I'm trying to get fizzbuzz working in D but I have no idea for the life of me what the problem is. I have tried reversing the logic and it does write both words when it is not appropriate, but when it is it just writes nothing.
Here's a screenshot of what the output looks like: http://puu.sh/p67Hd/2a5a598b1b.png
import std.stdio;
void main() {
for (uint i = 0; i < 100; i++) {
if (i%5 && i%3) write(i);
if (!i%3) write("Fizz");
if (!i%5) write("Buzz");
writeln();
}
}
The ! operator has precedence over % so your if statements look like
if ((!i) % 3) write("Fizz");
if ((!i) % 5) write("Buzz");
and because all of i is non-zero (besides the first time), !i is always 0 and 0 % 5 and 0 % 3 is always 0 (false).
To fix, all you need to do is add parenthesis around the % operations
if (!(i % 3)) write("Fizz");
if (!(i % 5)) write("Buzz");
The reason is the operator precedence in D.
The if(!i%3) will actually be interpreted as if((!i)%3), which results in either 0%3 (which is false), or 1%3 (which is true). !n will result in 1 if n is 0, otherwise it will always be 0. Because it's going from 0 to 100, (!i) % 3 will only be true once at the beginning. This is the reason why there is a FizzBuzz at the start of the output.
So instead your code should look like this:
import std.stdio;
void main() {
for (uint i = 0; i < 100; i++) {
if (i%5 && i%3) write(i);
if (!(i%3)) write("Fizz");
if (!(i%5)) write("Buzz");
writeln();
}
}
Related
How do I print a pattern with number of rows is equal to number of stars such that if 1 star then 1 row of 1 star, 2 star then 2 rows of 2 stars and so on, provided with good logic?
Here's the output for the logic mentioned above
*
**
**
***
***
***
using mod operator you can get proper output.Java has one important arithmetical operator you may not be familiar with, %, also known as the modulus or remainder operator. The % operator returns the remainder of two numbers. For instance 10 % 3 is 1 because 10 divided by 3 leaves a remainder of 1. You can use % just as you might use any other more common operator like + or -.
I write code in java.
Execution Steps:-
for loop for declare i value to 1.loop start with value 1 and last value 5.
for loop for declare i value to 1.in this we gave condition for j is i*i.
if j value is 1 then it print star.
if j value is not 1 then it goes to else part.
in this there are again if and else part in if part we use mod operator and if its value gives 0 value then print star otherwise goto else condition.
for(int i = 1; i <= 5; i++) {
for(int j = 1; j <= i * i; j++) {
if(j == 1) {
System.out.println();
System.out.print("*");
} else if(j % i == 0) {
System.out.print("*");
System.out.println();
} else {
System.out.print("*");
}
}
}
I am looking for a simple way to increment/decrement a number away from n, without using if statements, or creating a function. Here is an example:
Increment x from 9 to 10, n is 6
Decrement x from 3 to 2, n is 6
An obvious way to do this is with if statements, but that seems like too much code in my opinion. Here is a function that I could imagine using:
x += 1 * GetSign(6, 9) //GetSign(A, B) returns 1 or -1 depending on what would
Be necessary to move farther away from 6. The made up function above would look something like:
int GetSign(A, B)
{
if( A < B) return -1;
else return 1;
}
You can use the ternary operator:
int A = 6;
int B = 9;
x += (A < B) ? (-1) : (1);
I want something done for every second run of a for-loop. Is there a nice way to do this without using tons of lines?
And while we're at it, I would prefer a generic "trick", so doing something every third or forth time is possible as well?
Example:
for( int i = 0; i < 10; i++ )
{
do something;
if( i == odd number )
{
do something every alternating time;
}
}
Hope you can help.
Cheers.
