Recently I have some across this solution on codeforces site. I couldn't understand the condition in the if statement and also the condition of the ternary operator in this question can someone please help me with this ?
#include<stdio.h>
int main(void)
{
int a[1000]={0},ans,k;
while((k=getchar())!='\n'){
if(!a[k]){
ans++;
a[k]=1;
}
}
puts(ans&1 ? "IGNORE HIM!":"CHAT WITH HER!");
return 0;
}
It's a bizarre piece of code, I'm not sure entirely why it does what it does but it's easy enough to understand from a technical viewpoint (i.e., what it does).
The while loop gets a series of characters from standard input and, for each unique one, adds one to the ans variable. The expression !a[k] will be true if a[k] is zero (they're all initialised to this state).
When that condition is true, ans is incremented and a[k] is set to 1, meaning any more characters of that value will not affect the outcome (the condition won't ever be true again).
In terms of the if statement, the expression ans&1 will be true if ans is odd (has its lower-order bit set).
So it appears it tells you to ignore people who have an odd number of unique characters in their names and talk to the others. Of course, the whole thing falls apart since you don't actually initialise ans, meaning it can be an arbitrary value (not necessarily zero) and that this program is therefore pretty much able to tell you whatever it wants.
Once you've fixed that little issue, I'll be happy to chat further - paxdiablo has eight unique characters. I should warn you in advance though, I'm not a "HER" :-)
int a[1000] = {0};
This is a way to fill a[1000] with 0s.
If an array is partially initialized, elements that are not initialized receive the value 0 of the appropriate type (https://stackoverflow.com/a/2589751/3235496).
if (!a[k])
checks if k is an "already seen character", thus skipping further increments of ans.
x & 1
Finds if x is even or odd (checking the last bit). You can take a look at What is the fastest way to find if a number is even or odd? for further details.
So
puts(ans&1 ? "IGNORE HIM!" : "CHAT WITH HER!");
prints IGNORE HIM when ans is odd.
Related
I am getting an output of 24 which is the factorial for 4, but I should be getting the output for 5 factorial which is 120
#include <stdio.h>
int factorial(int number){
if(number==1){
return number;
}
return number*factorial(--number);
}
int main(){
int a=factorial(5);
printf("%d",a);
}
Your program suffers from undefined behavior.
In the first call to factorial(5), where you have
return number * factorial(--number);
you imagine that this is going to compute
5 * factorial(4);
But that's not guaranteed!
What if the compiler looks at it in a different order?
What it if works on the right-hand side first?
What if it first does the equivalent of:
temporary_result = factorial(--number);
and then does the multiplication:
return number * temporary_result;
If the compiler does it in that order, then temporary_result will be factorial(4), and it'll return 4 times that, which won't be 5!. Basically, if the compiler does it in that order -- and it might! -- then number gets decremented "too soon".
You might not have imagined that the compiler could do things this way.
You might have imagined that the expression would always be "parsed left to right".
But those imaginations are not correct.
(See also this answer for more discussion on order of evaluation.)
I said that the expression causes "undefined behavior", and this expression is a classic example. What makes this expression undefined is that there's a little too much going on inside it.
The problem with the expression
return number * factorial(--number);
is that the variable number is having its value used within it, and that same variable number is also being modified within it. And this pattern is, basically, poison.
Let's label the two spots where number appears, so that we can talk about them very clearly:
return number * factorial(--number);
/* A */ /* B */
At spot A we take the value of the variable number.
At spot B we modify the value of the variable number.
But the question is, at spot A, do we get the "old" or the "new" value of number?
Do we get it before or after spot B has modified it?
And the answer, as I already said, is: we don't know. There is no rule in C to tell us.
Again, you might have thought there was a rule about left-to-right evaluation, but there isn't. Because there's no rule that says how an expression like this should be parsed, a compiler can do anything it wants. It can parse it the "right" way, or the "wrong" way, or it can do something even more bizarre and unexpected. (And, really, there's no "right" or "wrong" way to parse an undefined expression like this in the first place.)
The solution to this problem is: Don't do that!
Don't write expressions where one variable (like number) is both used and modified.
