This question already has answers here:
Is there a difference between x++ and ++x in java?
(18 answers)
What is x after "x = x++"?
(18 answers)
Closed 4 years ago.
I want to get next character and I am writing the code like this
char c = 'A';
c = c++;
System.out.println(c);
The printed character is A. But if I use preincrement operator with 'c' then I get next character (B). Here is the code with preincrement operator.
char c = 'A';
c = ++c;
System.out.println(c);
Can someone explain the difference?
The increment operator doesn't make sense if you assign that result back to the variable. Doing
c = c++;
takes the return value of c++, which is 'A', and assigns that to c. Instead, simply do
c++; // or ++c
In your case, you probably want to do
System.out.println(++c); // prints 'B', and |c| is now 'B'
c = c++; means, first the current value will be used and then it will be incremented. Therefore it's first printing the current value
c = c++;
System.out.println(c);
translates to:
c = c; // since the actual value is returned first
System.out.println(c);
and hence, the value A get printed.
Whereas
c = ++c;
System.out.println(c);
translates to
c = c+1;
System.out.println(c);
Related
This question already has answers here:
Adding and subtracting chars, why does this work? [duplicate]
(4 answers)
Parentheses around data type?
(5 answers)
What does a 'int' in parenthesis mean when giving a value to an int?
(7 answers)
What is the purpose and meaning of (char)i or (int)i in java?
(3 answers)
Closed 2 years ago.
I encountered a similar piece of code:
public class print {
public static void main(String[] args) {
for (int i = 0; i < 6; i++) {
System.out.print((char) (i + 'a'));
}
}
}
If I run it, I get "abcdef".
My question regards this expression: (char) (i + 'a').
I kind of intuitively get what's going on, but I want a rigorous step-by-step explanation of how the computer translates it. As indicated in some answers, the char is simply a number displayed as a character. Fine, but what does this syntax with parentheses actually do? Is it a conversion? Can I use it for other types as well?
The ASCII value of a is 97.
When i = 0, i + 'a' = 0 + 97 => When cast into char, it will be a
When i = 1, i + 'a' = 1 + 97 => When cast into char, it will be b
When i = 2, i + 'a' = 2 + 97 => When cast into char, it will be c
...and so on
Java char is a 16-bit integral type. 'a' is the same as 97, which you can see with System.out.println((int) 'a'); - it follows that 98 is 'b' and so on across the entire ASCII table.
You can use char as an integer value and vise versa
Below code may help:
char aChar = 'a'; // a char
int aCharAscii = aChar; // 97
char bChar = 'a' + 1; // b char
int bCharAscii = aChar + 1; // 98
This question already has answers here:
Short hand assignment operator, +=, True Meaning?
(2 answers)
Difference between variable += value and variable = variable+value;
(4 answers)
Closed 3 years ago.
My code is as following:
char ch = t.charAt(t.length() - 1);
// result of XOR of two char is Integer.
for(int i = 0; i < s.length(); i++){
ch = ch^s.charAt(i);
ch = ch^t.charAt(i);
}
return ch;
it throws error
Line 6: error: incompatible types: possible lossy conversion from int to char
ch = ch^s.charAt(i);
Line 7: error: incompatible types: possible lossy conversion from int to char
ch = ch^t.charAt(i);
2 errors
However, When I change
ch = ch^s.charAt(i);
ch = ch^t.charAt(i);
to
ch ^= s.charAt(i);
ch ^= t.charAt(i);
Then, my code can work.
Are '^=' and '* = ^' different?? Why I search this question about '^=', it says they are same??
What is the '^=' operator?
From the functionality, they are the same: both perform an exclusive OR.
But the data types are different:
if I do x = x ^ y, the data type of x ^ y is always int or greater. When the result is assigned to something smaller, you have to cast.
if I do x ^= y, the data type doesn't grow as the assignment "knows its type".
See the language specification at https://docs.oracle.com/javase/specs/jls/se8/html/jls-15.html#jls-15.26.2 for more details.
You need to declare ch to be an int, or store the result of the bitwise xor ^= in an int field. Right now it's trying to store it in a char which is where the error is coming from.
I am trying to reform a string from ascii characters. I noticed when I break down ascii it matches the number with the ascii char, this changes when I tried to do (c+=c+0) on it as seen below, I tried to revert back to the original string but it doesn't seems to work. I was wondering does the ascii number change when you append it to a string?
StringBuilder b = new StringBuilder();
char c = "e".charAt(0);
System.out.println(c); //'e'
System.out.println(c+0); //101 -ascii
b.append(c+=c+0);
char d = b.charAt(0);
System.out.println(d-=d+0); //blank
When you use + to sum a char with and int, the result will be promoted
to int.
