Hello I have some experience with javascript but I would really like to learn how to program in C and one of the ways I am trying to learn is by converting some simple javascript code into C. My current attempt at converting a simple program compiles without any errors however doesn`t produce the answer I want it to. The javscript code produces the correct answer and I wrote it to solve project euler problem number 5 which can be found here: https://projecteuler.net/problem=5
Here is the working js code:
var number = 2520;
var count = 1;
var solved = false;
while (!solved) {
if (number % count === 0) {
if (count === 20) {
solved = true;
console.log(number);
} else {
count++;
}
} else {
number++;
count = 1;
}
}
Here is the C conversion which does not work:
#include <stdio.h>
int main () {
unsigned int number = 2520;
unsigned int count = 1;
unsigned int solved = 0;
while ((solved = 0)) {
if (number % count == 0) {
if (count == 20) {
solved = 1;
printf("%number");
} else {
count++;
}
} else {
number++;
count = 1;
}
}
return 0;
}
while ((solved = 0)) {
You can use the same syntax you would use in js here, namely, while (!solved), or ==, but just = is an assignment.
printf("%number");
Doesn't mean what you think it means, which is why it's not an actual error (%n is a distinct specifier, and with no corresponding input, you'd get umber as the output). To reproduce console.log() you'd want:
printf("%d\n", (int)number);
Or
printf("%u\n", number);
Notice the explicit \n, since printf() does not add a newline otherwise.
Replace
while (solved = 0)
with
while (!solved)
and
print("%number")
with
print("%u\n",number)
I know its already answered, but you should know why.
while ((solved = 0))
Will actually set solved to 0 AND return 0 (which is interpreted as false). So the while loop is exited right away.
printf also takes a pretty strictly formatted string for the first one (just typeing what makes sense is guarenteed to be wrong). The compiler has know way to know what is inside the string is anything other than a string (C++ has (almost) NO reflection, unlike javascript: your written code dissapears into ones and zeros). printf needs to take number in as the second argument. Try printf("%i\n",number);. That says "Print an integer followed by a newline. The integer's value is number."
Welcome to C! Your biggest problem going into is is going to be my biggest problem with Java Script: C is strictly typed with no reflection, while javascript has no types with almost everything relying on some sort of reflection.
Related
Greetings and happy holidays,
My question involves problem "1.2 Check Permutation" from "Cracking The Coding Interview" 6th Ed.
When we compare the two strings, we check for any values in the array that are less than 0. This would indicate different char counts of our two strings. However, shouldn't the comparison be !== 0 instead of < 0? This would catch both over and undercounts of the chars. I didn't see this in the books Errata nor on any related search results.
Below is the provided code solution. (I mainly work in JS, so perhaps I'm reading the code incorrectly)
Many thanks in advance.
boolean permutation(String s, String t) {
if (s.length() != t.length()) {
return false;
}
int[] letters = new int[128];
char[] s_array = s.toCharArray();
for (char c : s_array) {
letters[c]++;
}
for (int i = 0; i < t.length(); i++) {
int c = (int) t.charAt(i);
letters[c]--;
if (letters[c] < 0) { // this is the line in question
return false;
}
}
return true;
}
The comparison for != 0 only makes sense after all the differences are computed. The way the code is structured allows an early detection.
The very first test is for s.length() != t.length(). Once the code passed it, we know that there are equal number of characters. It means that - if the strings are not permutations - there is a character which appears more in t than in s. As soon as such character is found, we conclude that t is not a permutation.
I need to create a sequence of numbers using while or for that consists of the sum of the symbols of the number.
For example, I have a sequence from 1 to 10. In console (if I've already written a code) will go just 1, 2,3,4,5,6,7,8,9,1. If I take it from 30 to 40 in the console would be 3,4,5,6,7,8,9,10,11,12,13.
I need to create a code that displays a sum that goes from 1 to 100. I don't know how to do it but in console I need to see:
1
2
3
4
5
5
6
7
8
9
1
2
3
4
etc.
I've got some code but I got only NaN. I don't know why. Could you explain this to me?
for (let i = '1'; i <= 99; i++) {
let a = Number(i[0]);
let b = Number(i[1])
let b1 = Boolean(b)
if (b1 == false) {
console.log ('b false', a)
}
else {
console.log ('b true', a + b)
}
}
I hope you get what I was speaking about.