In this case, as you've already discovered, there's a simple fix:
return number * factorial(number - 1);
Now, we're not actually trying to modify the value of the variable number (as the expression --number did), we're just subtracting 1 from it before passing the smaller value off to the recursive call.
So now, we're not breaking the rule, we're not using and modifying number in the same expression.
We're just using its value twice, and that's fine.
For more (much more!) on the subject of undefined behavior in expressions like these, see Why are these constructs using pre and post-increment undefined behavior?
How to find the factorial of a number;
function factorial(n) {
if(n == 0 || n == 1 ) {
return 1;
}else {
return n * factorial(n-1);
}
//return newnum;
}
console.log(factorial(3))
I saw this question in a test in which we have to tell the output of the following code.
#include<stdio.h>
int main(){
int k = 0;
while(+(+k--)!=0)
k=k++;
printf("%d\n", k);
return 0;
}
The output is -1. I am unsure why this is the answer, though.
What does the expression +(+k--) mean in C?
This code is deeply, perhaps deliberately, confusing. It contains a narrowly-averted instance of the dread undefined behavior. It's hard to know whether the person who constructed this question was being very, very clever or very, very stupid. And the "lesson" this code might purport to teach or quiz you about -- namely, that the unary plus operator doesn't do much -- is not one that's important enough, I would think, to deserve this kind of subversive misdirection.
There are two confusing aspects of the code, the strange condition:
while(+(+k--)!=0)
and the demented statement it controls:
k=k++;
I'm going to cover the second part first.
If you have a variable like k that you want to increment by 1, C gives you not one, not two, not three, but four different ways to do it:
k = k + 1
k += 1
++k
k++
Despite this bounty (or perhaps because of it), some programmers get confused and cough out contortions like
k = k++;
If you can't figure out what this is supposed to do, don't worry: no one can. This expression contains two different attempts to alter k's value (the k = part, and the k++ part), and because there's no rule in C to say which of the attempted modifications "wins", an expression like this is formally undefined, meaning not only that it has no defined meaning, but that the whole program containing it is suspect.
Now, if you look very carefully, you'll see that in this particular program, the line k = k++ doesn't actually get executed, because (as we're about to see) the controlling condition is initially false, so the loop runs 0 times. So this particular program might not actually be undefined -- but it's still pathologically confusing.
See also these canonical SO answers to all questions concerning Undefined Behavior of this sort.
But you didn't ask about the k=k++ part. You asked about the first confusing part, the +(+k--)!=0 condition. This looks strange, because it is strange. No one would ever write such code in a real program. So there's not much reason to learn how to understand it. (Yes, it's true, exploring the boundaries of a system can help you learn about its fine points, but there's a line in my book between imaginative, thought-provoking explorations versus dunderheaded, abusive explorations, and this expression is pretty clearly on the wrong side of that line.)
Anyway, let's examine +(+k--)!=0. (And after doing so, let's forget all about it.) Any expression like this has to be understood from the inside out. I presume you know what
k--
does. It takes k's current value and "returns" it to the rest of the expression, and it more or less simultaneously decrements k, that is, it stores the quantity k-1 back into k.
But then what does the + do? This is unary plus, not binary plus. It's just like unary minus. You know that binary minus does subtraction: the expression
a - b
subtracts b from a. And you know that unary minus negates things: the expression
-a
gives you the negative of a. What unary + does is... basically nothing. +a gives you a's value, after changing positive values to positive and negative values to negative. So the expression
+k--
gives you whatever k-- gave you, that is, k's old value.
But we're not done, because we have
+(+k--)
This just takes whatever +k-- gave you, and applies unary + to it again. So it gives you whatever +k-- gave you, which was whatever k-- gave you, which was k's old value.
So in the end, the condition
while(+(+k--)!=0)
does exactly the same thing as the much more ordinary condition
while(k-- != 0)
would have done. (It also does the same thing as the even more complicated-looking condition while(+(+(+(+k--)))!=0) would have done. And those parentheses aren't really necessary; it also does the same thing as while(+ +k--!=0) would have done.)
Even figuring out what the "normal" condition
while(k-- != 0)
does is kind of tricky. There are sort of two things going on in this loop: As the loop runs potentially multiple times, we're going to:
keep doing k--, to make k smaller and smaller, but also
keep doing the body of the loop, whatever that does.
