Given the following trade.proto file:
message Trade {
message ProductInfo {
optional uint64 product_id = 1;
optional String product_name = 2;
repeated int other_field = 3;
}
optional string id = 1;
optional uint64 partition_time = 2;
optional ProductInfo product_info= 3;
}
Given the following codeļ¼
def getOtherFieldSize(trade: Trade): Int = {
if (20221205 == trade.getPartitionTime) {
trade.getProductInfo.getOtherFieldList.size
} else {
0
}
}
How to get the fields of trade.proto used in the code? For example:
partition_time
product_info.other_field
Related
I want to replace n occurrence of a substring in a string.
myString = "I have a mobile. I have a cat.";
How I can replace the second have of myString
hope this simple function helps. You can also extract the function contents if you don't wish a function. It's just two lines with some
Dart magic
void main() {
String myString = 'I have a mobile. I have a cat.';
String searchFor='have';
int replaceOn = 2;
String replaceText = 'newhave';
String result = customReplace(myString,searchFor,replaceOn,replaceText);
print(result);
}
String customReplace(String text,String searchText, int replaceOn, String replaceText){
Match result = searchText.allMatches(text).elementAt(replaceOn - 1);
return text.replaceRange(result.start,result.end,replaceText);
}
Something like that should work:
String replaceNthOccurrence(String input, int n, String from, String to) {
var index = -1;
while (--n >= 0) {
index = input.indexOf(from, ++index);
if (index == -1) {
break;
}
}
if (index != -1) {
var result = input.replaceFirst(from, to, index);
return result;
}
return input;
}
void main() {
var myString = "I have a mobile. I have a cat.";
var replacedString = replaceNthOccurrence(myString, 2, "have", "had");
print(replacedString); // prints "I have a mobile. I had a cat."
}
This would be a better solution to undertake as it check the fallbacks also. Let me list down all the scenarios:
If position is 0 then it will replace all occurrence.
If position is correct then it will replace at same location.
If position is wrong then it will send back input string.
If substring does not exist in input then it will send back input string.
void main() {
String input = "I have a mobile. I have a cat.";
print(replacenth(input, 'have', 'need', 1));
}
/// Computes the nth string replace.
String replacenth(String input, String substr, String replstr,int position) {
if(input.contains(substr))
{
var splittedStr = input.split(substr);
if(splittedStr.length == 0)
return input;
String finalStr = "";
for(int i = 0; i < splittedStr.length; i++)
{
finalStr += splittedStr[i];
if(i == (position - 1))
finalStr += replstr;
else if(i < (splittedStr.length - 1))
finalStr += substr;
}
return finalStr;
}
return input;
}
let's try with this
void main() {
var myString = "I have a mobile. I have a cat.I have a cat";
print(replaceInNthOccurrence(myString, "have", "test", 1));
}
String replaceInNthOccurrence(
String stringToChange, String searchingWord, String replacingWord, int n) {
if(n==1){
return stringToChange.replaceFirst(searchingWord, replacingWord);
}
final String separator = "#######";
String splittingString =
stringToChange.replaceAll(searchingWord, separator + searchingWord);
var splitArray = splittingString.split(separator);
print(splitArray);
String result = "";
for (int i = 0; i < splitArray.length; i++) {
if (i % n == 0) {
splitArray[i] = splitArray[i].replaceAll(searchingWord, replacingWord);
}
result += splitArray[i];
}
return result;
}
here the regex
void main() {
var myString = "I have a mobile. I have a cat. I have a cat. I have a cat.";
final newString =
myString.replaceAllMapped(new RegExp(r'^(.*?(have.*?){3})have'), (match) {
return '${match.group(1)}';
});
print(newString.replaceAll(" "," had "));
}
Demo link
Here it is one more variant which allows to replace any occurrence in subject string.
void main() {
const subject = 'I have a dog. I have a cat. I have a bird.';
final result = replaceStringByOccurrence(subject, 'have', '*have no*', 0);
print(result);
}
/// Looks for `occurrence` of `search` in `subject` and replace it with `replace`.
///
/// The occurrence index is started from 0.
String replaceStringByOccurrence(
String subject, String search, String replace, int occurence) {
if (occurence.isNegative) {
throw ArgumentError.value(occurence, 'occurrence', 'Cannot be negative');
}
final regex = RegExp(r'have');
final matches = regex.allMatches(subject);
if (occurence >= matches.length) {
throw IndexError(occurence, matches, 'occurrence',
'Cannot be more than count of matches');
}
int index = -1;
return subject.replaceAllMapped(regex, (match) {
index += 1;
return index == occurence ? replace : match.group(0)!;
});
}
Tested on dartpad.
This is a function if the endValueFixed is equal for example 12.0 I want to print the number without zero so I want it to be 12.
void calculateIncrease() {
setState(() {
primerResult = (startingValue * percentage) / 100;
endValue = startingValue + primerResult;
endValueFixe`enter code here`d = roundDouble(endValue, 2);
});
}
This may be an overkill but it works exactly as you wish:
void main() {
// This is your double value
final end = 98.04;
String intPart = "";
String doublePart = "";
int j = 0;
for (int i = 0; i < end.toString().length; i++) {
if (end.toString()[i] != '.') {
intPart += end.toString()[i];
} else {
j = i + 1;
break;
}
}
for (int l = j; l < end.toString().length; l++) {
doublePart += end.toString()[l];
}
if (doublePart[0] == "0" && doublePart[1] != "0") {
print(end);
} else {
print(end.toString());
}
}
You may use this code as a function and send whatever value to end.
if (endValueFixed==12) {
print('${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 wrote a program for a binary addition in java. But the result is sometimes not right.
