A friend just throw some code similar to following C# code:
int i = ...;
return i < 0 ? 0 : i;
That made me think. There's any "different" way to return zero for negative integers, or current positive value? More specifically I'm looking for bitwise operations, if possible.
BTW, I'm aware of Math.Max(0, i);
What's wrong with Math.Max?
You can do the equivalent without a branch using bitwise operations:
r = x ^ ((x ^ y) & -(x < y)); // == max(x, y)
If you substitute zero, it collapses to:
r = (y & -(0 < y)); // == max(0, y)
(Source: this list of bitwise tricks.)
If branches were extremely expensive on your platform, that might be worthwhile in some inner loop, I suppose, but it's pretty obscure and not the kind of thing I'd like to come across outside of an extremely time-sensitive function.
How about:
int i = ...;
return i & ~(i >> 31);
The below will do the trick and the code reads so well it practically don't need a comment ;)
((((0x80000000 & i) >> 31)^1) * 0xFFFFFFFF) & i
then again
int temp = (0x80000000 & i); //get the most significant bit (1 for negative 0 for positive)
temp = (temp>>31)^1; //more it to the least significant and not it (we now have 0 for negative numbers and one for positive)
temp *= 0xFFFFFFFF; //get a lof of F's if it's positive or a lot of zeros if the number was negative
temp = temp & i; //and the F's/zeros with the original number
and voila zero for all negative number and all positive are left unchanged
Short answer: No.
Bit operators do something very different, or rather are used for different problems.
If you know the size of your integers, you could test the highest (most significant) bit; if it's 1, the number is negative and you can act on that. But that would be a heck of a lot more work than the simple "<" test.
Not bitwise but different:
return (i + Math.abs(i))/2
EDIT:
return (int)(i/2f + Math.abs(i)/2f)
Related
For example, is every 4th bit set.
1000.1000 true
1010.1000 true
0010.1000 false
with offset of 1
0100.0100 true
0101.0100 true
0001.0100 false
Currently I am doing this by looping through every 4 bits
int num = 170; //1010.1010
int N = 4;
int offset = 0; //[0, N-1]
bool everyNth = true;
for (int i = 0; i < intervals ; i++){
if(((num >> (N*i)) & ((1 << (N - 1)) >> offset)) == 0){
every4th = false;
break;
}
}
return everyNth;
EXPLANATION OF CODE:
num = 1010.1010
The loop makes it so I look at each 4 bits as a block by right shifting * 4.
num >> 4 = 0000.1010
Then an & for a specific bit that can be offset.
And to only look at a specific bit of the chunk, a mask is created by ((1 << (N - 1)) >> offset)
0000.1010
1000 (mask >> offset0)
OR 0100 (mask >> offset1)
OR 0010 (mask >> offset2)
OR 0001 (mask >> offset3)
Is there a purely computational way to do this? Like how you can XOR your way through to figure out parity. I am working with 64 bit integers for my case, but I am wondering this in a more general case.
Additionally, I am under the assumption that bit operators are one of the fastest methods for calculations or math in general. If this is not true, please feel free to correct me on what the time and place is for bit operators.
If we had a mask M in which every Nth bit is set, then testing whether every Nth bit in a given integer x is set could be calculated as (x & M) == M. Or with offset, you could use ((x << offset) & M) == M. Shifting M right is fine too.
If N is constant, that's all there is to it, just use the right M.
If N is variable, the question becomes, how do we get a mask in which every Nth bit is set.
Here is a simple way to do that:
Start by setting the Nth bit
"Double" the mask until done
For example,
ulong M = 1UL << (N - 1);
do
{
M |= M << N;
N += N;
} while (N < 64);
That is clearly still a loop. But it's not a bit-by-bit loop, it makes only a logarithmic number of iterations.
You could precompute the masks and store them in a small array, the range of N is necessarily small.
There may also be a way based on ulong.MaxValue / ((1UL << N) - 1) but that needs something more to "align" the mask and 64-bit division is not so great anyway. Perhaps there is a smarter way to get the mask.
I am under the assumption that bit operators are one of the fastest methods for calculations or math in general
Bitwise operations are some of the fastest operations, but addition is equally fast, and multiplication is not that far behind (and a multiplication can do a lot more work at once, compared to how much more it costs).
What is the fastest way to find if a number is even or odd?
