for example, if I enter 12, I want to get 81 41 as the set bits in 12 are 1100
This is what I have for now, I do not think I am implementing the for loop correctly
#include <stdio.h>
void bin(unsigned n)
{
char list[6];
int x = 0, y = 1;
/* step 1 */
if (n > 1)
bin(n / 2);
/* step 2 */
list[x] = n % 2;
x++;
/*for(int i = 0; i < x; i++) {
printf("%d\n",list[i]);
}*/
for(int i = 0; i < 5; i++) {
if(list[i] == 1 && i == 5) {
printf("32%i",y);
}
if(list[i] == 1 && i == 4) {
printf("16%i",y);
}
if(list[i] == 1 && i == 3) {
printf("8%i",y);
}
if(list[i] == 1 && i == 2) {
printf("4%i",y);
}
if(list[i] == 1 && i == 1) {
printf("2%i",y);
}
if(list[i] == 1 && i == 0) {
printf("1%i",y);
}
}
}
I checked that I was correctly storing the bytes in the array, and it outputted correctly, but when I try to look for them one at a time in a loop, it seems to get stuck on the 32 bit integer, so for 12, it would print 321 321
This program has Undefined Behaviour from accessing uninitialized values of list. I'm going to refactor this code so its easier to talk about, but know this refactored code is still incorrect.
x is always 0. y is always 1. x++ has no effect. This function can be rewritten as:
void bin(unsigned n)
{
char list[6];
if (n > 1)
bin(n / 2);
list[0] = n % 2;
for (int i = 0; i < 5; i++) {
if (list[i] == 1) {
switch (i) {
case 5: printf("321"); break;
case 4: printf("161"); break;
case 3: printf("81"); break;
case 2: printf("41"); break;
case 1: printf("21"); break;
case 0: printf("11"); break;
}
}
}
}
There are some problems here.
Firstly, list is not shared between calls to bin, nor are any other variables.
In every call to bin, only list[0] is assigned a value - all others indices contain uninitialized values. You are (un)lucky in that these values are seemingly never 1.
With your example of 12 as the starting value:
When you initially call bin(12), what happens is:
bin(12) calls bin(6), bin(6) calls bin(3), bin(3) calls bin(1).
Starting from the end and working backwards, in bin(1):
n = 1, so list[0] = n % 2; assigns 1. The loop checks each element of list for the value 1, finds it when the index (i) equals 0, and prints 11.
This is repeated in bin(3), as 3 % 2 is also 1, and again this result is assigned to the first element of list. Again, we print 11.
In bin(6), 6 % 2 is 0. The loop finds no elements of list that equal 1. Nothing is printed.
And again, this is repeated in bin(12), as 12 % 2 is 0. Nothing is printed.
To reiterate, it is pure luck that this program appears to work. Accessing list[1] through list[4] (i < 5 ensures you never access the last element) in each function call is Undefined Behaviour. It is generally not worth reasoning about a program once UB has been invoked.
When dealing with bits, it would be a good time to use some bitwise operators.
Here is a program that more-or-less does what you have described.
It assumes 32-bit unsigned (consider using fixed width types from <stdint.h> to be more precise).
This program works by repeatedly shifting the bits of our initial value to the right b number of places and testing if the rightmost bit is set.
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char **argv)
{
unsigned num = argc > 1 ? atoi(argv[1]) : 42;
unsigned b = 32;
while (b--)
if ((num >> b) & 1)
printf("%u1 ", 1 << b);
putchar('\n');
}
$ ./a.out 12
81 41
I am currently implementing a code that runs Fibonacci sequence (up to the 60th number) in riscv32I, which means that the memory address that I can use only has 32bits.
I have first implemented the code in C, and then in Assembly, but I am curious if the algorithm I used has a name so I can do more research. The code is as such,
#include <stdint.h>
#include <stdio.h>
#include <inttypes.h>
int main() {
uint32_t n1 = 0; // first pre number (n - 2)
uint32_t n2 = 1; // second pre number (n - 1)
uint32_t add = 0; // current number
uint32_t store_hi = 0;
uint32_t store_lo = 0;
uint32_t result; // result
uint32_t carry; // carry bit
for (int i = 2; i < 61; i++) {
carry = 0; // reset carry bit
add = (uint32_t)(n2 + n1); // calculate current fib number
if (add < n1) { // if overflow
carry = 1; // set carry bit
}
result = store_hi + store_lo; // keeping track of higher bits
result = result + carry; // add carry bit
store_lo = store_hi; //
n1 = n2; // update first pre number
store_hi = result; //
n2 = add; // update second pre number
}
printf("Result32: 0x%08" PRIx32 " 0x%08" PRIx32 "\n", result, add);
uint64_t result64 = ((uint64_t)result << 32) | add;
printf("Result64: 0x%016" PRIx64 " -> %" PRId64 "\n", result64, result64);
}
Running the code gives
Result32: 0x00000168 0x6c8312d0
Result64: 0x000001686c8312d0 -> 1548008755920
The basic concept is that because the Fibonacci number gets too big to fit within a single 32bit memory address, we have to split it into 32bit memory address, one holding the upper bit, and one holding the lower bit.
