this is part of code in javascript for a Tic Tac Toe game. this part is checking win for diagonal in both directions. The diagonals work in X's winning but they do not work in O's winning. I basically wrote the same thing so I'm not sure what's the mistake I made.
Did I make some syntactic mistake?
function CheckWin_Diagonal(rowClicked, colClicked) {
if (rowClicked == colClicked) {
for (c = 0; c < rows; c++) {
if (intBoard[c][c] != X_CLASS)
break;
if (c == rows - 1) {
alert("PLAYER X WIN!!");
document.location.reload()
}
}
}
else {
if (rowClicked == colClicked) {
for (c = 0; c < rows; c++) {
if (intBoard[c][c] != CIRCLE_CLASS)
break;
if (c == rows - 1) {
alert("PLAYER O WIN!!");
}
}
}
}
}
function CheckWin_Diagonal2(rowClicked, colClicked) {
if (rowClicked + colClicked == cols - 1) {
for (r = 0; r < cols; r++) {
if (intBoard[r][(cols - 1) - r] != X_CLASS)
break;
if (r == cols - 1) {
alert("PLAYER X WIN!!");
document.location.reload()
}
}
}
else {
if (rowClicked + colClicked == cols - 1) {
for (r = 0; r < cols; r++) {
if (intBoard[r][(cols - 1) - r] != CIRCLE_CLASS)
break;
if (r == cols - 1) {
alert("PLAYER O WIN!!");
}
}
}
}
}
I haven't seen the logic at all, but you are doing:
function CheckWin_Diagonal(rowClicked, colClicked) {
if (rowClicked == colClicked) {
/** logic **/
}
else {
if (rowClicked == colClicked) {
/** logic **/
}
}
}
Obviously the code in the second if will never execute because it's the same condition as the first one (so it'll never get there).
You are basically saying:
IF SOMETHING HAPPENED DO THIS, OTHERWISE IF THE SAME THING HAPPENED DO THAT
So THAT will never happen, because you trapped the condition in the first IF and there's no OTHERWISE for the exact same condition
Your else blocks run the same if condition that triggered the else in the first place, guaranteeing they'll never run.
Related
Had to walk away from this one for the weekend. Would love to know my nemesis is giving me the answer plus some extra stuff? i want it to display 0 - 10
for (let i=-1; i++< 100; i*5) {
if (i < 0) {
continue;
} else if ( i > 50) {
break;
} else {
console.log(i/5);
}
}
Languages like Python will drop the remainder when using the / operator.
JavaScript won't, but you can get the same result with Math.floor().
for (let i = -1; i++ < 100; i * 5) {
if (i < 0) {
continue;
} else if (i > 50) {
break;
} else {
console.log(Math.floor(i / 5));
}
}
I looked for a function to determine if a number is prime and found this
for (var i = 2; i <= Math.sqrt(num); i++)
if (num % i === 0) {
return false;
}
return true;
and I don't understand why that works, yet this doesn't
for (var i = 2; i <= Math.sqrt(num); i++) {
if (num % i === 0) {
return false;
}
return true;
}
What is it about the (lack of the) block statement that is functioning differently
Your first code looks like this:
for (var i = 2; i <= Math.sqrt(num); i++){
if (num % i === 0) {
return false;
}
}
return true;
Notice how return true is on the outside. Your second code doesn't work because it prematurely returns true when there are more numbers to check. Your entire for loop is equivalent to
return num % 2 !== 0;
which is clearly incorrect.
Let me tell you something about blocks that you might not have known (It took me a while to discover this at least).
When you use a loop or if-else statement, you can ignore the using braces { and }.
Example 1
if (a === b){
c = 0;
}
is actually the same as
if (a === b)
c = 0;
Example 2
for (i = 0; i < 10; i++){
a += 1;
}
is actually the same as
for (i = 0; i < 10; i++)
a += 1;
However 1
if (a === b){
c = 0;
d = 1;
}
is not the same with
if (a === b)
c = 0;
d = 1;
However 2
for (i = 0; i < 10; i++){
a += 1;
b += 1;
}
is not the same with
for (i = 0; i < 10; i++)
a += 1;
b += 1;
Explanation
In loops and if-else statement, the block statement (codes surrounded by { and } groups the codes within it and execute it.
However, in the absence of { and }, the loops or if-else statement will only execute the single line of code after it.
Meaning,
var a = 0,
b = 0;
for (i = 0; i < 10; i++)
a += 1;
b += 1;
In this case, a === 10 but b === 1.
Also, in the following case,
var a = 0,
b = 0;
if (false)
a = 10;
b = 10;
a === 0 but b === 10.
This question already has answers here:
How does (A == B == C) comparison work in JavaScript?