It is pretty well known that
static inline int is_odd_A(int x) { return x & 1; }
is more efficient than
static inline int is_odd_B(int x) { return x % 2; }
But with the optimizer on, will is_odd_B be no different from is_odd_A? No — with gcc-4.2 -O2, we get, (in ARM assembly):
_is_odd_A:
and r0, r0, #1
bx lr
_is_odd_B:
mov r3, r0, lsr #31
add r0, r0, r3
and r0, r0, #1
rsb r0, r3, r0
bx lr
We see that is_odd_B takes 3 more instructions than is_odd_A, the main reason is because
((-1) % 2) == -1
((-1) & 1) == 1
However, all the following versions will generate the same code as is_odd_A:
#include <stdbool.h>
static inline bool is_odd_D(int x) { return x % 2; } // note the bool
static inline int is_odd_E(int x) { return x % 2 != 0; } // note the !=
What does this mean? The optimizer is usually sophisticated enough that, for these simple stuff, the clearest code is enough to guarantee best efficiency.
Usual way to do it:
int number = ...;
if(number % 2) { odd }
else { even }
Alternative:
int number = ...;
if(number & 1) { odd }
else { even }
Tested on GCC 3.3.1 and 4.3.2, both have about the same speed (without compiler optimization) as both result in the and instruction (compiled on x86) - I know that using the div instruction for modulo would be much slower, thus I didn't test it at all.
if (x & 1) is true then it's odd, otherwise it's even.
bool is_odd = number & 1;
int i=5;
if ( i%2 == 0 )
{
// Even
} else {
// Odd
}
int is_odd(int n)
{
if (n == 0)
return 0;
else if (n == 1)
return 1;
else
return !is_odd(n - 1);
}
Oh wait, you said fastest way, not funniest. My bad ;)
Above function only works for positive numbers of course.
Check to see if the last bit is 1.
int is_odd(int num) {
return num & 1;
}
If it's an integer, probably by just checking the least significant bit. Zero would be counted as even though.
The portable way is to use the modulus operator %:
if (x % 2 == 0) // number is even
If you know that you're only ever going to run on two's complement architectures, you can use a bitwise and:
if (x & 0x01 == 0) // number is even
Using the modulus operator can result in slower code relative to the bitwise and; however, I'd stick with it unless all of the following are true:
You are failing to meet a hard performance requirement;
You are executing x % 2 a lot (say in a tight loop that's being executed thousands of times);
Profiling indicates that usage of the mod operator is the bottleneck;
Profiling also indicates that using the bitwise-and relieves the bottleneck and allows you to meet the performance requirement.
Your question is not completely specified. Regardless, the answer is dependent on your compiler and the architecture of your machine. For example, are you on a machine using one's complement or two's complement signed number representations?
I write my code to be correct first, clear second, concise third and fast last. Therefore, I would code this routine as follows:
/* returns 0 if odd, 1 if even */
/* can use bool in C99 */
int IsEven(int n) {
return n % 2 == 0;
}
This method is correct, it more clearly expresses the intent than testing the LSB, it's concise and, believe it or not, it is blazing fast. If and only if profiling told me that this method were a bottleneck in my application would I consider deviating from it.
Check the least significant bit:
if (number & 0x01) {
// It's odd
} else {
// It's even
}
Can't you just look at the last digit and check if its even or odd if the input is in base 10?
{1, 3, 5, 7, 9} is odd
{0, 2, 4, 6, 8} is even
Additional info: The OP states that a number is a given, so I went with that when constructing this answer. This also requires the number to be in base 10. This answer is mathematically correct by definition of even/odd in base 10. Depending on the use case, you have a mathematically consistent result just by checking the last digit.
Note: If your input is already an int, just check the low bit of that. This answer is only useful for numbers represented as a sequence of digits. You could convert int->string to do this, but that would be much slower than n % 2 == 0.
Checking the last digit does work for a string of digits in any even base, not just 10. For bases lower than 10, like base 8 (octal), 9 and 8 aren't possible digits, but the low digit being odd or even still determines whether the whole number is.
For bases higher than 10, there will be extra possibilities, but you don't want to search a list anyway, just check if the digit as an integer is odd or even using the normal i % 2 == 0 or !=0 check.
For ASCII hex using 'a' .. 'f' to represent digits values 10 through 15, the low bit of ASCII code does not represent odd or even, because 'a' == 0x61 (odd) but represents 10 aka 0xa (even). So you'd have to convert the hex digit to an integer, or do some bit-hack on the ASCII code to flip the low bit according to some other bit or condition.