To do something every n-th time:
if (i % n == 0) …
Note that this will trigger on the first, the n+1-th, the 2n+1-th, etc. iterations. If you want to trigger on the nth, 2n-th, etc., do this:
if (i % n == n - 1) …
if (i % 2)
is true for odd number i, false otherwise. If you prefer, you can compare with 1 explicitly:
if (i % 2 == 1)
Doing something n times is similar. For instance, every 4 times, starting from 1
if (i % 4 == 1)
I am trying to implement Long Hand Multiplication method for 8 bit binary numbers stored in two arrays BeforeDecimal1 and BeforeDecimal2. The problem is I always get the wrong result. I tried to figure out the issue but couldn't do it. Here is the code:
This is a much more refined code then previous one. It is giving me result but the result is not correct.
int i=0,carry=0;
while(true)
{
if(BeforeDecimal2[i]!=0)
for(int j=7;j>=0;j--)
{
if(s[j]==1 && BeforeDecimal1[j]==1 && carry==0)
{
cout<<"Inside first, j= "<<j<<endl;
carry=1;
s[j]=0;
}
else
if(s[j]==1 && BeforeDecimal1[j]==0 && carry==1)
{
cout<<"Inside second, j= "<<j<<endl;
carry=1;
s[j]=0;
}
else
if(s[j]==0 && BeforeDecimal1[j]==0 && carry==1)
{
cout<<"Inside third, j= "<<j<<endl;
carry=0;
s[j]=1;
}
else
if(s[j]==0 && BeforeDecimal1[j]==0 && carry==0)
{
cout<<"Inside fourth, j= "<<j<<endl;
carry=0;
s[j]=0;
}
else
if(s[j]==0 && BeforeDecimal1[j]==1 && carry==0)
{
cout<<"Inside fifth, j= "<<j<<endl;
carry=0;
s[j]=1;
}
else
if(s[j]==1 && BeforeDecimal1[j]==1 && carry==1)
{
//cout<<"Inside fifth, j= "<<j<<endl;
carry=1;
s[j]=1;
}
else
if(s[j]==1 && BeforeDecimal1[j]==0 && carry==0)
{
//cout<<"Inside fifth, j= "<<j<<endl;
carry=0;
s[j]=1;
}
else
if(s[j]==0 && BeforeDecimal1[j]==1 && carry==1)
{
//cout<<"Inside fifth, j= "<<j<<endl;
carry=1;
s[j]=0;
}
}
for(int h=7;h>=0;h--)
{
if(h==0)
{
BeforeDecimal1[0]=0; // that is inserting zeros from the right
}
else
{
BeforeDecimal1[h]=BeforeDecimal1[h-1];
BeforeDecimal1[h-1]=0;
}
}
if(i==3)
break;
i++;
}
Regards
Maybe it would be easiest to back up and start with 8-bit binary numbers stored as 8-bit binary numbers. Much like when we do decimal multiplication, we start with a number of digits. We take the values of multiplying by those individual digits, and add them together to get the final result. The difference (or one obvious difference) is this since we're working in binary, all our digits represent powers of two, so we can get each intermediate result by simply bit shifting the input.
Since it's binary, we have only two possibilities for each digit: if it's a 0, then we need to add 0 times the other number shifted left the appropriate number of places. Obviously, 0 time whatever is still 0, so we simply do nothing in this case. The other possibility is that we have a 1, in which case we add 1 times the other number shifted left the appropriate number of places.
For example, let's consider something like 17 x 5, or (in binary) 10001 x 101.
10001
101
------
10001
+ 1000100
--------
= 1010101
Converting that to something more recognizable, we get 0x55, or 85d.
In code, that process comes out fairly short and simple. Start with a result of 0. Check whether the least significant bit in one operand is set. If so, add the other operand to the result. Shift the one operand right a bit and the other left a bit, and repeat until the operand you're shifting to the right equals 0:
unsigned short mul(unsigned char input1, unsigned char input2) {
unsigned short result = 0;
while (input2 != 0) {
if (input2 & 1)
result += input1;
input1 <<= 1;
input2 >>= 1;
}
return result;
}
If you want to deal with signed numbers, it's generally easiest to figure up the sign of the result separately, and do the multiplication on the absolute values.
You have problem in following lines of code
if(reverse==0)
{
totalReverse=totalReverse-1;
reverse=totalReverse;
}
after some iterations of the inner for loop (index j based) the values of reverse goes should goes to negative and when reverse less than 3 then there should be exception thrown.