Let's generalize the above algorithm to a 4 bit memory space, to make it easier to follow the algorithm. This means that the maximum int can be 16. Let ss set n1 = 10, n2 = 10.
Loop 1:
add = 4 # (10 + 10 = 20, but overflow, so 20 % 16 = 4)
carry = 1
result = 1
store_lo = 0
store_hi = 1
n1 = 10
n2 = 4
# output: 0x14, 0x1 hi bit, 0x4 lo bit, which is 10 + 10 = 20
Loop 2:
add = 14
carry = 0
result = 1
store_lo = 1
store_hi = 1
n1 = 4
n2 = 14
# output: 0x1e, 0x1 hi bit, 0xe or 14, lo bit, which is 10 + 20 = 30
loop 3:
add = 2 (14 + 4 = 18, but overflow, so 18 % 16, 2)
carry = 1
result = 3
store_lo = 1
store_hi = 2
n1 = 14
n2 = 2
#output: 0x32, 0x3 hi bit, 0x2 low bit, which is 20 + 30 = 50
.... and so on.
This should work for any base, but I am curious what this algorithm is denoted as, or if it is simply related to modules and powers?
Thanks!
It's called Arbitrary-precision arithmetic, you can read more about it here.
Arbitrary-precision arithmetic, also called bignum arithmetic, multiple-precision arithmetic, or sometimes infinite-precision arithmetic, indicates that calculations are performed on numbers whose digits of precision are limited only by the available memory of the host system.
One of the troubles of venturing from 32-bit to 64-bit is that the now distant horizon rapidly becomes as constraining as the former, "closer" horizon was.
Below is a sketch of an "ASCII digit" Fibonacci calculator. Its two seeds are "0" and "1" at the right end of two 20 character buffers. (20 is arbitrary, but takes one beyond the 60th value in the Fibonacci sequence.) This could be optimised and improved in a hundred different ways. It is a "powers of ten" version that uses two ASCII strings for its storage. One could, if one was patient, use 1000 character buffer, or 10,000 to go deep into the Fibonacci realm...
I hope you find this interesting.
EDIT: #Chux has pointed out that the sequence generated below indexes from 1, whereas indexing from 0 is correct. The simple fix (not shown here) would be to change three instances of ++fn to fn++ (an exercise left to the reader.) Thank you again, #Chux!
#include <stdio.h>
int main() {
char f[2][20 + 1], *fmt = "%s %-3d";
int fn = 0;
sprintf( f[0], "%20s", "0" );
sprintf( f[1], "%20s", "1" );
printf( fmt, f[ 0 ], ++fn );
putchar( '\n' );
printf( fmt, f[ 1 ], ++fn );
for( bool evod = false; getchar() != EOF; evod = !evod ) {
for( int carry = 0, i = 20; --i >= 0; ) {
if( f[ evod][i] == ' ' && f[!evod][i] == ' ' && carry == 0 ) break;
int f1 = f[ evod][i] == ' ' ? 0 : f[ evod][i] - '0';
int f2 = f[!evod][i] == ' ' ? 0 : f[!evod][i] - '0';
f1 += f2 + carry; carry = f1 / 10; f[ evod][i] = f1%10 + '0';
}
printf( fmt, f[ evod], ++fn );
}
return 0;
}
Output
0 1
1 2
1 3
2 4
3 5
5 6
8 7
13 8
21 9
/* omitted */
591286729879 59
956722026041 60
1548008755920 61
2504730781961 62
4052739537881 63
without using %, / or * , I have to find the no. is divisible by 3 or not?
it might be an interview question.
Thanks.
There are various ways. The simplest is pretty obvious:
int isdivby3(int n) {
if (n < 0) n = -n;
while (n > 0) n -= 3;
return n == 0;
}
But we can improve that. Any number can be represented like this: ("," means range inclusive):
Base2 (AKA binary)
(0,1) + 2*(0,1) + 4*(0,1)
Base4
(0,3) + 4*(0,3) + 16*(0,3)
BaseN
(0,N-1) + N*(0,N-1) + N*N*(0,N-1)
Now the trick is, a number x is divisible by n-1 if and only if the digitsum of x in base n is divisible by n-1. This trick is well-known for 9:
1926 = 6 + 2*10 + 9*100 + 1*1000
6+2+9+1 = 8 + 1*10
8+1 = 9 thus 1926 is divisible by 9
Now we can apply that for 3 too in base4. And were lucky since 4 is a power of 2 we can do binary bitwise operations. I use the notation number(base).
