if(y-- > 0 && matriz[x][y--]!=9 && matriz[x][y--]!=0)
When i'm doing this, my variable y is not decremented, right? And i can do this comparison y-- > 0?
Imagine that i have y=0, i want to compare if -1 > 0.
Thanks
Assuming all tests pass, your code is effectively equivalent to this:
if (y > 0)
{
y--; // post-decrement
if (matriz[x][y] != 9)
{
y--; // post-decrement
if (matriz[x][y] != 0)
{
y--; // post-decrement
// all true
}
}
}
If any fail, you still decrement (equivalence would be in an else-clause), but I'm omitting that for brevity.
Hopefully this is enough to clarify. I find it quite unwieldy, and if y < 2 at the start you'll end up with a negative index, which is probably a bad thing.
If y=0, your final two conditions will not be evaluated, since the first condition will fail.
If you want your first comparison to be -1>0, you need a prefix decrement: --y > 0
Yes you can do the comparison. Why are you writing such code? Please Please write could that is readable and that is understandable.
Related
I have noticed that there is a precedence of assignment while using if-else conditionals in Verilog. For example as in the code below:
if(counter < 6)
z <= 1;
else if(counter < 12)
z <= 2;
else
z <= 3;
I noticed that until the counter is less than 6, the value of z is assigned the value of 1 (z <= 1) and as soon as the value of the counter exceeds 6 and is less than 12, z is assigned the value 2 (z <= 2).
What if there are different variables used inside the conditionals as in the code below?
if(wire1_is_enabled)
z <= 1;
else if(wire2_is_enabled)
z <= 0;
What happens when both conditions are true? What is the behaviour of the assignment operator here?
I believe this is poor programming habit.
Yes, nested if-else branching statements naturally assume a priority in the order of execution. Think about using a case statement instead of deeply nesting if statements which are much more readable.
There is nothing wrong with this coding style unless you have code like:
if(counter < 6)
z <= 1;
else if(counter < 2)
z <= 2; // unreachable
else
z <= 3;
Then the statement z <= 2; becomes unreachable because when the first condition is false, the second condition can never be true. Fortunately there are a number of tools that can flag this problem for you.
Both if and case statements assume priority logic. However, you have an option to explicitly add a priority or unique keyword before the if or case keywords to declare and check your intent. See sections 12.4 and 12.5 in the IEEE 1800-2017 SystemVerilog LRM.
The 2 if/else statements behave the same way; the first condition to be true has the highest priority. Once a condition evaluates to true, all the following else clauses are ignored. Therefore, z <= 1 if both wire1_is_enabled and wire2_is_enabled are true. This is easy to prove to yourself with a simple simulation.
This is not a poor coding habit. This situation is common in Verilog. When you say programming, perhaps you are thinking of software languages. Keep in mind that these are hardware signals instead of software variables.
Any idea why the else if statment will be never executed ? The value of difference is constantly changing when the program runs.
double difference = abs(reale_x[0] - reale_x[1]);
if (0 <= difference < 45) {
timer_counter += 1;
if (timer_counter == 30) {
cout << "CLICK" << '\n';
}
}
else if (difference > 50) {
timer_counter = 0;
}
That is not how comparation works in c++.
What this code
if (0 <= difference < 45) {
does is it first compares if 0 is smaller or equal to difference. It is then "replaced" by a bool value either true or false. And then a bool value (so either 1 or 0) is compared to 45. And it will always be smaller than 45. What you have there is an always true statement.
So the way you would write this if statement is
if (difference >= 0 && difference < 45){
Note that because of your else if statement it will not execute if the difference is >44 and <51
if (0 <= difference < 45) will be executed as if ((0 <= difference) < 45), which will be either 0<45 or 1<45 and will always be true. That's why the else part is not getting executed.
in mathematics, we see and write 0 <= x < 45 or something like that to define the range of the variable x. But in order to tell the computer the same thing, you have to tell more clearly. Saying, to have to tell the compiler, that the value of x is greater than or equal to zero and at the same time, that value will be less than 45, and you can tell the compiler by this statement: difference >= && difference < 45 . the && is an 'AND' operator in most of the languages.
#include <iostream>
int GCD()
{
int a,b,k;
cout<<"Enter a and b"<<endl;
cin>>a>>b;
cout<<endl;
if (a>b)
{
k=a;
}
else
{
k=b;
}
cout<<k<<endl;
do
{
k=k-1;
} while(a%k !=0 && b%k !=0);
cout<<k<<endl;
return 0;
}
Why programm like this doesnt work correctly? For example when i enter 125 and 5 answer is 25, but supposed to be 5? Am wrong with logic in while loop? As i understood problem is in modulus operator. When k hits 25 it says that 125%25=0 and 5%=25=0. How can i fix this?
