I have a piece of code like this...
char c = 'a';
When I ask for Character.getNumericValue(c), it gives 10 as the output.
How can I swap this problem around so that 10 is the input and a is the output?
Do realize that the 10 in your example is not the ASCII code but the value of a as a hex digit (or rather, digit in any base greater than 10). To reverse that:
char c = Character.forDigit(10, 16);
Which you could have found by looking at the "see also" section in the API doc.
char c = 'a';
int i = 10;
System.out.println("Character c = [" + c + "], numeric valule = [" + (int)c + "]");
System.out.println("int i = [" + i + "], character valule = [" + (char)i + "]");
you can try:
int i = 97;
char c = (char) i; //should yield 'a';
System.out.println( "Integer " + i + " = Character " + c );
//outputs: "Integer 10 = Character a"
Related
I know my code can be simpler and more efficient... My code is supposed to grab the biggest set of 5 digits. It works, except it only is grabbing 3 digits, what would i need to modify to change that?
public class thousandDigits {
public static void main(String[] args) {
int greatest = 0;
String num = ("73167176531330624919225119674426574742355349194934"
+ "96983520312774506326239578318016984801869478851843"
+ "85861560789112949495459501737958331952853208805511"
+ "12540698747158523863050715693290963295227443043557"
+ "66896648950445244523161731856403098711121722383113"
+ "62229893423380308135336276614282806444486645238749"
+ "30358907296290491560440772390713810515859307960866"
+ "70172427121883998797908792274921901699720888093776"
+ "65727333001053367881220235421809751254540594752243"
+ "52584907711670556013604839586446706324415722155397"
+ "53697817977846174064955149290862569321978468622482"
+ "83972241375657056057490261407972968652414535100474"
+ "82166370484403199890008895243450658541227588666881"
+ "16427171479924442928230863465674813919123162824586"
+ "17866458359124566529476545682848912883142607690042"
+ "24219022671055626321111109370544217506941658960408"
+ "07198403850962455444362981230987879927244284909188"
+ "84580156166097919133875499200524063689912560717606"
+ "05886116467109405077541002256983155200055935729725"
+ "71636269561882670428252483600823257530420752963450");
for (int n = 0; n < num.length() - 5; n++) {
greatest = ((num.charAt(n)) + (num.charAt(n+1)) + (num.charAt(n+2)) + (num.charAt(n+3))
+ (num.charAt(n+4)));
if (greatest > n) {
n = greatest;
}
}
System.out.print(greatest);
}
}
OUTPUT:
357
I think you want to use String.substring(int, int) to iterate all possible 5 character substrings, and then you might use Math.max(int, int) to update greatest. Something like
int greatest = Integer.MIN_VALUE;
for (int i = 0; i < num.length() - 4; i++) {
// int value = Integer.parseInt(num.substring(i, i + 5));
int value = Integer.parseInt(String.valueOf(num.charAt(i))
+ num.charAt(1 + i) + num.charAt(2 + i) + num.charAt(3 + i)
+ num.charAt(4 + i));
greatest = Math.max(greatest, value);
}
System.out.println(greatest);
I get 99890.
I think you are trying to add 5 consecutive characters to get sum, and store starting index of highest sum.
But you should be using Character.getNumricValue(char) to convert (num.charAt(n)) to numeric value and then add.
greatest = Character.getNumericValue((num.charAt(n)) + Character.getNumericValue((num.charAt(n+1)) + Character.getNumericValue((num.charAt(n+2)) +
Character.getNumericValue((num.charAt(n+3)) +
Character.getNumericValue((num.charAt(n+4));
You need a valirable to store old value to compare and index
if(greatest > oldGreatest) {
index = n;
}
Then finally print using index out side loop:
System.out.print((num.charAt(index)) + (num.charAt(index+1) + (num.charAt(index +2)) + (num.charAt(index +3)) + (num.charAt(index +)));
Although #ElliottFrisch and #dave provides more elegant answer, I tried to modify from your original version and here is my code (I have tested it):
public class ThousandDigits {
public static void main(String[] args) {
int greatest = 0;
String num = ("73167176531330624919225119674426574742355349194934"
+ "96983520312774506326239578318016984801869478851843"
+ "85861560789112949495459501737958331952853208805511"
+ "12540698747158523863050715693290963295227443043557"
+ "66896648950445244523161731856403098711121722383113"
+ "62229893423380308135336276614282806444486645238749"
+ "30358907296290491560440772390713810515859307960866"
+ "70172427121883998797908792274921901699720888093776"
+ "65727333001053367881220235421809751254540594752243"
+ "52584907711670556013604839586446706324415722155397"
+ "53697817977846174064955149290862569321978468622482"
+ "83972241375657056057490261407972968652414535100474"
+ "82166370484403199890008895243450658541227588666881"
+ "16427171479924442928230863465674813919123162824586"
+ "17866458359124566529476545682848912883142607690042"
+ "24219022671055626321111109370544217506941658960408"
+ "07198403850962455444362981230987879927244284909188"
+ "84580156166097919133875499200524063689912560717606"
+ "05886116467109405077541002256983155200055935729725"
+ "71636269561882670428252483600823257530420752963450");
int max = -1;
for (int n = 0; n < num.length() - 4; n++) {
greatest = ((num.charAt(n) - '0') * 10000 + (num.charAt(n + 1) - '0') * 1000
+ (num.charAt(n + 2) - '0') * 100 + (num.charAt(n + 3) - '0') * 10 + (num.charAt(n + 4) - '0'));
if (max < greatest) {
max = greatest;
}
}
System.out.print(max);
}
}
I think you'll find it's not grabbing three digits, but rather the sum of the six characters you are pulling out is a 3-digit number.