27(10) = 123(4)
Digitsum
12(4)
Digitsum again
3(4) = Divisible!
Now let's translate that to C:
int div3(int n) {
if (n < 0) n = -n;
else if (n == 0) return 1;
while (n > 3) {
int d = 0;
while (n > 0) {
d += n & 3;
n >>= 2;
}
n = d;
}
return n == 3;
}
Blazing fast.
Subtract 3 until you either
hit 0 - number was divisible by 3 (or)
get a number less than 0 - number wasn't divisible
if (number > 0)
{
while (number > 0)
{
number -= 3;
}
}
else if( number < 0)
{
while number < 0:
number += 3
}
return number == 0
Here is a reasonably efficient algorithm for large numbers. (Well not very efficient, but reasonable given the constraints.)
Use sprintf to convert it to a string, convert each digit back to a number. Add up the digits. If you come up with 3, 6, or 9, it is divisible by 3. Anything else less than 10, it is not. Anything over 9, recurse.
For instance to test the number 813478902 you'd stringify, then add the digits to get 42, add those digits to get 6, so it is divisible by 3.
just use a for loop subtracting 3 over and over and see if you get to 0. if you get to negative without getting to 0 then you know its not divisible by 3
To print a count sequence which is divisible by 3 without division or modulus operator.
Notice the count sequence:
00: 00(00)
01: 0001
02: 0010
03: 00(11)
04: 0100
05: 0101
06: 01(10)
07: 0111
08: 1000
09: 10(01)
10: 1010
11: 1011
12: 11(00)
13: 1101
14: 1110
15: 11(11)
16: 10000
17: 10001
18: 100(10)
19: 10011
20: 10100
21: 101(01)
Note that the last two bits of those numbers which are divisible by three (shown in brackets) are cycling through {00, 11, 10, 01} . What we need to check is if the last two bits of the count sequence has these bits in a sequence.
First we start matching with mask = 00 and loop while the first number is not encountered with the lower two bits 00. When a match is found we then do (mask + 03) & 0x03 which gets us the next mask in the set. And we continue to match the last two bits of the next count with 11. Which can be done by ((count & 3) == mask)
The code is
#include <stdio.h>
int main (void)
{
int i=0;
unsigned int mask = 0x00;
for (i=0; i<100;i++)
{
if ((i&0x03) == mask)
{
printf ("\n%d", i);
mask = (mask + 3) & 0x03;
}
}
printf ("\n");
return 0;
}
This is not a general one. Best is to use the solution which #nightcracker have suggested
Also if you really want to implement the division operation i without using the divide operations. I would tell you to have a look at the Non-Restoring Division Algorithm, this can be done in program with a lot of bit manipulations with bitwise operators. Here are some links and references for it.
Wikipedia Link
Here is a demo from UMass
Also have a look at Computer Organization by Carl Hamacher, Zvonko Vranesic, Safwat Zaky
number = abs(number)
while (number > 0)
{
number -= 3;
}
return number == 0
Suppose n is the number in question and it is non-negative.
If n is 0 it is divisible by 3; otherwise n = (2^p)*(2*n1+1) and n is divisible by 3 iff 2*n1+1 is, iff there is a k>=0 with 2*n1+1 = 3*(2*k+1) iff n1 = 3*k+1 iff n1=1 or n1> 1 and n1-1 is divisible by 3. So:
int ism3( int n)
{ for(;n;)
{ while( !(n & 1)) n >>= 1;
n >>= 1;
if ( n == 0) return 0;
n-= 1;
}
return 1;
}
The simplest way to know if a number is divisible by 3 is to sum all its digits and divide the result by 3. If the sum of the digits is divisible by 3, so the number itself is divisible by 3. For instance, 54467565687 is divisible by 3, because 5+4+4+6+7+5+6+5+6+8+7 = 63, and 63 is divisible by 3. So, no matter how big is the number, you can find if it is divisible by 3 just adding all its digits, and subtracting 3 from the value of this sum until you have a result smaller than 3. If this result is 0, the value of the sum is divisible by 3 (and so the original number), otherwise the sum is not divisible by 3 (and the original number is not divisible, either). It's done much more quickly than subtract 3 successively from the original number (of course, specially if it is a large number) and with no divisions. Um abraço a todos.
Artur
A number is divisible by three if its binary alternating digit sum is zero:
bool by3(int n) {
int s=0;
for (int q=1; n; s+=q*(n&1), n>>=1, q*=-1);
return !s;
}
You could use user feedback:
int isDivisibleBy3(int n)
{
int areDivisibleBy3[] = {};
for(int i = 0; i < 0; i++)
{
if(n == areDivisibleBy3[i])
{
return 1;
}
}
return 0;
}
When a user reports a bug stating that a number that is divisble by 3 is not giving the correct result, you simply add that number to the array and increase the number i is compared to in the for loop condition.