Just out of curiosity, is there a way to get the sign of a number, any kind (but obviously a signed type), not just integer using some bitwise/masking, or other kind of, operation?
That is without using any conditional statement or calling the Math.Sign() function.
Thanks in advance!
EDIT: I recognize it was a misleading question. What I had in mind more likely something like: "get the same output of the Math.Sign() or, simplifying get 0 if x <= 0, 1 otherwise".
EDIT #2: to all those asking for code, I didn't have any in mind when I posted the question, but here's an example I came up with, just to give a context of a possible application:
x = (x < 0) ? 0 : x;
Having the sign into a variable could lead to:
x = sign * x; //where sign = 0 for x <= 0, otherwise sign = 1;
The aim would be to achieve the same result as the above :)
EDIT #3: FOUND IT! :D
// THIS IS NOT MEANT TO BE PLAIN C#!
// Returns 0 if x <= 0, 1 otherwise.
int SignOf(x)
{
return (1+x-(x+1)%x)/x;
}
Thanks to everyone!
is there a way to get the sign of a number (any kind, not just integer)
Not for any number type, no. For an integer, you can test the most significant bit: if it's 1, the number is negative. You could theoretically do the same with a floating point number, but bitwise operators don't work on float or double.
Here's a "zero safe" solution that works for all value types (int, float, double, decimal...etc):
(value.GetHashCode() >> 31) + 1;
Output: 1 = 1, -1 = 0, 0.5 = 1, -0.5 = 0, 0 = 1
It's also roughly 10% cheaper than (1+x-(x+1)%x)/x; in C#. Additionally if "value" is an integer, you can drop the GetHashCode() function call in which case (1+x-(x+1)%x)/x; is 127% more expensive ((value >> 31) + 1; is 56% cheaper).
Since 0 is positive it is illogical for a result of 1 for positive numbers & a result of 0 for 0. If you could parametrise -0 you would get an output of 0.
I understand that GetHashCode() is a function call, but the inner workings of the function in the C# language implementation is entirely "arithmetic". Basically the GetHashCode() function reads the memory section, that stores your float type, as an integer type:
*((int*)&singleValue);
How the GetHashCode function works (best source I could find quickly) - https://social.msdn.microsoft.com/Forums/vstudio/en-US/3c3fde60-1b4a-449f-afdc-fe5bba8fece3/hash-code-of-floats?forum=netfxbcl
If you want the output value to be 1 with the same sign as the input, use:
((float.GetHashCode() >> 31) * 2) + 1;
The above floating-point method is roughly 39% cheaper than System.Math.Sign(float) (System.Math.Sign(float) is roughly 65% more expensive). Where System.Math.Sign(float) throws an exception for float.NaN, ((float.NaN.GetHashCode() >> 31) * 2) + 1; does not and will return -1 instead of crashing.
or for integers:
((int >> 31) * 2) + 1;
The above integer method is roughly 56% cheaper than System.Math.Sign(int) (System.Math.Sign(int) is roughly 125% more expensive).
It depends on the type of number value type you are targeting.
For signed Integers C# and most computer systems use the so called Ones' complement representation.
That means the sign is stored in the first bit of the value.
So you can extract the sign like this:
Int16 number = -2;
Int16 sign = (number & Int16.MinValue) >> 16;
Boolean isNegative = Convert.ToBoolean(sign);
Note that up until now we have not used any conditional operator (explicitly anyways)
But: You still don't know whether the number has a sign or not.
The logical equivalent of your question: "How do I know, if my number is negative?" explicitly requires the usage of a conditional operator as the question is, after all conditional.
So you won't be able to dodge:
if(isNegative)
doThis();
else
doThat();
to just get the sign, you can avoid conditional operators as you will see below in Sign extension of int32 struct. however to get the name I dont think you can avoid conditional operator
class Program
{
static void Main(string[] args)
{
Console.WriteLine(0.Sign());
Console.WriteLine(0.SignName());
Console.WriteLine(12.Sign());
Console.WriteLine(12.SignName());
Console.WriteLine((-15).Sign());
Console.WriteLine((-15).SignName());
Console.ReadLine();
}
}
public static class extensions
{
public static string Sign(this int signedNumber)
{
return (signedNumber.ToString("+00;-00").Substring(0, 1));
}
public static string SignName(this int signedNumber)
{
return (signedNumber.ToString("+00;-00").Substring(0, 1)=="+"?"positive":"negative");
}
}
if x==0 you will have a divby0 exception with this code you posted:
int SignOf(x) {
return (1+x-(x+1)%x)/x; }
I want to make
BigInteger.ModPow(1/BigInteger, 2,5);
but 1/BigInteger always return 0, which causes, that the result is 0 too. I tried to look for some BigDecimal class for c# but I have found nothing. Is there any way how to count this even if there is no BigDecimal?