(6 answers)
Closed 6 years ago.
today I decided I wanted to make a simple js code that would accept a number (in meters), and decide what the appropriate metric unit to use would be. The code turned out to be a little more complicated than I had expected, but I was able to figure out most of the bugs as I found them (even if it meant rearranging all of my code). However, when it came to my if/else statement I could not figure it out. If I put in a number that is less than 1 nothing happens. If I put in a number more than 9 it logs the same thing every time. The structure itself may need some work, but if someone could help me with the if/else statement I would be very thankful. Here is the code (init is called when the body loads):
function init() {
var x = prompt("How many meters?");
convertMetricMeters(x);
function convertDown(x) {
if (0.1 >= x >= 0.99) {
console.log("deci");
}
else if (0.100 >= x >= 0.999) {
console.log("centi");
}
else if (0.1000 >= x) {
console.log("milli");
}
else {
console.log("error");
}
}
function convertUp(x) {
if (1 <= x <= 99) {
console.log("deca");
}
else if (100 <= x <= 999) {
console.log("hecto");
}
else if (1000 <= x) {
console.log("kilo");
}
else {
console.log("error");
}
}
function convertMetricMeters(x) {
if (x < 1) {
convertDown(x);
}
else if (x > 9) {
convertUp(x);
}
else {
console.log("Appropriate Metric Unit");
}
}
}
Use && as AND operator in javascript
Convert these 100 <= x <= 999 to 100 <= x && x <= 999
You could simplify the check a bit and return if a condition is true.
function convertDown(x) {
if (x < 0.01) {
console.log("milli");
return;
}
if (x < 0.1) {
console.log("centi");
return;
}
if (x < 1) {
console.log("deci");
return;
}
console.log("error");
}
Your code has 2 sort of errors. One was simple to fix, that you have to add && between two conditions in if statement.
Now coming to the other part, the less than 1 items. It needed a different logic. Well, your maths seems to be needing bit attention. 0.1 is same as 0.100 and is same as 0.1000
I have updated the code to look for the number of digits after the decimal point and then console.log accordingly.
The updated code will be:
function init() {
var x = prompt("How many meters?");
convertMetricMeters(x);
function convertDown(x) {
// checks the number of digits after decimal point
decimals = (x.split('.')[1] || []).length
if (decimals == 1 || decimals == 2) {
console.log("deci");
}
else if (decimals == 3) {
console.log("centi");
}
else if (decimals == 4) {
console.log("milli");
}
else {
console.log("error");
}
}
function convertUp(x) {
if (1 <= x && x <= 99) {
console.log("deca");
}
else if (100 <= x && x <= 999) {
console.log("hecto");
}
else if (1000 <= x) {
console.log("kilo");
}
else {
console.log("error");
}
}
function convertMetricMeters(x) {
if (x < 1) {
convertDown(x);
}
else if (x > 9) {
convertUp(x);
}
else {
console.log("Appropriate Metric Unit");
}
}
}
Working jsfiddle example: https://jsfiddle.net/w7pf3moL/
A simplified version with only 1 method convert(float x):
function init() {
var x = prompt("How many meters?");
convertMetricMeters(x);
function convert(x) {
if (x < 0.01) console.log("milli");
else if (x < 0.1) console.log("centi");
else if (x < 1) console.log("deci");
else if (x < 10) console.log("meter");
else if (x < 100) console.log("deca");
else if (x < 1000) console.log("hecto");
else console.log("kilo");
}
function convertMetricMeters(x) {
if (x > 0) {
convert(x);
} else {
console.log("Appropriate Metric Unit");
}
}
}
init();
When will else condition executed ?? is there any possibility of i value which will not satisfy any if , else if condition and reach else block in javascript.
random = Math.random();
i=0; //0,0.1,0.001
console.log(random);
if(random < i) {
console.log("<");
} else if(random > i) {
console.log(">");
} else if(random == i) {
console.log("=");
}else{
console.log("nothing");
}
Else will never be reached. The Result of Math.random() is between 0 and 1 (see W3Schools)
Doing some practice runs on codecademy and came across the following problem: So the objective is to print "Fizz" if the numbers are divisible by 3. "Buzz" if the numbers are divisible by 5. And "FizzBuzz" if the numbers are divisible by both 3 and 5.
Here is my code, and I thought I had it right, but when I run it they tell me that my code is not 100% accurate. Looking to see any alternatives to this code, or what might be the issue...
Code:
for ( i = 0; i < 21; i++)
{
if (i % 3 == 0 )
{
console.log("Fizz");
}
if (i % 5 == 0)
{
console.log ("Buzz");
}
if ( i % 5 == 0 && i % 3 === 0)
{
console.log("FizzBuzz");
}
else
{
console.log(i);
}
}
You need to use else if to stop from other conditions executing:
for (var i = 1; i <= 20; i++) {
if (i % 5 === 0 && i % 3 === 0) {
console.log("FizzBuzz");
} else if (i % 3 === 0) {
console.log("Fizz");
} else if (i % 5 === 0) {
console.log("Buzz");
} else {
console.log(i);
}
}
It can easily be done in a one liner.
for (i = 1; i <= 20; i++) {
console.log(i%3?(i%5?i:'buzz'):(i%5?'fizz':'fizzbuzz'));
};
For a nice formatting also output i on each iteration:
for (i = 1; i < 21; ++i) {
console.log(i+": "+(i%3?(i%5?i:'buzz'):(i%5?'fizz':'fizzbuzz')));
};