This is great because then you never have to worry about numbers the user never uses.
Don't forget to add a unit test for whenever a user reports a bug!
Closed. This question needs to be more focused. It is not currently accepting answers.
Closed 4 years ago.
Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
How can I check if a given number is even or odd in C?
Use the modulo (%) operator to check if there's a remainder when dividing by 2:
if (x % 2) { /* x is odd */ }
A few people have criticized my answer above stating that using x & 1 is "faster" or "more efficient". I do not believe this to be the case.
Out of curiosity, I created two trivial test case programs:
/* modulo.c */
#include <stdio.h>
int main(void)
{
int x;
for (x = 0; x < 10; x++)
if (x % 2)
printf("%d is odd\n", x);
return 0;
}
/* and.c */
#include <stdio.h>
int main(void)
{
int x;
for (x = 0; x < 10; x++)
if (x & 1)
printf("%d is odd\n", x);
return 0;
}
I then compiled these with gcc 4.1.3 on one of my machines 5 different times:
With no optimization flags.
With -O
With -Os
With -O2
With -O3
I examined the assembly output of each compile (using gcc -S) and found that in each case, the output for and.c and modulo.c were identical (they both used the andl $1, %eax instruction). I doubt this is a "new" feature, and I suspect it dates back to ancient versions. I also doubt any modern (made in the past 20 years) non-arcane compiler, commercial or open source, lacks such optimization. I would test on other compilers, but I don't have any available at the moment.
If anyone else would care to test other compilers and/or platform targets, and gets a different result, I'd be very interested to know.
Finally, the modulo version is guaranteed by the standard to work whether the integer is positive, negative or zero, regardless of the implementation's representation of signed integers. The bitwise-and version is not. Yes, I realise two's complement is somewhat ubiquitous, so this is not really an issue.
You guys are waaaaaaaay too efficient. What you really want is:
public boolean isOdd(int num) {
int i = 0;
boolean odd = false;
while (i != num) {
odd = !odd;
i = i + 1;
}
return odd;
}
Repeat for isEven.
Of course, that doesn't work for negative numbers. But with brilliance comes sacrifice...
Use bit arithmetic:
if((x & 1) == 0)
printf("EVEN!\n");
else
printf("ODD!\n");
This is faster than using division or modulus.
[Joke mode="on"]
public enum Evenness
{
Unknown = 0,
Even = 1,
Odd = 2
}
public static Evenness AnalyzeEvenness(object o)
{
if (o == null)
return Evenness.Unknown;
string foo = o.ToString();
if (String.IsNullOrEmpty(foo))
return Evenness.Unknown;
char bar = foo[foo.Length - 1];
switch (bar)
{
case '0':
case '2':
case '4':
case '6':
case '8':
return Evenness.Even;
case '1':
case '3':
case '5':
case '7':
case '9':
return Evenness.Odd;
default:
return Evenness.Unknown;
}
}
[Joke mode="off"]
EDIT: Added confusing values to the enum.
In response to ffpf - I had exactly the same argument with a colleague years ago, and the answer is no, it doesn't work with negative numbers.
The C standard stipulates that negative numbers can be represented in 3 ways:
2's complement
1's complement
sign and magnitude
Checking like this:
isEven = (x & 1);
will work for 2's complement and sign and magnitude representation, but not for 1's complement.
However, I believe that the following will work for all cases:
isEven = (x & 1) ^ ((-1 & 1) | ((x < 0) ? 0 : 1)));
Thanks to ffpf for pointing out that the text box was eating everything after my less than character!
A nice one is:
/*forward declaration, C compiles in one pass*/
bool isOdd(unsigned int n);
bool isEven(unsigned int n)
{
if (n == 0)
return true ; // I know 0 is even
else
return isOdd(n-1) ; // n is even if n-1 is odd
}
bool isOdd(unsigned int n)
{
if (n == 0)
return false ;
else
return isEven(n-1) ; // n is odd if n-1 is even
}
Note that this method use tail recursion involving two functions. It can be implemented efficiently (turned into a while/until kind of loop) if your compiler supports tail recursion like a Scheme compiler. In this case the stack should not overflow !
A number is even if, when divided by two, the remainder is 0. A number is odd if, when divided by 2, the remainder is 1.
// Java
public static boolean isOdd(int num){
return num % 2 != 0;
}
/* C */
int isOdd(int num){
return num % 2;
}
Methods are great!
i % 2 == 0
I'd say just divide it by 2 and if there is a 0 remainder, it's even, otherwise it's odd.
Using the modulus (%) makes this easy.
eg.
