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can we assign value 0 in index x in array
I am trying to assign 0 at x place in array but if I write a[x]=0; then, it will take the value 0 at every index and I want to give 0 at only x th index( since x is Dynamics), if you know how to write it, please let me know
11 Réponses
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array[x] = 0; should only assign zero to the element having <x> as index, except you use <x> in a loop counter. But what did you mean by "x is dynamic"?
I suggest to share your code link in Description to be clear 👍
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#include <iostream>
#include<cmath>
using namespace std;
int main()
{ int s;
cin>>s;
while(s>0)
{ int temp,n,l,x,t,d,p,i,y;
cin>>n;
p =ceil(log2(n+1));
x=p;
if(p=1){
x=p+1;}
else if (p%2!=0){
x=p+1;}
bool arr[x];
bool ar[x];
temp=n,i=0;
while(temp>1)
{arr[i]=temp%2;
temp=temp/2;
i++;
}
arr[i]=1;
if(p%2!=0)
ar[p+1]=0;
for(l=0;l<=x;l=l+2)
{
t=arr[l];
ar[l]=arr[l+1];
ar[l+1]=t;
}
for(d=0;d<=x;d++)
{
y=y+(pow(2,d)*ar[d]);
}
cout<<y<<endl;
y=0;
s--;
}return 0;
}
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I posted a picture of elaborated question, can you please check that
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vishal shivhare
Check the code again bro, and see if this is really what you meant to do, it looks odd to me ...
if (p = 1)// p == 1?
{
x = p + 1;
}
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Ya bro that is the mistake, thanks for that, but the question I asked remains same ?
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Bro arr is storing the bits but in reverse order
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Yeah bro, I see that. It seems to be the problem. Because that way, the bits to be swapped are from wrong index.
And the way a number is formed by the lines
for (d = 0; d <= x; d++)
{
y = y + (pow(2, d) * ar[d]);
}
Starts from 0th bit, while actually it was not the 0th bit in the number.
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This is the code to swipe the bits of no. Then print the no.( Of changed bits)
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Swipe the bits? can you elaborate on that please? this is first time I hear bit swiping. An example might help to understand better 👌
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Can you tell me in which part of the code the problem is bro?
I am checking the code, but I see the <arr> isn't storing the bits as they should be. I think this is where the problem begins. You should check whether the bits in <arr> are correct. After this is good we can move on to swapping the bits 👍