But we do the k-- part right away, before (or in the process of) deciding whether to take another trip through the loop. And remember that k-- "returns" the old value of k, before decrementing it. In this program, the initial value of k is 0. So k-- is going to "return" the old value 0, then update k to -1. But then the rest of the condition is != 0 -- and it's not true that 0 != 0. That is, 0 is equal to 0, so we won't make any trips through the loop, so we won't try to execute the problematic statement k=k++ at all.
In other words, in this particular loop, although I said that "there are sort of two things going on", it turns out that thing 1 happens one time, but thing 2 happens zero times.
At any rate, I hope it's now adequately clear why this poor excuse for a program ends up printing -1 as the final value of k. Normally, I don't like to answer quiz questions like this -- it feels like cheating -- but in this case, since I fundamentally disagree with the whole point of the exercise, I don't mind.
At first glance it looks like this code invokes undefined behavior however that is not the case.
First let's format the code correctly:
#include<stdio.h>
int main(){
int k = 0;
while(+(+k--)!=0)
k=k++;
printf("%d\n", k);
return 0;
}
So now we can see that the statement k=k++; is inside of the loop.
Now let's trace the program:
When the loop condition is first evaluated, k has the value 0. The expression k-- has the current value of k, which is 0, and k is decremented as a side effect. So after this statement the value of k is -1.
The leading + on this expression has no effect on the value, so +k-- evaluated to 0 and similarly +(+k--) evaluates to 0.
Then the != operator is evaluated. Since 0!=0 is false, the body of the loop is not entered. Had the body been entered, you would invoke undefined behavior because k=k++ both reads and writes k without a sequence point. But the loop is not entered, so no UB.
Finally the value of k is printed which is -1.
Here's a version of this that shows operator precedence:
+(+(k--))
The two unary + operators don't do anything, so this expression is exactly equivalent to k--. The person that wrote this most likely was trying to mess with your mind.
I got the task in university to realize an input of a maximum of 10 integers, which shall be stored in a one dimensional vector. Afterwards, every integer of the vector needs to be displayed on the display (via printf).
However, I don't know how to check the vector for each number. I thought something along the lines of letting the pointer of the vector run from 0 to 9 and comparing the value of each element with all elements again, but I am sure there is a much smarter way. I don't in any case know how to code this idea since I am new to C.
Here is what I have tried:
#include <stdio.h>
int main(void)
{
int vector[10];
int a;
int b;
int c;
a = 0;
b = 0;
c = 0;
printf("Please input 10 integers.\n\n");
while (a <= 10);
{
for (scanf_s("%lf", &vektor[a]) == 0)
{
printf("This is not an integer. Please try again.\n");
fflush(stdin);
}
a++;
}
for (b <= 10);
{
if (vector[b] != vector[c]);
{
printf("&d", vector[b]);
c++;
}
b++;
}
return 0;
}
Your code has several problems, some syntactic and some semantic. Your compiler will help with many of the former kind, such as
misspelling of variable name vector in one place (though perhaps this was a missed after-the-fact edit), and
incorrect syntax for a for loop
Some compilers will notice that your scanf format is mismatched with the corresponding argument. Also, you might even get a warning that clues you in to the semicolons that are erroneously placed between your loop headers and their intended bodies. I don't know any compiler that would warn you that bad input will cause your input loop to spin indefinitely, however.
But I guess the most significant issue is that the details of your approach to printing only non-duplicate elements simply will not serve. For this purpose, I recommend figuring out how to describe in words how the computer (or a person) should solve the problem before trying to write C code to implement it. These are really two different exercises, especially for someone whose familiarity with C is limited. You can reason about the prose description without being bogged down and distracted by C syntax.
For example, here are some words that might suit:
Consider each element, E, of the array in turn, from first to last.
Check all the elements preceding E in the array for one that contains the same value.
If none of the elements before E contains the same value as E then E contains the first appearance of its value, so print it. Otherwise, E's value was already printed when some previous element was processed, so do not print it again.
Consider the next E, if any (go back to step 1).
I am a beginner in C. I am trying to print out the length and the number of words in a string inputted by user. I have trouble dealing with the counting of words. I attempt to scan the number of the space character , then add 1 to the result. It is guaranteed that each word is only separated by one space.