For example if i add 1110+111. The result should be 10101.
But my program throws out 10001.
Maybe one of you find the mistake.
import java.util.Scanner;
public class BinaryAdder {
public static String add(String binary1, String binary2) {
int a = binary1.length()-1;
int b = binary2.length()-1;
int sum = 0;
int carry = 0;
StringBuffer sb = new StringBuffer();
while (a >= 0 || b >= 0) {
int help1 = 0;
int help2 = 0;
if( a >=0){
help1 = binary1.charAt(a) == '0' ? 0 : 1;
a--;
} if( b >=0){
help2 = binary2.charAt(b) == '0' ? 0 : 1;
b--;
}
sum = help1 +help2 +carry;
if(sum >=2){
sb.append("0");
carry = 1;
} else {
sb.append(String.valueOf(sum));
carry = 0;
}
}
if(carry == 1){
sb.append("1");
}
sb.reverse();
String s = sb.toString();
s = s.replaceFirst("^0*", "");
return s;
}
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.print("First: ");
String input1 = scan.next("(0|1)*");
System.out.print("Second: ");
String input2 = scan.next("(0|1)*");
scan.close();
System.out.println("Result: " + add(input1, input2));
}
}
this function is much simpler :
public static String binaryAdd(String binary1,String binary2){
return Long.toBinaryString(Long.parseLong(binary1,2)+Long.parseLong(binary2,2));
}
you can change Long.parseLong into Integer.parseInt if you don't expect very large numbers, you can also replace parse(Long/Int) with parseUnsigned(Long/Int) since you don't expect your strings to have a minus sign do you ?
You are not considering the case when
help1 + help2 = 3
So your method String add(String binary1, String binary2) should be like this:
public static String add(String binary1, String binary2) {
int a = binary1.length()-1;
int b = binary2.length()-1;
int sum = 0;
int carry = 0;
StringBuffer sb = new StringBuffer();
while (a >= 0 || b >= 0) {
int help1 = 0;
int help2 = 0;
if( a >=0){
help1 = binary1.charAt(a) == '0' ? 0 : 1;
a--;
} if( b >=0){
help2 = binary2.charAt(b) == '0' ? 0 : 1;
b--;
}
sum = help1 +help2 +carry;
if (sum == 3){
sb.append("1");
carry = 1;
}
else if(sum ==2){
sb.append("0");
carry = 1;
} else {
sb.append(String.valueOf(sum));
carry = 0;
}
}
if(carry == 1){
sb.append("1");
}
sb.reverse();
String s = sb.toString();
s = s.replaceFirst("^0*", "");
return s;
}
I hope this could help you!
sum = help1 +help2 +carry;
if(sum >=2){
sb.append("0");
carry = 1;
} else {
sb.append(String.valueOf(sum));
carry = 0;
}
If sum is 2 then append "0" and carry = 1
What about when the sum is 3, append "1" and carry = 1
Will never be 4 or greater
Know I'm a bit late but I've just done a similar task so to anyone in my position, here's how I tackled it...
import java.util.Scanner;
public class Binary_Aids {
public static void main(String args[]) {
System.out.println("Enter the value you want to be converted");
Scanner inp = new Scanner(System.in);
int num = inp.nextInt();
String result = "";
while(num > 0) {
result = result + Math.floorMod(num, 2);
num = Math.round(num/2);
}
String flippedresult = "";
for(int i = 0; i < result.length(); i++) {
flippedresult = result.charAt(i) + flippedresult;
}
System.out.println(flippedresult);
}
}
This took an input and converted to binary. Once here, I used this program to add the numbers then convert back...
import java.util.Scanner;
public class Binary_Aids {
public static void main(String args[]) {
Scanner inp = new Scanner(System.in);
String decimalToBinaryString = new String();
System.out.println("First decimal number to be added");
int num1 = inp.nextInt();
String binary1 = decimalToBinaryString(num1);
System.out.println("Input decimal number 2");
int num2 = inp.nextInt();
String binary2 = decimalToBinaryString(num2);
int patternlength = Math.max[binary1.length[], binary2.length[]];
while(binary1.length() < patternlength) {
binary1 = "0" + binary2;
}
System.out.println(binary1);
System.out.println(binary2);
int carry = 0;
int frequency_of_one;
String result = "";
for(int i = patternlength -i; i >= 0; i--) {
frequency_of_one = carry;
if(binary1.charAt(i) == '1') {
frequency_of_one++;
}
if(binary2.charAt(i) == '1') {
frequency_of_one++;
}
switch(frequency_of_one) {
case 0 ;
carry = 0;
result = "1" + result;
break;
case 1 ;
carry = 0;
result = "1" + result;
break;
case 2;
carry = 1;
result = "0" + result;
breake;
case 3;
carry = 1;
result = "1" + result;
breake;
}
}
if(carry == 1) {
result = "1" + result;
}
System.out.println(result);
}
public static String decimalToBinaryString(int decimal1) {
String result = "";
while(decimal1 > 0) {
result = result + Math.floorMod(decimal1, 2);
decimal = Math.round(decimal1/2);
}
String flipresult = "";
for(int i = 0; i < result.length[]; i++) {
flipresult = result.charAt(i) + flippedresult;
}
return flippedresult;
}
}
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