You have some mistakes here:
The GCD is lower or equal to the lower number. Currently, you start checking with the larger number. You need to flip the if block to if (a<b). (not exactly an error, but you check much more numbers than needed)
You need to check if the inital k is the GCD. When using a do {} while() the first number you check is k-1. Use a simple while instead. Also the loop condition has a logic flaw.
while (!((a % k == 0) && (b % k == 0)))
{
k--;
}
Note that the brackets around the modulo are not neccessary, but improve readability a bit.
Your code will not compile under all compilers and you should not omit the namespace std::.
Your while statement has logic error. It needs to be
while(!(a%k == 0 && b%k == 0));
When k is equal to 25, 125%25==0 so in your while statement a%k !=0 part is equals to false so it exit your do-while but it needs to test if b%k is equal to 0 or not!
Also your implementation tends to execute slow when a and b is big. You can take a look efficent solutions.
I am attempting to solve a hw problem in which I need to write down what the program will output. I am stuck however, on the syntax "if ( !(i%3)). What does that really mean? Does it mean that the program is checking for any i that is divisible by three? aka, is the if statement only runs if i is divisible by three?
int main () {
for (int i=0; i<10; (i<3?i++;i+=2)) {
if (!(i%3)) {
continue;
}
else if (i%7 ==0) {
break;
}
cout << i<< endl;
}
Does it mean that the program is checking for any i that is divisible by three? aka, is the if statement only runs if i is divisible by three?
Correct. The longer version of that check would be
if (i % 3 == 0)
continue;
The most common use case for such branching is probably FizzBuzz.
İt means if i is not(!) divisible by 3 continue.
For example if i is 3,6,9 it won't continue otherwise it will continue.
if (x) where x is int implicitly compared with zero. I.e. if (x) equals to if (x != 0). ! is negation. So if (!x) equals to if (x == 0). And the last step is if (!(i%3)) equals to if ((i%3) == 0) what is the same with check, that i deivisible by 3
The if() statement is false only if the result inside the parentheses is 0 (false). Take a look at your program:
i%3 may return 0 (false), 1 (true), or 2 (true)
The negation operator ! changes the result of the operation (i%3). So, if the i is divisible with 3 the statement will return 0 (false). Being negate, ! it will result in True. Otherwise the result of (i%3) will be true and with the operator ! the result of the hole statement will be false. Basically this code is checking if the value of i is divisible with 3.
Other options will be:
if (0==i%3)
{
/*code*/
}
Your code can be simplified as below
int main() {
for (int i=0; i<10;)
{
if (i % 3 == 0) {
continue;
}
else if (i % 7 == 0) {
break;
}
cout << i << endl;
i = i<3 ? i+1 : i+2;
}
}
When you write a integer variable like i as a condition, what happens is that if i==0 then the result of the condition is false, otherwise it would be true.
Let's check it out in your program, if(!(x%3)), let's name condition= !(x%3), when this condition is true? when x%3 == 0, note that the negation operator ! is behind x%3, so in this case the condition would be equal to true, more formally the condition is equal to :
if(x%3==0)
these kinds of conditions are common, check this example out :
int t = 10;
while(t--){
cout<<t<<endl;
}
The above condition i.e if(!(i%3)) will true when " i is not disvisable by 3"
Hope this helps.
In java and other languages there is a special type to represent booleans and evaluate expressions; in c and its variants there is no such thing, instead an expression is considered "true" if -taken as a integer number- is equal to 0; false for every other value. (Fun fact: this is why you usually end the code by return 0)
So if(x%3) in c is equivalent to if(x%3==0) in other languages. That said, if(x%3) execute the if body when the remainder of x/3 is 0, that is when x is a multiple of 3.
In your code you have the ! before, that -as you may know- "inverts" the expression. That means that if(!(x%3)) can be read as "If the remainder of the integer division of x by 3 is not 0", or alternatively: "If x is not a multiple of 3".
So, basically, you saw it right.
Hope I helped!
while ( (i=t-i%10 ? i/10 : !printf("%d\n",j)) || (i=++j<0?-j:j)<101 );
I came across this on codegolf
Please explain the usage of ? and : and why is there no statement following the while loop? As in why is there a ; after the parenthesis.
There is a boolean operation going on inside the parentheses of the while loop:
while (boolean);
Since the ternary operator is a boolean operator, it's perfectly legal.
So what's this doing? Looks like modular arithmetic, printing going on over a range up to 101.
I'll agree that it's cryptic and obscure. It looks more like a code obfuscation runner up. But it appears to be compilable and runnable. Did you try it? What did it do?
The ?: is a ternary operator.
An expression of form <A> ? <B> : <C> evaluates to:
If <A> is true, then it evaluates to <B>
If <A> is false, then it evaluates to <C>
The ; after the while loop indicates an empty instruction. It is equivalent to writing
while (<condition>) {}
The code you posted seems like being obfuscated.