1/a is 0 for |a|>1, since BigIntegers use integer division where the fractional part of a division is ignored. I'm not sure what result you're expecting for this.
I assume you want to modular multiplicative inverse of a modulo m, and not a fractional number. This inverse exists iff a and m are co-prime, i.e. gcd(a, m) = 1.
The linked wikipedia page lists the two standard algorithms for calculating the modular multiplicative inverse:
Extended Euclidean algorithm, which works for arbitrary moduli
It's fast, but has input dependent runtime.
I don't have C# code at hand, but porting the pseudo code from wikipedia should be straight forward.
Using Euler's theorem:
This requires knowledge of φ(m) i.e. you need to know the prime factors of m. It's a popular choice when m is a prime and thus φ(m) = m-1 when it simply becomes . If you need constant runtime and you know φ(m), this is the way to go.
In C# this becomes BigInteger.ModPow(a, phiOfM-1, m)
The overload of the / operator chosen, is the following:
public static BigInteger operator /(
BigInteger dividend,
BigInteger divisor
)
See BigInteger.Division Operator. If the result is between 0 and 1 (which is likely when dividend is 1 as in your case), because the return value is an integer, 0 is returned, as you see.
What are you trying to do with the ModPow method? Do you realize that 2,5 are two arguments, two and five, not "two-point-five"? Is your intention "take square modulo 5"?
If you want floating-point division, you can use:
1.0 / (double)yourBigInt
Note the cast to double. This may lose precision and even "underflow" to zero if yourBigInt is too huge.
For example you need to get d in the next:
3*d = 1 (mod 9167368)
this is equally:
3*d = 1 + k * 9167368, where k = 1, 2, 3, ...
rewrite it:
d = (1 + k * 9167368)/3
Your d must be the integer with the lowest k.
Let's write the formula:
d = (1 + k * fi)/e
public static int MultiplicativeInverse(int e, int fi)
{
double result;
int k = 1;
while (true)
{
result = (1 + (k * fi)) / (double) e;
if ((Math.Round(result, 5) % 1) == 0) //integer
{
return (int)result;
}
else
{
k++;
}
}
}
let's test this code:
Assert.AreEqual(Helper.MultiplicativeInverse(3, 9167368), 6111579); // passed
I want to ensure that a division of integers is always rounded up if necessary. Is there a better way than this? There is a lot of casting going on. :-)
(int)Math.Ceiling((double)myInt1 / myInt2)
UPDATE: This question was the subject of my blog in January 2013. Thanks for the great question!
Getting integer arithmetic correct is hard. As has been demonstrated amply thus far, the moment you try to do a "clever" trick, odds are good that you've made a mistake. And when a flaw is found, changing the code to fix the flaw without considering whether the fix breaks something else is not a good problem-solving technique. So far we've had I think five different incorrect integer arithmetic solutions to this completely not-particularly-difficult problem posted.
The right way to approach integer arithmetic problems -- that is, the way that increases the likelihood of getting the answer right the first time - is to approach the problem carefully, solve it one step at a time, and use good engineering principles in doing so.
Start by reading the specification for what you're trying to replace. The specification for integer division clearly states:
The division rounds the result towards zero
The result is zero or positive when the two operands have the same sign and zero or negative when the two operands have opposite signs
If the left operand is the smallest representable int and the right operand is –1, an overflow occurs. [...] it is implementation-defined as to whether [an ArithmeticException] is thrown or the overflow goes unreported with the resulting value being that of the left operand.
If the value of the right operand is zero, a System.DivideByZeroException is thrown.
What we want is an integer division function which computes the quotient but rounds the result always upwards, not always towards zero.
So write a specification for that function. Our function int DivRoundUp(int dividend, int divisor) must have behaviour defined for every possible input. That undefined behaviour is deeply worrying, so let's eliminate it. We'll say that our operation has this specification:
operation throws if divisor is zero
operation throws if dividend is int.minval and divisor is -1
if there is no remainder -- division is 'even' -- then the return value is the integral quotient
Otherwise it returns the smallest integer that is greater than the quotient, that is, it always rounds up.