4 % 2 = 0 therefore 4 is even
5 % 2 = 1 therefore 5 is odd
One more solution to the problem
(children are welcome to vote)
bool isEven(unsigned int x)
{
unsigned int half1 = 0, half2 = 0;
while (x)
{
if (x) { half1++; x--; }
if (x) { half2++; x--; }
}
return half1 == half2;
}
I would build a table of the parities (0 if even 1 if odd) of the integers (so one could do a lookup :D), but gcc won't let me make arrays of such sizes:
typedef unsigned int uint;
char parity_uint [UINT_MAX];
char parity_sint_shifted [((uint) INT_MAX) + ((uint) abs (INT_MIN))];
char* parity_sint = parity_sint_shifted - INT_MIN;
void build_parity_tables () {
char parity = 0;
unsigned int ui;
for (ui = 1; ui <= UINT_MAX; ++ui) {
parity_uint [ui - 1] = parity;
parity = !parity;
}
parity = 0;
int si;
for (si = 1; si <= INT_MAX; ++si) {
parity_sint [si - 1] = parity;
parity = !parity;
}
parity = 1;
for (si = -1; si >= INT_MIN; --si) {
parity_sint [si] = parity;
parity = !parity;
}
}
char uparity (unsigned int n) {
if (n == 0) {
return 0;
}
return parity_uint [n - 1];
}
char sparity (int n) {
if (n == 0) {
return 0;
}
if (n < 0) {
++n;
}
return parity_sint [n - 1];
}
So let's instead resort to the mathematical definition of even and odd instead.
An integer n is even if there exists an integer k such that n = 2k.
An integer n is odd if there exists an integer k such that n = 2k + 1.
Here's the code for it:
char even (int n) {
int k;
for (k = INT_MIN; k <= INT_MAX; ++k) {
if (n == 2 * k) {
return 1;
}
}
return 0;
}
char odd (int n) {
int k;
for (k = INT_MIN; k <= INT_MAX; ++k) {
if (n == 2 * k + 1) {
return 1;
}
}
return 0;
}
Let C-integers denote the possible values of int in a given C compilation. (Note that C-integers is a subset of the integers.)
Now one might worry that for a given n in C-integers that the corresponding integer k might not exist within C-integers. But with a little proof it is can be shown that for all integers n, |n| <= |2n| (*), where |n| is "n if n is positive and -n otherwise". In other words, for all n in integers at least one of the following holds (exactly either cases (1 and 2) or cases (3 and 4) in fact but I won't prove it here):
Case 1: n <= 2n.
Case 2: -n <= -2n.
Case 3: -n <= 2n.
Case 4: n <= -2n.
Now take 2k = n. (Such a k does exist if n is even, but I won't prove it here. If n is not even then the loop in even fails to return early anyway, so it doesn't matter.) But this implies k < n if n not 0 by (*) and the fact (again not proven here) that for all m, z in integers 2m = z implies z not equal to m given m is not 0. In the case n is 0, 2*0 = 0 so 0 is even we are done (if n = 0 then 0 is in C-integers because n is in C-integer in the function even, hence k = 0 is in C-integers). Thus such a k in C-integers exists for n in C-integers if n is even.
A similar argument shows that if n is odd, there exists a k in C-integers such that n = 2k + 1.
Hence the functions even and odd presented here will work properly for all C-integers.
// C#
bool isEven = ((i % 2) == 0);
Here is an answer in
Java:
public static boolean isEven (Integer Number) {
Pattern number = Pattern.compile("^.*?(?:[02]|8|(?:6|4))$");
String num = Number.toString(Number);
Boolean numbr = new Boolean(number.matcher(num).matches());
return numbr.booleanValue();
}
Try this: return (((a>>1)<<1) == a)
Example:
a = 10101011
-----------------
a>>1 --> 01010101
a<<1 --> 10101010
b = 10011100
-----------------
b>>1 --> 01001110
b<<1 --> 10011100
Reading this rather entertaining discussion, I remembered that I had a real-world, time-sensitive function that tested for odd and even numbers inside the main loop. It's an integer power function, posted elsewhere on StackOverflow, as follows. The benchmarks were quite surprising. At least in this real-world function, modulo is slower, and significantly so. The winner, by a wide margin, requiring 67% of modulo's time, is an or ( | ) approach, and is nowhere to be found elsewhere on this page.
static dbl IntPow(dbl st0, int x) {
UINT OrMask = UINT_MAX -1;
dbl st1=1.0;
if(0==x) return (dbl)1.0;
while(1 != x) {
if (UINT_MAX == (x|OrMask)) { // if LSB is 1...
//if(x & 1) {
//if(x % 2) {
st1 *= st0;
}
x = x >> 1; // shift x right 1 bit...
st0 *= st0;
}
return st1 * st0;
}
For 300 million loops, the benchmark timings are as follows.