Although I like the accepted answer however from question I gather you were asking something else, that is;
30 become 3+0=3
31 become 3+1=4
37 becomes 3+7=10
Why are we checking for boolean is beyond the scope of the question
Here is simple snnipet does exactly what you ask for
for (let i = 30; i <= 40; i++) {
let x=i.toString();
console.log( 'numbers from ' +i + ' are added together to become '+ (Number(x[0])+Number((x[1])||0)))
}
what er are doing is exactly what Maskin stated begin with for loop then in each increment convert it to string so we can split it, this takes care of NAN issue.
you don't need to call to string just do it once as in let x then simply call the split as x[0] and so on.
within second number we have created a self computation (x[1])||0) that is if there is second value if not then zero. following would work like charm
for (let i = 1; i <= 10; i++) {
let x=i.toString();
console.log( 'numbers from ' +i + ' are added together to become '+ (Number(x[0])+Number((x[1])||0)))
}
Did you observe what happens to ten
here is my real question and solution what if you Don't know the length of the digits in number or for what ever reason you are to go about staring from 100 on wards. We need some form of AI into the code
for (let i = 110; i <= 120; i++) {
let x= Array.from(String(i), Number);
console.log(
x.reduce(function(a, b){ return a + b;})
);
};
You simply make an array with Array.from function then use simple Array.reduce function to run custom functions that adds up all the values as sum, finally run that in console.
Nice, simple and AI
You got NaN because of "i[0]". You need to add toString() call.
for (let i = '1'; i <= 99; i++) {
let a = Number(i.toString()[0]);
let b = Number(i.toString()[1])
let b1 = Boolean(b)
if (b1 == false) {
console.log('b false', a)
} else {
console.log('b true', a + b)
}
}
So the way a for loop works is that you declare a variable to loop, then state the loop condition and then you ask what happens at the end of the loop, normally you increment (which means take the variable and add one to it).
When you say let i = '1', what you're actually doing, is creating a new string, which when you ask for i[0], it gives you the first character in the string.
You should look up the modulo operator. You want to add the number of units, which you can get by dividing by 10 and then casting to an int, to the number in the tens, which you get with the modulo.
As an aside, when you ask a question on StackOverflow, you should ask in a way that means people who have similar questions to you can find their answers.
I am trying to solve a kata that seems to be simple on codewars but i seem to not be getting it right.
The instruction for this is as simple as below
Given the string representations of two integers, return the string representation of the sum of those integers.
For example:
sumStrings('1','2') // => '3'
A string representation of an integer will contain no characters besides the ten numerals "0" to "9".
And this is what i have tried
function sumStrings(a,b) {
return ((+a) + (+b)).toString();
}
But the results solves all except two and these are the errors i get
sumStrings('712569312664357328695151392', '8100824045303269669937') - Expected: '712577413488402631964821329', instead got: '7.125774134884027e+26'
sumStrings('50095301248058391139327916261', '81055900096023504197206408605') - Expected: '131151201344081895336534324866', instead got: '1.3115120134408189e+29'
I don't seem to understand where the issues is from. Any help would help thanks.
The value you entered is bigger than the int type max value. You can try changing your code to:
function sumStrings(a,b) {
return ((BigInt(a)) + BigInt(b)).toString();
}
This way it should return the right value
You could pop the digits and collect with a carry over for the next digit.
function add(a, b) {
var aa = Array.from(a, Number),
bb = Array.from(b, Number),
result = [],
carry = 0,
i = Math.max(a.length, b.length);
while (i--) {
carry += (aa.pop() || 0) + (bb.pop() || 0);
result.unshift(carry % 10);
carry = Math.floor(carry / 10);
}
while (carry) {
result.unshift(carry % 10);
carry = Math.floor(carry / 10);
}
return result.join('');
}
console.log(add('712569312664357328695151392', '8100824045303269669937'));
console.log(add('50095301248058391139327916261', '81055900096023504197206408605'));
The problem is that regular javascript integers are not having enough space to store that much big number, So it uses the exponential notation to not lose its precision
what you can do is split each number into parts and add them separately,
one such example is here SO answer
My solution is:
function sumStrings(a,b) {
return BigInt(a) + BigInt(b) + ''
}
Converting from a string to a number or vice versa is not perfect in any language, they will be off by some digits. This doesn't seem to affect small numbers, but it affects big numbers a lot.