Please explain the usage of ? and :
That's the conditional operator. a ? b : c evaluates a and converts it to a boolean value. Then it evaluates b if its true, or c if its false, and the overall value of the expression is the result of evaluating b or c.
So the first sub-expression:
assigns t-i%10 to i. The result of that expression is the new value of i.
if i is not zero, the result of the expression is i/10
otherwise, print j, and the result of the expression is zero (since printf returns a non-zero count of characters printed, which ! converts to zero).
Then the second sub-expression, after ||, is only evaluated if the result of the first expression was zero. I'll leave you to figure out what that does.
why is there no statement following the while loop?
There's an empty statement, ;, so the loop body does nothing. All the action happens in the side effects of the conditional expression. This is a common technique when the purpose of the code is to baffle the reader; but please don't do this sort of thing when writing code that anyone you care about might need to maintain.
This is the Conditional Operator (also called ternary operator).
It is a one-line syntax to do the same as if (?) condition doA else (:) doB;
In your example:
(i=t-i%10 ? i/10 : !printf("%d\n",j)
Is equivalent to
if (i=t-i%10)
i/10;
else
!printf("%d\n",j);
?: is the short hand notation for if then else
(i=t-i%10 ? i/10 : !printf("%d\n",j)<br>
equals to
if( i= t-i%10 )
then { i/10 }
else { !printf("%d\n",j) }
Your while loop will run when the statement before the || is true OR the statement after the || is true.
notice that your code does not make any sense.
while ( (i=t-i%10 ? i/10 : !printf("%d\n",j)) || (i=++j<0?-j:j)<101 );
in the most human-readable i can do it for u, it's equivalent to:
while (i < 101)
{
i = (t - i) % 10;
if (i > 0)
{
i = i / 10;
}
else
{
printf("%d\n",j);
}
i = ++j;
if (i < 0)
{
i = i - j;
}
else
{
i = j;
}
}
Greetings.
I am the proud perpetrator of that code. Here goes the full version:
main()
{
int t=getchar()-48,i=100,j=-i;
while ((i=t-i%10?i/10:!printf("%d\n",j)) || (i=++j<0?-j:j)<101 );
}
It is my submission to a programming challenge or "code golf" where you are asked to create the tinniest program that would accept a digit as a parameter and print all the numbers in the range -100 to 100 that include the given digit. Using strings or regular expressions is forbidden.
Here's the link to the challenge.
The point is that it is doing all the work into a single statement that evaluates to a boolean. In fact, this is the result of merging two different while loops into a single one. It is equivalent to the following code:
main()
{
int i,t=getchar()-'0',j=-100;
do
{
i = j<0? -j : j;
do
{
if (t == i%10)
{
printf("%d\n",j);
break;
}
}
while(i/=10);
}
while (j++<100);
}
Now lets dissect that loop a little.
First, the initialisation.
int t=getchar()-48,i=100,j=-i;
A character will be read from the standard input. You are supposed to type a number between 0 and 9. 48 is the value for the zero character ('0'), so t will end up holding an integer between 0 and 9.
i and j will be 100 and -100. j will be run from -100 to 100 (inclusive) and i will always hold the absolute value of j.
Now the loop:
while ((i=t-i%10?i/10:!printf("%d\n",j)) || (i=++j<0?-j:j)<101 );
Let's read it as
while ( A || B ) /* do nothing */ ;
with A equals to (i=t-i%10?i/10:!printf("%d\n",j)) and B equals to (i=++j<0?-j:j)<101
The point is that A is evaluated as a boolean. If true, B won't be evaluated at all and the loop will execute again. If false, B will be evaluated and in turn, if B is true we'll repeat again and once B is false, the loop will be exited.
So A is the inner loop and B the outer loop. Let's dissect them
(i=t-i%10?i/10:!printf("%d\n",j))
It's a ternary operator in the form i = CONDITION? X : Y; It means that first CONDITION will be evaluated. If true, i will be set to the value of X; otherwise i will be set to Y.
Here CONDITION (t-i%10) can be read as t - (i%10). This will evaluate to true if i modulo 10 is different than t, and false if i%10 and t are the same value.
If different, it's equivalent to i = i / 10;
If same, the operation will be i = !printf("%d\n",j)
If you think about it hard enough, you'll see that it's just a loop that checks if any of the decimal digits in the integer in i is equal to t.
The loop will keep going until exhausting all digits of i (i/10 will be zero) or the printf statement is run. Printf returns the number of digits printed, which should always be more than zero, so !printf(...) shall always evaluate to false, also terminating the loop.
Now for the B part (outer loop), it will just increment j until it reaches 101, and set i to the absolute value of j in the way.
Hope I made any sense.
Yes, I found this thread by searching for my code in google because I couldn't find the challenge post.