3.962 the | and mask approach
4.851 the & approach
5.850 the % approach
For people who think theory, or an assembly language listing, settles arguments like these, this should be a cautionary tale. There are more things in heaven and earth, Horatio, than are dreamt of in your philosophy.
This is a follow up to the discussion with #RocketRoy regarding his answer, but it might be useful to anyone who wants to compare these results.
tl;dr From what I've seen, Roy's approach ((0xFFFFFFFF == (x | 0xFFFFFFFE)) is not completely optimized to x & 1 as the mod approach, but in practice running times should turn out equal in all cases.
So, first I compared the compiled output using Compiler Explorer:
Functions tested:
int isOdd_mod(unsigned x) {
return (x % 2);
}
int isOdd_and(unsigned x) {
return (x & 1);
}
int isOdd_or(unsigned x) {
return (0xFFFFFFFF == (x | 0xFFFFFFFE));
}
CLang 3.9.0 with -O3:
isOdd_mod(unsigned int): # #isOdd_mod(unsigned int)
and edi, 1
mov eax, edi
ret
isOdd_and(unsigned int): # #isOdd_and(unsigned int)
and edi, 1
mov eax, edi
ret
isOdd_or(unsigned int): # #isOdd_or(unsigned int)
and edi, 1
mov eax, edi
ret
GCC 6.2 with -O3:
isOdd_mod(unsigned int):
mov eax, edi
and eax, 1
ret
isOdd_and(unsigned int):
mov eax, edi
and eax, 1
ret
isOdd_or(unsigned int):
or edi, -2
xor eax, eax
cmp edi, -1
sete al
ret
Hats down to CLang, it realized that all three cases are functionally equal. However, Roy's approach isn't optimized in GCC, so YMMV.
It's similar with Visual Studio; inspecting the disassembly Release x64 (VS2015) for these three functions, I could see that the comparison part is equal for "mod" and "and" cases, and slightly larger for the Roy's "or" case:
// x % 2
test bl,1
je (some address)
// x & 1
test bl,1
je (some address)
// Roy's bitwise or
mov eax,ebx
or eax,0FFFFFFFEh
cmp eax,0FFFFFFFFh
jne (some address)
However, after running an actual benchmark for comparing these three options (plain mod, bitwise or, bitwise and), results were completely equal (again, Visual Studio 2005 x86/x64, Release build, no debugger attached).
Release assembly uses the test instruction for and and mod cases, while Roy's case uses the cmp eax,0FFFFFFFFh approach, but it's heavily unrolled and optimized so there is no difference in practice.
My results after 20 runs (i7 3610QM, Windows 10 power plan set to High Performance):
[Test: Plain mod 2 ] AVERAGE TIME: 689.29 ms (Relative diff.: +0.000%)
[Test: Bitwise or ] AVERAGE TIME: 689.63 ms (Relative diff.: +0.048%)
[Test: Bitwise and ] AVERAGE TIME: 687.80 ms (Relative diff.: -0.217%)
The difference between these options is less than 0.3%, so it's rather obvious the assembly is equal in all cases.
Here is the code if anyone wants to try, with a caveat that I only tested it on Windows (check the #if LINUX conditional for the get_time definition and implement it if needed, taken from this answer).
#include <stdio.h>
#if LINUX
#include <sys/time.h>
#include <sys/resource.h>
double get_time()
{
struct timeval t;
struct timezone tzp;
gettimeofday(&t, &tzp);
return t.tv_sec + t.tv_usec*1e-6;
}
#else
#include <windows.h>
double get_time()
{
LARGE_INTEGER t, f;
QueryPerformanceCounter(&t);
QueryPerformanceFrequency(&f);
return (double)t.QuadPart / (double)f.QuadPart * 1000.0;
}
#endif
#define NUM_ITERATIONS (1000 * 1000 * 1000)
// using a macro to avoid function call overhead
#define Benchmark(accumulator, name, operation) { \
double startTime = get_time(); \
double dummySum = 0.0, elapsed; \
int x; \
for (x = 0; x < NUM_ITERATIONS; x++) { \
if (operation) dummySum += x; \
} \
elapsed = get_time() - startTime; \
accumulator += elapsed; \
if (dummySum > 2000) \
printf("[Test: %-12s] %0.2f ms\r\n", name, elapsed); \
}
void DumpAverage(char *test, double totalTime, double reference)
{
printf("[Test: %-12s] AVERAGE TIME: %0.2f ms (Relative diff.: %+6.3f%%)\r\n",
test, totalTime, (totalTime - reference) / reference * 100.0);
}
int main(void)
{
int repeats = 20;
double runningTimes[3] = { 0 };
int k;
for (k = 0; k < repeats; k++) {
printf("Run %d of %d...\r\n", k + 1, repeats);
Benchmark(runningTimes[0], "Plain mod 2", (x % 2));
Benchmark(runningTimes[1], "Bitwise or", (0xFFFFFFFF == (x | 0xFFFFFFFE)));
Benchmark(runningTimes[2], "Bitwise and", (x & 1));
}
{
double reference = runningTimes[0] / repeats;
printf("\r\n");
DumpAverage("Plain mod 2", runningTimes[0] / repeats, reference);
DumpAverage("Bitwise or", runningTimes[1] / repeats, reference);
DumpAverage("Bitwise and", runningTimes[2] / repeats, reference);
}
getchar();
return 0;
}
I know this is just syntactic sugar and only applicable in .net but what about extension method...
