Incomprehensible infinite backward loop C++ - c++

I am new to C++ and I am trying to achieve a backward for loop, I founded solutions which works very well but I want to know what my version is not correct and making an infinite loop.
Here is a working version that I founded (I don't understand how we can decrease i in the condition..) :
for (unsigned i = size ; i-- > 0 ; )
{
// do stuff with i
}
Here is a version I wrote which works but don't go down to 0 (this way seems more logical to me) :
for (unsigned i = size-1 ; i > 0 ; i--)
{
// do stuff with i
}
If I say for exemple n=10, I will get this if I print i in the loop :
9
8
7
6
5
4
3
2
1
And here is the version which for me is the more logical and should go down to zero but is providing an infinite loop.
for (unsigned i = size-1 ; i >= 0 ; i--)
{
// do stuff with i
}
Could someone explain to me why the last version isn't working and what is the best choice to make ?

An unsigned number is always >= 0. (When it reaches zero a further decrement sets it to std::numeric_limits<unsigned>::max())
So your final for loop is equivalent to
for (unsigned i = size-1 ; true ; i--)
which, of course, loops forever.
In your first loop, you have i-- > 0 as the stopping condition. When i is zero, i-- is an expression with value zero (so the loop halts), despite the fact that i is then set to std::numeric_limits<unsigned>::max(). Some folk like (me included; cue the downvotes) to write i-->0 and regard --> as the slide operator. See What is the "-->" operator in C++?

The statement i >= 0 is always true because i is unsigned which means that is never below zero. If you decrease the value while i is zero, there will occur a so-called underflow and it will have a very high number.
The first version certainly gets the job done, so I would stick to it.

Related

Inifinite loop makes variable come out as 0

I have this piece of code in my school book.
#include<iostream>
using namespace std;
int main() {
int x=10,c=1;
while (c < 5) {
x += x*c;
c *= 2;
c++;
c -= 2;
cout << "X=" << x<<'\n';
}
system("pause");
return 0;
}
As you can see it's an infinite loop, when logically traced, it should show 20,40,80 and so on.
However it always shows 0.
when adding system("pause") after each loop cycle it shows the correct values, but when left as shown above (infinitely looping) it shows zero.
Any ideas of the reason?
c is always 1 no matter what. The loop becomes infinite. Eventually, X becomes 0 due to integer overflow.
c = 1
c *= 2; c = 2
c++; c = 3
c -= 2; c = 1 <-- infinite
Here is my answer for your questions:
Why do you get infinitely looping?
awesomeyi did answer you above, because the condition of the while loop is always true, so it is never ended.
Why does X always equal to 0?
Please pay your attention on X varable, its value will be increased after ending one loop x += x*c. Because you are in the infinitely loop, x's value will be increased forever until greater than the limited value of an integer variable. Then, the value will be set as zero. Please see my output when running your code.
Removing the pause doesn't cause it to always show zero. It just prints output so quickly that zeroes are all you see at the bottom. Add the pause back in and click through about 30-40 iterations and see if it helps you understand what is happening.

Having trouble understanding a portion of code (bit operation)