When you use compound assignment operators +=, the result will be converted to left
operand data type(in this case, char)
This is what happens with your code:
StringBuilder b = new StringBuilder();
char c = "e".charAt(0);
System.out.println(c); //'e'
System.out.println(c+0); //101 -ascii
b.append(c+=c+0); // result of c+c+0 is int 202, it is converted to char Ê
char d = b.charAt(0); // char d = Ê
System.out.println(d-=d+0); // result of d-(d+0) is int 0, it will be converted to null
It's not clear what you're trying to accomplish, but hopefully this example helps you figure out what's going on:
public static void main(String[] args) {
StringBuilder b = new StringBuilder();
char c = "e".charAt(0);
System.out.println(c); // e
System.out.println((int) c); // 101
char doubleC = (char) (c + c);
b.append(doubleC);
char d = b.charAt(0);
System.out.println((int) d); // 202
System.out.println(d - c); // 101
System.out.println((char) (d - c)); // e
}
Whether a character is printed as a letter or its ASCII value depends on its type. You can see that (int) c casts the character to its int ASCII value. Your approach of c+0 also does this but in a less intuitive way.
You also apparently want to "undo" your modification to c, but subtracting d-d always gives you 0. You need to store your value of c and subtract that from d.
+0s don't do anything because the character is underneath a number (so it's essentially like saying 101 + 0).
d-=d+0 evaluates to d = d - d + 0, so (because we're manipulating chars) d - d = 0
If you wanted to get c from d, you'd have to do d = d - c, because 'c' is the difference between the initial and the final value.
So no, the ASCII number of a character doesn't change, you're just not doing the reverse calculation as you should be.
It's like saying: x + y = z,
and then wondering why z - z != y.
You should be doing: z - x = y
#Edit
To add to your comment under pkpnd's answer - if you have only String b to work with, you'd still need to know the difference (i.e. by how much you increased/decreased the number) to reverse to the original chars.
This question already has answers here:
How do I shift array characters to the right in Java?
(2 answers)
Closed 5 years ago.
This is what I have right now:
class encoded{
public static void main(String[] args){
int shift = In.getInt();
String s1 = "bool";
char[] ch=s1.toCharArray();
for(int i=0;i<ch.length;i++){
char c = (char) (((ch[i] - 'a' - shift) % 26) + 'a');
System.out.print(c);
}
}
}
This code works to shift all characters in the string left by however much I want it to shift. The problem arises when, for example, I shift 'abc' 1 left, it returns '`ab'. What I want is for the characters to wrap back around to 'z' instead, so that the shifted 'abc' becomes 'zab' instead. How would I go about doing this? The characters need to always wrap back around to z when they're shifted left of 'a', and need to wrap back around to a when they're shifted right of 'z' (this would be done by changing "'a' - shift" to "'a' + shift".
Thanks so much in advance!
Not doing your homework for you, but giving you some hints:
char c = (char) (((ch[i] - 'a' - shift) % 26) + 'a');
You see, you are doing the shift operation unconditionally. Instead, read about using the if statement. You want to do your program different things, compared on the value of ch[i]. For some cases, just "shift", for others to replace values.
The problem is that the modulo operation can yield negative results.
In Java8 you can use Math.floorMod as an alternative:
class Main{
public static void main(String[] args){
int shift = 1;
String s1 = "abool";
char[] ch=s1.toCharArray();
for(int i=0;i<ch.length;i++){
char c = (char) ((Math.floorMod((ch[i] - 'a' - shift), 26)) + 'a');
System.out.print(c);
}
}
}
An alternative would be to use char c = (char) ( (((ch[i] - 'a' - shift) % 26) + 26) % 26 + 'a');
char c = '0';
int i = 0;
System.out.println(c == i);
Why does this always returns false?
Although this question is very unclear, I am pretty sure the poster wants to know why this prints false:
char c = '0';
int i = 0;
System.out.println(c == i);
The answer is because every printable character is assigned a unique code number, and that's the value that a char has when treated as an int. The code number for the character 0 is decimal 48, and obviously 48 is not equal to 0.
Why aren't the character codes for the digits equal to the digits themselves? Mostly because the first few codes, especially 0, are too special to be used for such a mundane purpose.
The char c = '0' has the ascii code 48. This number is compared to s, not '0'. If you want to compare c with s you can either do:
if(c == s) // compare ascii code of c with s
This will be true if c = '0' and s = 48.
or
if(c == s + '0') // compare the digit represented by c
// with the digit represented by s
This will be true if c = '0' and s = 0.
The char and int value can not we directly compare we need to apply casting. So need to casting char to string and after string will pars into integer
char c='0';
int i=0;
Answer is like
String c = String.valueOf(c);
System.out.println(Integer.parseInt(c) == i)
It will return true;
Hope it will help you
Thanks
You're saying that s is an Integer and c (from what I see) is a Char.. so there you, that's the problem: Integer vs. Char comparation.