@Saurabh:
You are right. I was supposed there were infinite 0's on the left.
For 32-bit number, the MSB should also be checked in addition to LSB.
Change the first line to: c=1-(N&1)-((N>>31)&1); will fix this case.
On 2010-12-22 14:44, Saurabh Koar wrote:
@Terence: I think the above fails fo
@Terence: I think the above fails for 0X.Correct me if I m wrong.
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zero_group(N) {
c=1-(N&1); // For even N, zero groups is one more than 1 groups.
while(N) {
d = (N&(-N)); // Get the least significiant bit.
N &= N+d; // Clear the last 1-group bits
c++; // inc counter.
}
return c;
}
For more bit manipulations, refer to
How do you count the number of zero group bits in a number? group bits
is any consecutive zero or one bits, for example, 2 is represented
as 010 has two zero bits groups the least
significant bit and the group starts after one.
Also, I am in a bad need for algorithms on bits ma
hmm... for question number 1.. .. u can understand more easliy with the
following examlple
#include
// 1. #include
#pragma pack(1)
struct value
{
int a;
int bit3: 4;
int bit4: 4;
};
#pragma pack()
int main()
{
struct value bit;
printf("%d",sizeof(bit));
system("pause");
return
1.
i think if the total no of bits is within no of bits in a int , the
size will b same as int .
here total bits 3 < 32 (bits in int according to gcc, in Turbo c its 16)
so size of structure will be 4(in gcc), 2 (in TC)
if total no of bits is 40 (suppose) than size will be 8(in GCC) , 4(in TC)
tell the o/p of following with explanations
1. #include
int main()
{
struct value
{
int bit1:1;
int bit3:4;
int bit4:4;
}bit;
printf("%d\n",sizeof(bit));
return 0;
}
2.
#include
int main()
{
struct value
{
int bit1: 1;
int bit3: 4;
int bit4: 4;
} bit={1,2,2};
printf("%d %d %d\n",bit.bit1,bit.b
Given two integers A & B. Determine how many bits required to convert
A to B.how to write a function int BitSwapReqd(int A, int B);
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Richa Gupta
(IT-BHU,India)
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