a: " + isScrambledPalindrome("a")); // true
Using a hash map, I found a cool way to only track the odd character counts and still determine the answer.
Fun javascript hash map solution:
function isScrambledPalindrome( input ) {
var chars = {};
input.split("").forEach(function(char) {
if (chars[char]) {
delete chars[char]
} else {
chars[char] = "odd" }
});
return (Object.keys(chars).length <= 1);
}
isScrambledPalindrome("aba"); // true
isScrambledPalindrome("abba"); // true
isScrambledPalindrome("abca"); // false
Any string can be palindrome only if at most one character occur odd no. of times and all other characters must occur even number of times.
The following program can be used to check whether a palindrome can be string or not.
vector<int> vec(256,0); //Vector for all ASCII characters present.
for(int i=0;i<s.length();++i)
{
vec[s[i]-'a']++;
}
int odd_count=0,flag=0;
for(int i=0;i<vec.size();++i)
{
if(vec[i]%2!=0)
odd_count++;
if(odd_count>1)
{
flag=1;
cout<<"Can't be palindrome"<<endl;
break;
}
}
if(flag==0)
cout<<"Yes can be palindrome"<<endl;
My code check if can it is palindrome or can be manipulated to Palindrome
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
//Tested on windows 64 bit arhc by using cygwin64 and GCC
bool isPalindrome (char *text);
int main()
{
char text[100]; // it could be N with defining N
bool isPal,isPosPal = false;
printf("Give me a string to test if it is Anagram of Palindrome\n");
gets(text);
isPal = isPalindrome(text);
isPosPal = isAnagramOfPalindrome(text);
if(isPal == false)
{
printf("Not a palindrome.\n");
}
else
{
printf("Palindrome.\n");
}
if(isPosPal == false)
{
printf("Not Anagram of Palindrome\n");
}
else
{
printf("Anagram of Palindrome\n");
}
return 0;
}
bool isPalindrome (char *text) {
int begin, middle, end, length = 0;
length = getLength(text);
end = length - 1;
middle = length/2;
for (begin = 0; begin < middle; begin++)
{
if (text[begin] != text[end])
{
return false;
}
end--;
}
if (begin == middle)
return true;
}
int getLength (char *text) {
int length = 0;
while (text[length] != '\0')
length++;
printf("length: %d\n",length);
return length;
}
int isAnagramOfPalindrome (char *text) {
int length = getLength(text);
int i = 0,j=0;
bool arr[26] = {false};
int counter = 0;
//char string[100]="neveroddoreven";
int a;
for (i = 0; i < length; i++)
{
a = text[i];
a = a-97;
if(arr[a])
{
arr[a] = false;
}
else
{
arr[a] = true;
}
}
for(j = 0; j < 27 ; j++)
{
if (arr[a] == true)
{
counter++;
}
}
printf("counter: %d\n",counter);
if(counter > 1)
{
return false;
}
else if(counter == 1)
{
if(length % 2 == 0)
return false;
else
return true;
}
else if(counter == 0)
{
return true;
}
}
as others have posted, the idea is to have each character occur an even number of times for an even length string, and one character an odd number of times for an odd length string.
The reason the books suggest using a hash table is due to execution time. It is an O(1) operation to insert into / retrieve from a hash map. Yes a list can be used but the execution time will be slightly slower as the sorting of the list will be O(N log N) time.
Pseudo code for a list implementation would be:
sortedList = unsortedList.sort;
bool oddCharFound = false;
//if language does not permit nullable char then initialise
//current char to first element, initialise count to 1 and loop from i=1
currentChar = null;
currentCharCount = 0;
for (int i=0; i <= sortedList.Length; i++) //start from first element go one past end of list
{
if(i == sortedList.Length
|| sortedList[i] != currentChar)
{
if(currentCharCount % 2 = 1)
{
//check if breaks rule
if((sortedList.Length % 2 = 1 && oddCharFound)
|| oddCharFound)
{
return false;
}
else
{
oddCharFound = true;
}
}
if(i!= sortedList.Length)
{
currentCharCount = 1;
currentChar = sortedList[i];
}
}
else
{
currentCharCount++;
}
}
return true;
Here is a simple solution using an array; no sort needed
#include <stdio.h>
#include <stdlib.h>
int main()
{
int a[256] = { 0 };
unsigned char i[] = {"aaBcBccc"};
unsigned char *p = &i[0];
int c = 0;
int j;
int flag = 0;
while (*p != 0)
{
a[*p]++;
p++;
}
for(j=0; j<256; j++)
{
if(a[j] & 1)
{
c++;
if(c > 1)
{
flag = 1;
break;
}
}
}
if(flag)
printf("Nope\n");
else
printf("yup\n");
return 0;
}
C#:
bool ok = s.GroupBy(c => c).Select(g => g.Count()).Where(c => c == 1).Count() < 2;
This solution, however, does use hashing.
Assuming all input characters are lower case letters.
#include<stdio.h>
int main(){
char *str;
char arr[27];
int j;
int a;
j = 0;
printf("Enter the string : ");
scanf("%s", str);
while (*str != '\0'){
a = *str;
a = a%27;
if(arr[a] == *str){
arr[a]=0;
j--;
}else{
arr[a] = *str;
j++;
}
*str++;
}
if(j==0 || j== -1 || j==1){
printf ("\nThe string can be a palindrome\n");
}
}