public static class RudiGroblerExtensions
{
public static bool IsOdd(this int i)
{
return ((i % 2) != 0);
}
}
Now you can do the following
int i = 5;
if (i.IsOdd())
{
// Do something...
}
In the "creative but confusing category" I offer:
int isOdd(int n) { return n ^ n * n ? isOdd(n * n) : n; }
A variant on this theme that is specific to Microsoft C++:
__declspec(naked) bool __fastcall isOdd(const int x)
{
__asm
{
mov eax,ecx
mul eax
mul eax
mul eax
mul eax
mul eax
mul eax
ret
}
}
The bitwise method depends on the inner representation of the integer. Modulo will work anywhere there is a modulo operator. For example, some systems actually use the low level bits for tagging (like dynamic languages), so the raw x & 1 won't actually work in that case.
IsOdd(int x) { return true; }
Proof of correctness - consider the set of all positive integers and suppose there is a non-empty set of integers that are not odd. Because positive integers are well-ordered, there will be a smallest not odd number, which in itself is pretty odd, so clearly that number can't be in the set. Therefore this set cannot be non-empty. Repeat for negative integers except look for the greatest not odd number.
Portable:
i % 2 ? odd : even;
Unportable:
i & 1 ? odd : even;
i << (BITS_PER_INT - 1) ? odd : even;
As some people have posted, there are numerous ways to do this. According to this website, the fastest way is the modulus operator:
if (x % 2 == 0)
total += 1; //even number
else
total -= 1; //odd number
However, here is some other code that was bench marked by the author which ran slower than the common modulus operation above:
if ((x & 1) == 0)
total += 1; //even number
else
total -= 1; //odd number
System.Math.DivRem((long)x, (long)2, out outvalue);
if ( outvalue == 0)
total += 1; //even number
else
total -= 1; //odd number
if (((x / 2) * 2) == x)
total += 1; //even number
else
total -= 1; //odd number
if (((x >> 1) << 1) == x)
total += 1; //even number
else
total -= 1; //odd number
while (index > 1)
index -= 2;
if (index == 0)
total += 1; //even number
else
total -= 1; //odd number
tempstr = x.ToString();
index = tempstr.Length - 1;
//this assumes base 10
if (tempstr[index] == '0' || tempstr[index] == '2' || tempstr[index] == '4' || tempstr[index] == '6' || tempstr[index] == '8')
total += 1; //even number
else
total -= 1; //odd number
How many people even knew of the Math.System.DivRem method or why would they use it??
int isOdd(int i){
return(i % 2);
}
done.
To give more elaboration on the bitwise operator method for those of us who didn't do much boolean algebra during our studies, here is an explanation. Probably not of much use to the OP, but I felt like making it clear why NUMBER & 1 works.
Please note like as someone answered above, the way negative numbers are represented can stop this method working. In fact it can even break the modulo operator method too since each language can differ in how it deals with negative operands.
However if you know that NUMBER will always be positive, this works well.
As Tooony above made the point that only the last digit in binary (and denary) is important.
A boolean logic AND gate dictates that both inputs have to be a 1 (or high voltage) for 1 to be returned.
1 & 0 = 0.
0 & 1 = 0.
0 & 0 = 0.
1 & 1 = 1.
If you represent any number as binary (I have used an 8 bit representation here), odd numbers have 1 at the end, even numbers have 0.
For example:
1 = 00000001
2 = 00000010
3 = 00000011
4 = 00000100
If you take any number and use bitwise AND (& in java) it by 1 it will either return 00000001, = 1 meaning the number is odd. Or 00000000 = 0, meaning the number is even.
E.g
Is odd?
1 & 1 =
00000001 &
00000001 =
00000001 <— Odd
2 & 1 =
00000010 &
00000001 =
00000000 <— Even
54 & 1 =
00000001 &
00110110 =
00000000 <— Even
This is why this works:
if(number & 1){
//Number is odd
} else {
//Number is even
}
Sorry if this is redundant.
Number Zero parity | zero http://tinyurl.com/oexhr3k
Python code sequence.
