Is there a way to retrieve the name of a system call being made while in syscall()? I see that we have access to the number; can I use that somehow?
if you want a string you have to hold an array of syscall names.
you can do it in syscall.c just like syscalls[]:
static char* syscallnames[] = {
[SYS_fork] "fork",
[SYS_exit] "exit",
[SYS_wait] "wait",
...
};
and use it in syscall() like this:
void
syscall(void)
{
int num;
num = proc->tf->eax;
if(num > 0 && num < NELEM(syscalls) && syscalls[num]) {
cprintf("calling sys call: %s", syscallnames[num]); // <<< code addition
proc->tf->eax = syscalls[num]();
} else {
cprintf("%d %s: unknown sys call %d\n",
proc->pid, proc->name, num);
proc->tf->eax = -1;
}
}
Related
This is a function if the endValueFixed is equal for example 12.0 I want to print the number without zero so I want it to be 12.
void calculateIncrease() {
setState(() {
primerResult = (startingValue * percentage) / 100;
endValue = startingValue + primerResult;
endValueFixe`enter code here`d = roundDouble(endValue, 2);
});
}
This may be an overkill but it works exactly as you wish:
void main() {
// This is your double value
final end = 98.04;
String intPart = "";
String doublePart = "";
int j = 0;
for (int i = 0; i < end.toString().length; i++) {
if (end.toString()[i] != '.') {
intPart += end.toString()[i];
} else {
j = i + 1;
break;
}
}
for (int l = j; l < end.toString().length; l++) {
doublePart += end.toString()[l];
}
if (doublePart[0] == "0" && doublePart[1] != "0") {
print(end);
} else {
print(end.toString());
}
}
You may use this code as a function and send whatever value to end.
if (endValueFixed==12) {
print('${endValueFixed.toInt()}');
}
conditionally cast it to an int and print it then :)
I have some difficulties when fixing PMD warnings, this was my simplified method:
public String rank(String keyword, int pageSize, int totalPage)
{
String result = "0"; // A: DataflowAnomalyAnalysis: Found 'DD'-anomaly for variable 'result'
if (isNotBlank(keyword))
{
boolean find = false; // B: DataflowAnomalyAnalysis: Found 'DD'-anomaly for variable 'find'
for (int page = 1; page < totalPage; page++)
{
int rank = getRank(keyword, pageSize, totalPage);
if (rank != 0)
{
find = true; // B(1)
result = String.valueOf(rank); // A(1)
break;
}
}
if (!find)
{
result = format("{0}+", totalPage * pageSize - 1); // A(2)
}
}
return result;
}
I tried this and got "OnlyOneReturn" warnings:
public String rank(String keyword, int pageSize, int totalPage)
{
if (isNotBlank(keyword))
{
for (int page = 1; page < totalPage; page++)
{
int rank = getRank(keyword, pageSize, totalPage);
if (rank != 0)
{
return String.valueOf(rank); // OnlyOneReturn
}
}
return format("{0}+", totalPage * pageSize - 1); // OnlyOneReturn
}
return "0";
}
How do I have to write this code please?
A 'DD'-anomaly an a dataflow analysis tells you that you assign a variable more than once without using it between the assignments. So all but the last assignment are unnecessary. It usually indicates that you didn't separate your scenarios properly. In your case you have three scenarios:
If the keyword is blank then the return value is "0".
Otherwise loop through all pages and if getRank() returns a rank other than zero then this is the return value.
Otherwise the return value is "totalPage * pageSize - 1+"
If you implement those scenarios one by one you end up with a method that has not any dataflow or other PMD issues:
public String rank(String keyword, int pageSize, int totalPage) {
String result;
if (isNotBlank(keyword)) {
result = "0";
} else {
int rank = 0;
for (int page = 1; page < totalPage && rank == 0; page++) {
rank = getRank(keyword, pageSize, totalPage);
}
if (rank != 0) {
result = String.valueOf(rank);
} else {
result = format("{0}+", totalPage * pageSize - 1);
}
}
return result;
}
If you take a closer look at the for loop you see that page is only used for looping. It is not used inside the loop. This indicates that the for loop is probably not necessary. getRank(keyword, pageSize, totalPage) should always return the same value as its arguments never change during the loop. So it might be enough to call getRank(...) just once.
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");
}
}
I am facing an issue in reading char values.
See my program below. I want to evaluate an infix expression.
As you can see I want to read '10' , '*', '20' and then use them...but if I use string indexer s[0] will be '1' and not '10' and hence I am not able to get the expected result.
Can you guys suggest me something? Code is in c#
class Program
{
static void Main(string[] args)
{
string infix = "10*2+20-20+3";
float result = EvaluateInfix(infix);
Console.WriteLine(result);
Console.ReadKey();
}
public static float EvaluateInfix(string s)
{
Stack<float> operand = new Stack<float>();
Stack<char> operator1 = new Stack<char>();
int len = s.Length;
for (int i = 0; i < len; i++)
{
if (isOperator(s[i])) // I am having an issue here as s[i] gives each character and I want the number 10
operator1.Push(s[i]);
else
{
operand.Push(s[i]);
if (operand.Count == 2)
Compute(operand, operator1);
}
}
return operand.Pop();
}
public static void Compute(Stack<float> operand, Stack<char> operator1)
{
float operand1 = operand.Pop();
float operand2 = operand.Pop();
char op = operator1.Pop();
if (op == '+')
operand.Push(operand1 + operand2);
else
if(op=='-')
operand.Push(operand1 - operand2);
else
if(op=='*')
operand.Push(operand1 * operand2);
else
if(op=='/')
operand.Push(operand1 / operand2);
}
public static bool isOperator(char c)
{
bool result = false;
if (c == '+' || c == '-' || c == '*' || c == '/')
result = true;
return result;
}
}
}
You'll need to split the string - which means working out exactly how you want to split the string. I suspect you'll find Regex.Split to be the most appropriate splitting tool in this case, as you're dealing with patterns. Alternatively, you may want to write your own splitting routine.
Do you only need to deal with integers and operators? How about whitespace? Brackets? Leading negative numbers? Multiplication by negative numbers (e.g. "3*-5")?
Store the numerical value in a variable, and push that when you encounter an operator or the end of the string:
int num = 0;
foreach (char c in s) {
if (isOperator(c)) {
if (num != 0) {
operand.Push(num);
num = 0;
}
operator1.Push(c);
if (operand.Count == 2) {
Compute(operand, operator1);
}
} else {
num = num * 10 + (int)(c - '0');
}
}
if (num != 0) {
operand.Push(num);
}