Now we have a specification, so we know we can come up with a testable design. Suppose we add an additional design criterion that the problem be solved solely with integer arithmetic, rather than computing the quotient as a double, since the "double" solution has been explicitly rejected in the problem statement.
So what must we compute? Clearly, to meet our spec while remaining solely in integer arithmetic, we need to know three facts. First, what was the integer quotient? Second, was the division free of remainder? And third, if not, was the integer quotient computed by rounding up or down?
Now that we have a specification and a design, we can start writing code.
public static int DivRoundUp(int dividend, int divisor)
{
if (divisor == 0 ) throw ...
if (divisor == -1 && dividend == Int32.MinValue) throw ...
int roundedTowardsZeroQuotient = dividend / divisor;
bool dividedEvenly = (dividend % divisor) == 0;
if (dividedEvenly)
return roundedTowardsZeroQuotient;
// At this point we know that divisor was not zero
// (because we would have thrown) and we know that
// dividend was not zero (because there would have been no remainder)
// Therefore both are non-zero. Either they are of the same sign,
// or opposite signs. If they're of opposite sign then we rounded
// UP towards zero so we're done. If they're of the same sign then
// we rounded DOWN towards zero, so we need to add one.
bool wasRoundedDown = ((divisor > 0) == (dividend > 0));
if (wasRoundedDown)
return roundedTowardsZeroQuotient + 1;
else
return roundedTowardsZeroQuotient;
}
Is this clever? No. Beautiful? No. Short? No. Correct according to the specification? I believe so, but I have not fully tested it. It looks pretty good though.
We're professionals here; use good engineering practices. Research your tools, specify the desired behaviour, consider error cases first, and write the code to emphasize its obvious correctness. And when you find a bug, consider whether your algorithm is deeply flawed to begin with before you just randomly start swapping the directions of comparisons around and break stuff that already works.
All the answers here so far seem rather over-complicated.
In C# and Java, for positive dividend and divisor, you simply need to do:
( dividend + divisor - 1 ) / divisor
Source: Number Conversion, Roland Backhouse, 2001
The final int-based answer
For signed integers:
int div = a / b;
if (((a ^ b) >= 0) && (a % b != 0))
div++;
For unsigned integers:
int div = a / b;
if (a % b != 0)
div++;
The reasoning for this answer
Integer division '/' is defined to round towards zero (7.7.2 of the spec), but we want to round up. This means that negative answers are already rounded correctly, but positive answers need to be adjusted.
Non-zero positive answers are easy to detect, but answer zero is a little trickier, since that can be either the rounding up of a negative value or the rounding down of a positive one.
The safest bet is to detect when the answer should be positive by checking that the signs of both integers are identical. Integer xor operator '^' on the two values will result in a 0 sign-bit when this is the case, meaning a non-negative result, so the check (a ^ b) >= 0 determines that the result should have been positive before rounding. Also note that for unsigned integers, every answer is obviously positive, so this check can be omitted.
The only check remaining is then whether any rounding has occurred, for which a % b != 0 will do the job.
Lessons learned
Arithmetic (integer or otherwise) isn't nearly as simple as it seems. Thinking carefully required at all times.
Also, although my final answer is perhaps not as 'simple' or 'obvious' or perhaps even 'fast' as the floating point answers, it has one very strong redeeming quality for me; I have now reasoned through the answer, so I am actually certain it is correct (until someone smarter tells me otherwise -furtive glance in Eric's direction-).
To get the same feeling of certainty about the floating point answer, I'd have to do more (and possibly more complicated) thinking about whether there is any conditions under which the floating-point precision might get in the way, and whether Math.Ceiling perhaps does something undesirable on 'just the right' inputs.
The path travelled
Replace (note I replaced the second myInt1 with myInt2, assuming that was what you meant):
(int)Math.Ceiling((double)myInt1 / myInt2)
with:
(myInt1 - 1 + myInt2) / myInt2
The only caveat being that if myInt1 - 1 + myInt2 overflows the integer type you are using, you might not get what you expect.
Reason this is wrong: -1000000 and 3999 should give -250, this gives -249
EDIT:
Considering this has the same error as the other integer solution for negative myInt1 values, it might be easier to do something like:
int rem;
int div = Math.DivRem(myInt1, myInt2, out rem);
if (rem > 0)
div++;
That should give the correct result in div using only integer operations.
