## How to increment bits correctly in an array?

If I have a binary notation such as "1000010" which equals 66 and I want to increment it by one to "1000011" which equals 67. How is that done correctly in my array? Currently it's printing out "0100010" which is 34, but no where near the correct answer. I don't think my array is shifting correctly, nor will it increase size as the numbers get larger. Although, I can't make any assumptions about how big the array can be other than what's explicitly stated.

```public class math {

//=================================================================
// increment(A) returns an array of bits representing A+1.
//=================================================================

public static byte[] increment(byte[] A)
{
byte carry= 1;
for(int i = 0; i<A.length; i++){
byte b = A[i];
A [i] ^= carry;
carry &= b;
}
return A;
}

private static String toBinString (byte [] a)
{
String res = "";
for (int i = 0; i <a. length; i++)
{
res = (a [i] == 0 ? "0": "1") + res;
}
return res;
}

/**
* @param args
*/
public static void main(String[] args) {
byte [] A ={1,0,0,0,0,1,0};

increment(A);
System.out.println (toBinString (A));

}
}
```

This worked for me:

```public static void main(String[] args) {
byte [] A ={1,0,0,0,0,1,0};
increment(A);
System.out.println (toBinString (A));
}

public static byte[] increment(byte[] A) {
boolean carry = true;
for (int i = (A.length - 1); i >= 0; i--) {
if (carry) {
if (A[i] == 0) {
A[i] = 1;
carry = false;
}
else {
A[i] = 0;
carry = true;
}
}
}

return A;
}

private static String toBinString (byte [] a) {
String res = "";
for (int i = 0; i < a. length; i++) {
res += (a [i] == 0 ? "0": "1") ;
}
return res;
}
```

This worked for me: public static void main(String[] args) { byte [] A ={1,0,0,0,0,1,0​}; increment(A); System.out.println (toBinString (A)); } public static byte[]  In your declaration of A it looks like you want the leftmost (first) array be the most significant bit, in the rest of your program you consider the first element of the array as the least significant bit. Easiest solution would probably be to enter your number in reverse order in the array, or inverse the array – fvu Sep 7 '11 at 0:27

The lazy (and secure) way for increment by one :

```    String s = "1000010";
for (int i = 0; i < 5; i++) {
System.out.print(s);
System.out.println("\t" + Integer.valueOf(s, 2));
s = Integer.toBinaryString(Integer.valueOf(s, 2) + 1);
}
```

Output :

```1000010 66
1000011 67
1000100 68
1000101 69
1000110 70
```

(Edited for presentation)

Increment a number by one by manipulating the bits. Get the position of rightmost unset bit of n. Let this position be k. Set the k-th bit of n. Toggle the last k-1 bits of n. Finally, return n. Since you can't increment the offset (in this case, the 0), increment your base address by 4 on each loop (the number of bytes in a word) until you reach the end of the array. It looks like you're already incrementing your address by 4 each time, and your end condition looks correct, so just use the syntax above and it should work.

```//Function call
incrementCtr(ctr, ctr.length - 1);

//Increments the last byte of the array
private static void incrementCtr(byte[] ctr, int index) {

ctr[index]++;

//If byte = 0, it means I have a carry so I'll call
//function again with previous index
if(ctr[index] == 0) {
if(index != 0)
incrementCtr(ctr, index - 1);
else
return;
}
}
```

each pair of ciphertext it is necessary to guess the following key bits (and For detecting the correct key we maintain an array of counters of size 227 for each D1 and D2. A counter is incremented when it is correctly verified using a partially  Creating an array of given size and increment. Learn more about array creation, linspace, colon operator MATLAB

Late but concise:

```public static void increment(byte[] a) {
for (int i = a.length - 1; i >= 0; --i) {
if (++a[i] != 0) {
return a;
}
}
throw new IllegalStateException("Counter overflow");
}
```

an increment to the address such that when a carry occurs it skips over bits not bit to receive carry We have correctly incremented the address for dimension We can continue to do this and step through dimension number two of the array. To access the data in the array requires that we know the address of the data and then use the load word (lw) or store word (sw) instructions. Words (which is how integers are stored) in MIPS take up 32 bits or 4 bytes. Therefore, if we have a declaration such as: list: .word 3, 0, 1, 2, 6, -2, 4, 7, 3, 7

```public static byte[] increment(byte[] array) {
byte[] r = array.clone();
for ( int i = array.length - 1; i >= 0; i-- ) {
byte x = array[ i ];
if ( x == -1 )
continue;
r[ i ] = (byte) (x + 1);
Arrays.fill( r, i + 1, array.length, (byte) 0 );
return r;
}
throw new IllegalArgumentException( Arrays.toString( array ) );
}
```

exception if overflow

Incrementing a binary counter As another example of aggregate analysis, We use an array A[0 . . k — 1] of bits, where length[A] = k, as the counter. As with the stack example, a cursory analysis yields a bound that is correct but not tight. I think your while loop also.You want to compare the one element with the rest with in a string. Store the element in a temporary srting, iterate through the entire string comparing it.

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Declaring Array Variables; Creating Array Objects; Accessing Array Elements; Changing Array (loops and conditionals) to execute different bits of a program based on tests. (Actually, to be more technically correct, it throws an exception. Commonly, the increment is used to change the value of the loop index to bring​  It's because array is treated as a constant pointer in the function it is declared. There is a reason for it. Array variable is supposed to point to the first element of the array or first memory instance of the block of the contiguous memory loca