The function could go like this.
function sumStrings(a, b) {
return (BigInt(a) + BigInt(b)).toString() // or parseInt for both
}
However, it's still not perfect since if we try to do:
console.log((4213213124214211215421314213.0 + 124214321214213434213124211.0) === sumStrings('4213213124214211215421314213', '124214321214213434213124211'))
The output would be false.
Can someone please help implement something in my javascript project? #http://codepen.io/urketadic/pen/YpLgBX
I want number to turn red if its wrong and not in sequence with pi. Its really difficult for me to keep count and compare with everything.
I've tried a lot of this and at the end I've come up with this code:
var count = 0;
// color the mistake right away
$("#inputsm").keyup(function(event) {
var pressed = event.key;
answer = $("#inputsm").val();
pisub = pi.substr(input,answer.length)
if (pressed!=="Backspace"&&pressed!=="Delete") count++;
else count--;
console.log(count);
});
I'm just confused, i don't know how i can do this. Also does text area even allow numbers to turn red? I've tried adding jquery css as well but it doesn't work. Can someone write it in their own codepen and post a link?
Take a look at this fiddle: https://jsfiddle.net/ebv5n64j/2/
I've put together something that does basically what you are trying to do. You can modify it from there but that is the basic concept. I designed mine to not let the user continue until they get it right, but you could easily change that.
This is a snippet, but jsfiddle has the complete working version:
// If it's a delete command
if(code === 8){
if(!$("#wrong").length > 0)
inputCount = (inputCount === 0 ? 0 : --inputCount);
$("#pi span").last().remove();
console.log(inputCount);
} else if (code >= 48 && code <= 57) {
var inputNumber = code - 48;
var numSpan;
$("#wrong").remove();
numSpan = $("<span>"); // make a new one
// Append the number
numSpan.text(inputNumber);
numSpan.removeClass("incorrect");
if(String(inputNumber) === piDigits[inputCount]){
numSpan.addClass("correct");
inputCount++;
} else {
numSpan.attr("id", "wrong");
}
$("#pi").append(numSpan);
placeCaretAtEnd(this);
}
So I looked at your pen and thought that you were somewhat over-complicating the solution. I think a better way would be to compare the string of the input textarea with the substring of pi. Anyways, here's the fixed code and I've linked to the pen with the working version.
By the way, you mention in the description that if they want to start at the number 4 in 3.14 they should type in 1 (for index 1), but you take their input and subtract it by 1, so it essentially starts them off at 1 instead of 4, if they typed in 1.
$("#inputsm").keyup(function() {
var thisLength = parseInt(input) + $(this).val().length - 1;
if($(this).val().trim() === pi.substring(parseInt(input), thisLength)) {
console.log("good so far!");
$(this).removeAttr('style');
} else {
console.log("ahhh no good!");
$(this).css('background', 'red');
}
});
http://codepen.io/msafi/pen/dOKogK/
Beginning JavaScript learner here...
I'm working through Adrian Neumann's Simple Programming Problems and my question is about number 6 in the elementary exercises.
Write a program that asks the user for a number n and gives him the possibility to choose between computing the sum and computing the product of 1,…,n.
// var myArray = []; // for testing
var mySum = 0;
var userNum = prompt("What is your number? ");
var userChoice = prompt("Would you like to add up (+) or multiply (*) all the numbers from 1 to your number? Please enter (+) or (*): ");
if (userChoice == "+") {
for (var i = userNum; i > 0; i--) {
mySum += +i;
}
console.log("Your answer is " + mySum);
} else if (userChoice == "*") {
for (var i = userNum; i > 0; i--) {
mySum *= +i;
// myArray.push(i); // for testing
}
console.log("Your answer is " + mySum);
// console.log(myArray); // for testing
}
When entering values for multiplication, the answer is always 0. Obviously, I thought 0 was being included in the iteration, so I setup an empty array myArray, and pushed all the numbers to the array using myArray.push(i);... 0 was never included as a value in the array.
Other than some obvious form-validation considerations, can anyone tell me what I'm missing? Why is my answer always 0?
Note: The sum section of the code seems to work brilliantly.
Please note I'm a beginniner to JavaScript, so if you'd like to comment, let me know WHY you changed the code the way you do, rather than simply spitting back code to me. That's a big help, thanks.
Well, you initialize mySum to 0 so in every iteration of the loop you'll multiplying i by zero and save the result (again, zero) back into mySum. For multiplication you'd have to start at one.
You didn't set mySum to 1 before multiplication. 0*i = 0.