If you're after the largest five digit number, you need to extract five digits (not six) as you do and assign them a weight. So the first digit must be multiplied by 10,000, the second by 1,000 and so on.
But there's more: you're are getting the character at an index within your string. This is not what you want as it is not the same as the numeric value of that character. For that you need:
num.charAt(n) - '0'
These changes should allow you to correct your algorithm as it stands.
A more efficient approach would be to extract 5-digit sub-strings and convert them to integers. The first one would be:
Integer.parseInt(num.subString(0, 5));
You can iterate to get each one to find the greatest.
I can't figure out how to perform this for both lowercase and uppercase. The Ascii value that I'm changing is being shifted by a randomly generated integer between 3 and 7. The purpose of the code is to produce a random credit card number.
char character3 = lastName.charAt(0);
int ascii3 = (int) character3;
char character4 = lastName.charAt(firstName.length()-1);
int ascii4 = (int) character4;
//testing output
System.out.println(ascii1 + " " + ascii2 + " " + ascii3 + " " + ascii4 + " ");
int Ascii1 = ascii1 + finalRandom;
int Ascii2 = ascii2 + finalRandom;
int Ascii3 = ascii3 + finalRandom;
int Ascii4 = ascii4 + finalRandom;
//Wrapping value back to "a" if it's Ascii value exceeds "z"
char Wrap1 = (char) Ascii1;
char translated = (char)('A' + (Wrap1 -'A') % ('Z' - 'A' + 1));
char Wrap2 = (char) Ascii2;
char translated2 = (char)('A' + (Wrap2 - 'A') % ('Z' - 'A' + 1));
The cleanest way is to use if statements, but this should also work
ch = ( char ) ( ( ch + SHIFT - 'A' ) % 26 + 'A' );
The above only works on uppercase characters and fails on lowercase characters. Here is a solution that works on both upper and lowercase characters, completely free of all if/while/for statements.
public static char shift( char c, int shift ) {
int l = c - 'a';
int u = c - 'A';
// The '>> 31' operator fills all bits with bit 31 of l
int mask = l >> 31;
l = ( l + shift ) % 26;
u = ( u + shift ) % 26;
return ( char ) ( ( ( l + 'a' ) & ~mask ) + ( ( u + 'A' ) & mask ) );
}
The above solution works by calculating the wrapping shift for both lowercase and uppercase characters and masks out the invalid one. mask is -1 if it is a uppercase character or 0 if it is a lowercase character. This is generated by getting the sign bit after subtracting 'a' from the desired character which is negative if it is an uppercase character since 'a' > 'Z'.
System.out.println(7 + 5 + " ");
This prints out 12, but in another order
System.out.println(" " + 5 + 7);
it prints out 57. Why is this?
Firstly, this has nothing to do with System.out.println. You'll see exactly the same effect if you use:
String x = 7 + 5 + "";
String y = " " + 5 + 7;
It's got everything to do with associativity. The + operator is left-associative, so the above two statements are equivalent to:
String x = (7 + 5) + "";
String y = (" " + 5) + 7;
Now look at the results of the first expression in each case: 7 + 5 is just 12, as int... whereas " " + 5 is "5" (a string).
Or to break it down further:
int x1 = 7 + 5; // 12 (integer addition)
String x = x1 + ""; // "12" (conversion to string, then string concatenation)
String y1 = " " + 5; // "5" (conversion to string, then string concatenation)
String y = y1 + 7; // "57" (conversion to string, then string concatenation)
Justification: JLS 15.18 (additive operators):
The additive operators have the same precedence and are syntactically left-associative (they group left-to-right).
Easy. System.out.println(7 + 5 + " ") is viewed as a mathematical equation, whereas System.out.println(" " + 5 + 7) whereas having the space beforehand, Java (I'm assuming) views it as a string. Thus 'concatenating' the two.