I can't understand how to count number of 1's in binary representation.
I have my code, and I hope someone can explain it for me.
Code:
int count (int x)
{
int nr=0;
while(x != 0)
{
nr+=x%2;
x/=2;
}
return nr;
}
Why while ? For example if i have 1011, it wouldn't stop at 0?
Why nr += x%2 ?
Why x/=2 ?!
First:
nr += x % 2;
Imagine x in binary:
...1001101
The Modulo operator returns the remainder from a / b.
Now the last bit of x is either a 0, in which case 2 will always go into x with 0 remainder, or a 1, in which case it returns a 1.
As you can see x % 2 will return (if the last bit is a one) a one, thus incrementing nr by one, or not, in which case nr is unchanged.
x /= 2;
This divides x by two, and because it is a integer, drops the remainder. What this means is is the binary was
....10
It will find out how many times 2 would go into it, in this case 1. It effectively drops the last digit of the binary number because in base 2 (binary) the number of times 2 goes into a number is just the same as 'shifting' everything down a space (This is a poor explanation, please ask if you need elaboration). This effectively 'iterates' through the binary number, allowing the line about to check the next bit.
This will iterate until the binary is just 1 and then half that, drop the remainder and x will equal 0,
while (x != 0)
in which case exit the loop, you have checked every bit.
Also:
'count`is possibly not the most descriptive name for a function, consider naming it something more descriptive of its purpose.
nr will always be a integer greater or equal to zero, so you should probably have the return type unsigned int
int count (int x)
{
int nr=0;
while(x != 0)
{
nr+=x%2;
x/=2;
}
return nr;
}
This program basically gives the numbers of set bits in a given integer.
For instance, lets start with the example integer 11 ( binary representation - 1011).
First flow will enter the while loop and check for the number, if it is equal to zero.
while(11 != 0)
Since 11 is not equal to zero it enter the while loop and nr is assigned the value 1 (11%2 = 1).nr += 11%2;
Then it executes the second line inside the loop (x = x/2). This line of code assigns the value 5 (11/2 = 5 ) to x.
Once done with the body of the while loop, it then again checks if x ie 5 is equal to zero.
while( 5 != 0).
Since it is not the case,the flow goes inside the while loop for the second time and nr is assigned the value 2 ( 1+ 5%2).
After that the value of x is divided by 2 (x/2, 5/2 = 2 )and it assigns 2 to x.
Similarly in the next loop, while (2 != 0 ), nr adds (2 + 2%2), since 2%2 is 0, value of nr remains 2 and value of x is decreased to 1 (2/2) in the next line.
1 is not eqaul to 0 so it enters the while loop for the third time.
In the third execution of the while loop nr value is increased to 3 (2 + 1%2).
After that value of x is reduced to 0 ( x = 1/2 which is 0).
Since it fails the check (while x != 0), the flow comes out of the loop.
At the end the value of nr (Which is the number of bits set in a given integer) is returned to the calling function.
Best way to understand the flow of a program is executing the program through a debugger. I strongly suggest you to execute the program once through a debugger.It will help you to understand the flow completely.

For loop doesn't stop at zero index

Sorry for a probably stupid question.
My loop does not stop at 0 and I have no idea why.
This loop will stop at i = 1
for (unsigned int i = 3 ; i > 0 ; i--)
Whereas this will stop at i = 4294967295 instead
for (unsigned int i = 3 ; i >= 0 ; i--)
Why? Do I miss anything?
In fact this i >= 0 for unsigned int is always true. So you cannot use it as loop invariant.
Use this instead:
for (unsigned int i = 4 ; i-- > 0 ;)
println(i);
it will print 3,2,1,0
A for loop iterates until the break condition evaluates to false.
In your case the loop will repeatedly check:
is i>0 (i>=0 respectively) true? If so, execute body and decrement i, else skip the loop-body.
Your first loop will at some point reach i=1. i>0 will evaluate to true. The body will be executed, i will be decremented. i will then be 0. Hence i>0 will evaluate to false and the body will not be executed and you will break out of the loop.
Your second loop will not stop (tested with g++). Why? Your i will reach 0. i>=0 will evaluate to true. The body will be executed, i will be decremented. But since you are using an unsigned int this will then assign 4294967295 to i. Since 4294967295>=0, the loop will continue, reach 0 again and again assign 4294967295 to i thus executing indefinitely.
This happens since an unsigned int represents 0 as all 0s in binary and subtracting 1 from it will yield all 1s which is interpreted as 4294967295.
How to avoid the problem? As mentioned in the comments a simple way out of it would be to use an int instead of an unsigned int.