I executed the code, and inputted
I go to school by bus.
The second output was not as I expected
22
24
22 is correct since it is indeed the length of the string, but I do not understand why it prints 24 instead of 6 (result of 5 spaces plus 1).
The code is as follows
#include<stdio.h>
#include<string.h>
int main(){
char in[256];
int cl=0,cw,i;
scanf("%[^\n]",&in);
cw=strlen(in);
for(i=0;i<=cw;i++){
if(in[i]=' '){
cl=cl+1;
}
}
printf("%d\n",cw);
printf("%d\n",cl+1);
return 0;
}
What went wrong, and how can I get a correct output?
if(in[i]=' '){
will always be true, you need to change it to
if(in[i]==' '){
2 mistakes I see in your code.
if(in[i]=' ') should be if(in[i]==' ')
== Checks if the values of two operands are equal or not. If yes, then the condition becomes true.
= Simple assignment operator. Assigns values from right side operands to left side operand.
i<=cw should be i<cw
indexing starts with 0 in C.
As Pras said, on ifs you normally use ==, because you are comparing and it returns true or false
If you just use = it attributes (insert) a value on the right to the left.
You should use == instead of = in the if statement that detects the space.
The assignment operator, = gives is always true if the number is asssigned to a nonzero value.
The equivalence operator, == is what actually compares whether two numbers have the same value.
It may be surprising that the assignment operator can actually test whether or not a number is zero(at least on x860. This is because jz, one of the testing instructions, checks something called a flag register (a special memory location) that contains information about the arithmetic operation most recently executed. == does a cmp, which subtracts both numbers from each other to see if it is 0. Jz stands for "jump if zero", so two equal numbers result to the "jump" to the instructions in the if statement. When = is used, the compiler uses jz instead (in order to comply with the c idea that nonzero number in if is true), which results in the behavior described above.
I am trying to modify a value in an array using the C programming language and I seem to be hitting a blank wall with this seemingly easy operation. Please see code snippet below:
while(1) {
printf("Current prime candidate is %i\n",nextPrimeCandidate);
int innerSieve;//=2;
int currentPrimeCandidate=0;
for (innerSieve=2;innerSieve<SIEVELIMIT;innerSieve++) {
currentPrimeCandidate = nextPrimeCandidate * innerSieve;
//printf("Inner Sieve is b4 funny place %i,%i\n",innerSieve,currentPrimeCandidate);
//initArray[currentPrimeCandidate]=5;
//VERY UNIQUE LINE
myArray[currentPrimeCandidate] = 0;
//printf("Inner Sieve after funny place is %i,%i \n",innerSieve,currentPrimeCandidate);
}
nextPrimeCandidate=getNextPrimeCandidate(myArray,++nextPrimeCandidate);
if ((nextPrimeCandidate^2) > SIEVELIMIT ) break;
}
The problem is with the line highlighted with the VERY UNIQUE LINE comment. For some reason, when the innerSieve variable reaches 33 and gets to that line, it sets the contents of the innerSieve variable to the value of that line ( which currently is 0) and basically forces the loop into an infinite loop( the SIEVELIMIT variable is set at 50). It seems that there is some funny stuff going on in the registers when I checked using the Eclipse Debug facility but I am not too sure what I should be looking for.
If you need the whole code listing, this can be provided.( with a particular variable which is not yet initialised in the code being initialised at the precise point that the innerSieve variable hits 32)
Any help will be greatly appreciated.
Guessing that currentPrimeCandidate is greater than the maximum index of myArray, and you're overwriting innerSieve (which likely follows myArray on the stack).
#ruslik hit on it in the comment. The problem is this line:
if ((nextPrimeCandidate^2) > SIEVELIMIT ) break;
In C, the ^ operator is not the power operator, it is the bitwise xor operator. You're iterating far too many times than you intend, which is resulting in an array-index-out-of-bounds error, so you're overwriting random memory and getting strange results.
There is no power operator in C (though there is the pow function). Since you're just squaring a number, the simplest fix is to multiply the number by itself:
if ((nextPrimeCandidate * nextPrimeCandidate) > SIEVELIMIT ) break;