Reason this is wrong: -1 and -5 should give 1, this gives 0
EDIT (once more, with feeling):
The division operator rounds towards zero; for negative results this is exactly right, so only non-negative results need adjustment. Also considering that DivRem just does a / and a % anyway, let's skip the call (and start with the easy comparison to avoid modulo calculation when it is not needed):
int div = myInt1 / myInt2;
if ((div >= 0) && (myInt1 % myInt2 != 0))
div++;
Reason this is wrong: -1 and 5 should give 0, this gives 1
(In my own defence of the last attempt I should never have attempted a reasoned answer while my mind was telling me I was 2 hours late for sleep)
Perfect chance to use an extension method:
public static class Int32Methods
{
public static int DivideByAndRoundUp(this int number, int divideBy)
{
return (int)Math.Ceiling((float)number / (float)divideBy);
}
}
This makes your code uber readable too:
int result = myInt.DivideByAndRoundUp(4);
You could write a helper.
static int DivideRoundUp(int p1, int p2) {
return (int)Math.Ceiling((double)p1 / p2);
}
You could use something like the following.
a / b + ((Math.Sign(a) * Math.Sign(b) > 0) && (a % b != 0)) ? 1 : 0)
For signed or unsigned integers.
q = x / y + !(((x < 0) != (y < 0)) || !(x % y));
For signed dividends and unsigned divisors.
q = x / y + !((x < 0) || !(x % y));
For unsigned dividends and signed divisors.
q = x / y + !((y < 0) || !(x % y));
For unsigned integers.
q = x / y + !!(x % y);
Zero divisor fails (as with a native operation).
Cannot overflow.
Elegant and correct.
The key to understanding the behavior is to recognize the difference in truncated, floored and ceilinged division. C#/C++ is natively truncated. When the quotient is negative (i.e. the operators signs are different) then truncation is a ceiling (less negative). Otherwise truncation is a floor (less positive).
So, if there is a remainder, add 1 if the result is positive. Modulo is the same, but you instead add the divisor. Flooring is the same, but you subtract under the reversed conditions.
By round up, I take it you mean away form zero always. Without any castings, use the Math.DivRem() function
/// <summary>
/// Divide a/b but always round up
/// </summary>
/// <param name="a">The numerator.</param>
/// <param name="b">The denominator.</param>
int DivRndUp(int a, int b)
{
// remove sign
int s = Math.Sign(a) * Math.Sign(b);
a = Math.Abs(a);
b = Math.Abs(b);
var c = Math.DivRem(a, b, out int r);
// if remainder >0 round up
if (r > 0)
{
c++;
}
return s * c;
}
If roundup means always up regardless of sign, then
/// <summary>
/// Divide a/b but always round up
/// </summary>
/// <param name="a">The numerator.</param>
/// <param name="b">The denominator.</param>
int DivRndUp(int a, int b)
{
// remove sign
int s = Math.Sign(a) * Math.Sign(b);
a = Math.Abs(a);
b = Math.Abs(b);
var c = Math.DivRem(a, b, out int r);
// if remainder >0 round up
if (r > 0)
{
c+=s;
}
return s * c;
}
Some of the above answers use floats, this is inefficient and really not necessary. For unsigned ints this is an efficient answer for int1/int2:
(int1 == 0) ? 0 : (int1 - 1) / int2 + 1;
For signed ints this will not be correct
The problem with all the solutions here is either that they need a cast or they have a numerical problem. Casting to float or double is always an option, but we can do better.
When you use the code of the answer from #jerryjvl
int div = myInt1 / myInt2;
if ((div >= 0) && (myInt1 % myInt2 != 0))
div++;
there is a rounding error. 1 / 5 would round up, because 1 % 5 != 0. But this is wrong, because rounding will only occur if you replace the 1 with a 3, so the result is 0.6. We need to find a way to round up when the calculation give us a value greater than or equal to 0.5. The result of the modulo operator in the upper example has a range from 0 to myInt2-1. The rounding will only occur if the remainder is greater than 50% of the divisor. So the adjusted code looks like this:
int div = myInt1 / myInt2;
if (myInt1 % myInt2 >= myInt2 / 2)
div++;
Of course we have a rounding problem at myInt2 / 2 too, but this result will give you a better rounding solution than the other ones on this site.