Are you running this code without exception handling?
to me this smells like shift and add. is there a requirement that you may use operations simulating logical gates only?
for your full adder you have 3 inputs s(s[j]), b(BeforeDecimal1[j]), c(carry), and two outputs ns(new s[j]), nc (new carry)
the table looks like this
s b c ns nc
0 0 0 0 0 handled in v5 clause 4
0 0 1 1 0 handled in v5 clause 3
0 1 0 1 0 handled in v6 clause 5
0 1 1 0 1
1 0 0 1 0
1 0 1 0 1 handled in v5 clause 2
1 1 0 0 1 handled in v5 clause 1
1 1 1 1 1
your code covers only 4 (now 5) of these 8 clauses
to avoid the ugly if-else-if rake i recommend to use temporary result variables (carry and s still valid in the next if clause)
when you analyze the table you could also do (pseudo bool notation)
nc = s && b || s && c || b && c;
ns = s XOR b XOR c; // there is no XOR in C++: axb = a&&!b || !a&&b
arithmetic notation
nc = (s + b + c) / 2;
ns = (s + b + c) % 2;
// [...]
for(int j=7;j>=0;j--)
{
// start changed code
const int sum = s[j] + BeforeDecimal1[j] + carry;
s[j]=sum % 2;
carry=sum / 2;
// end changed code
}
// [...]
here is a nice simulation of your problem Sequential Multiplication
Unless your requirement precisely states otherwise, which isn't clear from your question or any of your comments so far, it is not necessary to process arrays of bits. Arrays of bytes are much more efficient in both space and time.
You don't need this exhaustive explosion of cases either. The only special case is where either operand is zero, i.e. a[i]|b[i] == 0, when
result[i] = carry;
carry = 0;
All other cases can be handled by:
result[i] = a[i]*b[i]+carry;
carry = (result[i] >>> 8) & 1;
result[i] &= 0xff;
I don't see much point in the names BeforeDecimal1 and BeforeDecimal2 either.
What would be the output of this program ?
#include<stdio.h>
#include<conio.h>
void main()
{
clrscr();
int x=20,y=30,z=10;
int i=x<y<z;
printf("%d",i);
getch();
}
Actually i=20<30<10, so the condition is false and the value of i should be 0 but i equals 1. Why?
This int i=x<y<z; doesn't work the way you intended.
The effect is int i=(x<y)<z;, where x<yis evaluated first, and the value true is then compared to z.
Pascal points out below that in C the result of the comparison is 1 instead of true. However, the C++ true is implicitly converted to 1 in the next comparison, so the result is the same.
The comparison operators don't work like that. Your program is equivalent to:
i = (x < y) < z;
which is equivalent to:
i = (x < y);
i = i < z;
After the first operation, i == 1. So the second operation is equivalent to:
i = 1 < 10;
You need to rewrite your statement as:
i = (x < y) && (y < z);
The < operator has left-to-right associativity. Therefore x<y<z will do (x<y)<z. The result of the first parenthesis is 1, 1 is smaller than 10, so you'll get 1.
That's not how it works. It's better to see with parenthesis:
int i = (x<y)<z;
Now, first x<y is evaluated. It's true, 20<30, and true is 1 as an integer. 1<z is then true again.
Its precedence is from left to right. Thats is why it is like
20<30 = true
1<10 TRUE
SO FINALLY TRUE
Actually < is left-associative, so first, 20<30 is evaluated (giving 1 usually), then 1 is less than 10.
The output of "1" is correct. This is evaluated as (20<30) < 10, which is 1 < 10, which is 1.
The problem is that you are comparing a boolean value to an integer value which in most cases doesn't make sense.
< is evaulated from left to right, so 20<30 is true, or one, which is less than 10.
The operator < associates from left to right.
So x < y < z is same as ( x < y ) < z
Your expression evaluates as:
( x < y ) < z
= ( 20 < 30 ) < 10
= ( 1 ) < 10
= 1