String secondLine = ...E......E..E.E;
String failures = "E";
String passed = ".";
int i = 0;
while ((i = (secondLine.indexOf(failures, i) + 1)) > 0) {
System.out.println(i);
feedbackString += "<strong style='color: red;'><li>Failed: </strong><strong>" + i + "</strong> - " + "out of " + resultString.length() + " tests.<br>";
}
The total number of the tests is the sum of the dots which is = 12. and the E's are the failures which in this case = 4.
The dots are the passes and the E's are the failures. Whenever this is ran and there is a failure, it adds an 'E' in front of the dot which becomes'.E'. I can get the E's on its own but I want a statement that says given a dot comes before the E and then it should print the '.E' and pass it into a variable called failedTest.
The code above has an output of: 4, 11, 14, 16 which is not what I want as its considering each character separately but I want it to consider '.E' as one that is if an E comes after a dot then it should consider it as one. if this should be ran considering the '.E' as failure, the expected output should be 3, 9, 11, 12. Thanks :)
I understand your problem better now: if i am right, you print a ., ran the test and if it do not pass, you will print an E.
The point if the final length of secondLine is the number of test plus the number of error, so the if the first error is the test T-i, the T-i dot would be at the position T-i, its E would be at the position (T-1) + 1. The following test would be displaced one position: the dot of the test T-(i+1) would be at the position (T-i) + 2. Do you see the point?
So, this is my suggestion
String secondLine = "...E......E..E.E";
String failures = "E";
String passed = ".";
int detectedErrors = 0;
int i = 0;
while ((i = (secondLine.indexOf(failures, i) + 1)) > 0) {
System.out.println(i);
feedbackString += "<strong style='color: red;'><li>Failed: </strong><strong>"
+ (i - detectedErrors) // here
+ "</strong> - " + "out of " + resultString.length() + " tests.<br>";
detectedErrors += 1; // and here
}
You need to keep track of the number of extra characters in the secondLine and subtract them from the match position. Also change your failure String to ".E"
String secondLine = "...E......E..E.E";
String failures = ".E";
String passed = ".";
int n = 0;
int i = 0;
while ((i = (secondLine.indexOf(failures, i) + 1)) > 0) {
System.out.println(i-n);
n++;
feedbackString += "<strong style='color: red;'><li>Failed: </strong><strong>" + (i-n) + "</strong> - " + "out of " + resultString.length() + " tests.<br>";
I am trying to convert an integer to a binary number using an integer array. The first conversion is toBinaryString where I get the proper conversion "11111111" then the next step to convert to an array. This is where it goes wrong and I think its the getChar line.
int x = 255;
string=(Integer.toBinaryString(x));
int[] array = new int[string.length()];
for (int i=0; i< string.length(); i++){
array[i] = string.getChar(i);
Log.d("TAG", " Data " + array[1] "," + array[2] + "," + array[3]);
Log displays ( Data 0,0,0 ) the results I am looking for is ( Data 1,1,1 )
Here is the final code and it works.
// NEW
int x = 128;
string=(Integer.toBinaryString(x));
int[] array = new int[string.length()];
for (int i=0; i < string.length(); i++) {
array[i] = Integer.parseInt(string.substring(i,i+1));
}
Log.d("TAG", "Data " + array[0] + "" + array[1]+ "" + array[2] + "" + array[3]+ " " + array[4]+ "" + array[5] + "" + array[6] + "" + array[7]);
// Take your input integer
int x = 255;
// make an array of integers the size of Integers (in bits)
int[] digits = new Integer[Integer.SIZE];
// Iterate SIZE times through that array
for (int j = 0; j < Integer.SIZE; ++j) {
// mask of the lowest bit and assign it to the next-to-last
// Don't forget to subtract one to get indicies 0..(SIZE-1)
digits[Integer.SIZE-j-1] = x & 0x1;
// Shift off that bit moving the next bit into place
x >>= 1;
}
how about :
array[i] = Integer.parseInt(string.substring(i,i+1));
First of all, an array starts from the index of 0 instead of 1, so change the following line of code:
Log.d("TAG", " Data " + array[1] "," + array[2] + "," + array[3]);
To:
Log.d("TAG", " Data " + array[0] "," + array[1] + "," + array[2]);
Next, the following line of code:
array[i] = string.getChar(i);
You are trying to get a character value to an Integer array, you may want to try the following instead and parse the value into an Integer (also, the "getChar" function does not exist):
array[i] = Integer.parseInt(String.valueOf(string.charAt(i)));
I hope I've helped.
P.S - Thanks to Hyrum Hammon for pointing out about the String.valueOf.