Need to understand for loop better - post increment operator

I was not really clear with the post increment operator that I always used with for loops.
My latest and newly acquired understanding of post increment operator is the following:
int a = 5
int b = a++ //a will increment and return back its old value 5
so b = 5
Armed with this new knowledge i decided to understand/apply it to the places where i commonly used the post increment operator as in a for loop . Now it seems like I am lost
since I am ending up with the wrong output theoretically
Consider the following code
for(int i=0 ; i< 3 ; i++)
{
std::cout << i;
}
First loop
i starts with 0 which is less than 3 so ( increment by 1 however since its i++ it returns old value 0)
so cout should display 1 // But it displays 0
Second Loop
i is now 1 which is less than 3 so i++ is applied - Now i is 2 and returns back 1
so cout should display 2 //But it display 1
Third Loop
i is now 2 which is less than 3 so i++ is applied - Now i is 3 and returns back 2
so cout should display 3 //But it display 2
Fourth Loop
i is now 3 which is not less than 3 so loop exits
Could anyone please clear my understanding and point me in the right direction.
The output should be 0,1,2 where am i going wrong ?
What you're missing is when each of those sections of the for statement happen:
for (int i = 0 ; i < 3 ; i++)
// 111111111 22222 333
The first bit happens once before any iterations are done.
The second expression is evaluated before each potential iteration and, if false, no further iterations are done.
The third bit is done at the end of each iteration, before returning to evaluate the second bit.
Now re-read that last bullet point carefully. The i++ is done at the end of an iteration, after the cout << i. And, immediately after that, the continuation condition is checked (the second part).
So the loop is effectively the same as:
{ // Outer braces just to limit scope of i, same as for loop.
int i = 0;
while (i < 3) {
cout << i;
i++;
}
}
That's why you get 0 1 2.
The semicolons in a for loop delimit three different expressions. The value of the third expression is irrelevant to the behavior of the loop. You could replace i++ with ++i and your loop would behave the same; either way, i is incremented by 1.
To observe the behavior of the increment operators, consider the following loops:
for(int i = 0 ; i++ < 3 ; ) cout << i;
/* ^^^ Nothing here! */
Note that the third expression is empty. The output is 123—0 is skipped because the test, including increment, occurs before every iteration of the loop.
How does the postincremented loop behave compared to a similar one with the preincrement operator?
for(int i = 0 ; ++i < 3 ; ) cout << i;
Now the output is 12—3 is not printed because the conditional test sees the value 3 as soon as it is incremented to that, so the loop exits.
You just need to know that the statement int i = 0 is executed first , i.e , i takes first the value 0 , checks if the condition is true and then what the loop body is executed .
Then , i is incremented.

No checking condition in for loop C++

char i;
for (i = 1; i < 10, i++;)
{
cout << (i+1) << endl;
}
return 0;
I understand that for loop has the following syntax:
for(initialization; condition; increment)
{
}
As I run the debug, why it never checks the condition and it eventually stops at i = 0?
(Thank you Damien and Andars, I don't know why the "on purpose" statement was removed, but you interpret my question correctly. Could someone explain why the complier skips the condition before the comma, and why the loop stop at i = 0 instead of looping forever? Thanks!)
I believe he is indicating that he wrote the code that way on purpose.
To answer, i++ will always return true. The last value is the value that matters with comma separated statements, so the loop will not stop.
Edit for elaborations:
It isn't simply using a logical or, it disregards what is before the comma and only takes the last value. The value of anything non-zero is considered true, and since i starts at 1, and goes up, it will always be non-zero (until it overflows and wraps back around, which explains why i ends at 0).
If you say:
x = 4, 5, 6;
x will be equal to 6, the last value.
Change
for (i = 1; i < 10, i++;)
to
for (i = 1; i < 10; i++)
Change to
for (i = 1; i < 10; i++) //Notice correct placement of ;
It seems to me also that the code was written incorrectly on purpose. As others have mentioned, the comma operator will discard the value of the i<10 and only i++ will be evaluated as condition. This will return true until i overflows (values only from -127 to 127) and ends up at -1, when i++ will return 0 and the loop exits. Thus the final value for i will be 0.
Because you used the comma operator instead of a semi-colon