# defining function for number parity check
def parity(number):
"""Parity check function"""
# if number is 0 (zero) return 'Zero neither ODD nor EVEN',
# otherwise number&1, checking last bit, if 0, then EVEN,
# if 1, then ODD.
return (number == 0 and 'Zero neither ODD nor EVEN') \
or (number&1 and 'ODD' or 'EVEN')
# cycle trough numbers from 0 to 13
for number in range(0, 14):
print "{0:>4} : {0:08b} : {1:}".format(number, parity(number))
Output:
0 : 00000000 : Zero neither ODD nor EVEN
1 : 00000001 : ODD
2 : 00000010 : EVEN
3 : 00000011 : ODD
4 : 00000100 : EVEN
5 : 00000101 : ODD
6 : 00000110 : EVEN
7 : 00000111 : ODD
8 : 00001000 : EVEN
9 : 00001001 : ODD
10 : 00001010 : EVEN
11 : 00001011 : ODD
12 : 00001100 : EVEN
13 : 00001101 : ODD
I execute this code for ODD & EVEN:
#include <stdio.h>
int main()
{
int number;
printf("Enter an integer: ");
scanf("%d", &number);
if(number % 2 == 0)
printf("%d is even.", number);
else
printf("%d is odd.", number);
}
For the sake of discussion...
You only need to look at the last digit in any given number to see if it is even or odd.
Signed, unsigned, positive, negative - they are all the same with regards to this.
So this should work all round: -
void tellMeIfItIsAnOddNumberPlease(int iToTest){
int iLastDigit;
iLastDigit = iToTest - (iToTest / 10 * 10);
if (iLastDigit % 2 == 0){
printf("The number %d is even!\n", iToTest);
} else {
printf("The number %d is odd!\n", iToTest);
}
}
The key here is in the third line of code, the division operator performs an integer division, so that result are missing the fraction part of the result. So for example 222 / 10 will give 22 as a result. Then multiply it again with 10 and you have 220. Subtract that from the original 222 and you end up with 2, which by magic is the same number as the last digit in the original number. ;-)
The parenthesis are there to remind us of the order the calculation is done in. First do the division and the multiplication, then subtract the result from the original number. We could leave them out, since the priority is higher for division and multiplication than of subtraction, but this gives us "more readable" code.
We could make it all completely unreadable if we wanted to. It would make no difference whatsoever for a modern compiler: -
printf("%d%s\n",iToTest,0==(iToTest-iToTest/10*10)%2?" is even":" is odd");
But it would make the code way harder to maintain in the future. Just imagine that you would like to change the text for odd numbers to "is not even". Then someone else later on want to find out what changes you made and perform a svn diff or similar...
If you are not worried about portability but more about speed, you could have a look at the least significant bit. If that bit is set to 1 it is an odd number, if it is 0 it's an even number.
On a little endian system, like Intel's x86 architecture it would be something like this: -
if (iToTest & 1) {
// Even
} else {
// Odd
}
If you want to be efficient, use bitwise operators (x & 1), but if you want to be readable use modulo 2 (x % 2)
Checking even or odd is a simple task.
We know that any number exactly divisible by 2 is even number else odd.
We just need to check divisibility of any number and for checking divisibility we use % operator
Checking even odd using if else
if(num%2 ==0)
{
printf("Even");
}
else
{
printf("Odd");
}
C program to check even or odd using if else
Using Conditional/Ternary operator
(num%2 ==0) printf("Even") : printf("Odd");
C program to check even or odd using conditional operator.
Using Bitwise operator
if(num & 1)
{
printf("Odd");
}
else
{
printf("Even");
}
+66% faster > !(i%2) / i%2 == 0
int isOdd(int n)
{
return n & 1;
}
The code checks the last bit of the integer if it's 1 in Binary
Explanation
Binary : Decimal
-------------------
0000 = 0
0001 = 1
0010 = 2
0011 = 3
0100 = 4
0101 = 5
0110 = 6
0111 = 7
1000 = 8
1001 = 9
and so on...
Notice the rightmost bit is always 1 for Odd numbers.
the & bitwise AND operator checks the rightmost bit in our return line if it's 1
Think of it as true & false
When we compare n with 1 which means 0001 in binary (number of zeros doesn't matter).
then let's just Imagine that we have the integer n with a size of 1 byte.
It'd be represented by 8-bit / 8-binary digits.
If the int n was 7 and we compare it with 1, It's like
7 (1-byte int)| 0 0 0 0 0 1 1 1
&
1 (1-byte int)| 0 0 0 0 0 0 0 1
********************************************
Result | F F F F F F F T
Which F stands for false and T for true.
It compares only the rightmost bit if they're both true. So, automagically 7 & 1 is True.
What if I want to check the bit before the rightmost?
Simply change n & 1 to n & 2 which 2 represents 0010 in Binary and so on.
I suggest using hexadecimal notation if you're a beginner to bitwise operations
return n & 1; >> return n & 0x01;.