CareerCup

XOR with 1's

well ths soultion will be like
1) let the number be A = 10101111
2) A1 = A & 10101010 = 10101010
3) A2 = A & 01010101 = 00000101
4) A1 = A1 >> 1 = 01010101
5) A2 = A2 << 1 = 00001010
6) Answer = A1 | A2 = 01011111

so u require 5 instructions

need only one ~ (not) operation.
Eg. A= 0x10101111;
B = ~A = 0x01010000;

here we r dng either rotate left with carrry or rotate with right
so if we consider rotate left
a1=num & 0x80000;//get the msb
a=a<<1; //where a is number //on which we r apllying
a=a |a1;
so we required here three instructions

not operator would not work in this case.

for example,

A=00000000
~A = 11111111

where swap(A) should return 00000000

I think it needs 3 instructions:
shl $0x1, $A
jnc 1f
or 0x1, $A
1f:

Sorry that there are two typos in the codes. The correct one should be:

shl $0x1, $A
jnc 1f
or $0x1, $A
1:


And it is gcc assembler syntax. For the intel assembler syntax, it should be
shl A, 1
jnc label1
or A, 1
label1:

You are right. I am cofused.

int swapBits(unsigned x)
{
return ((x & 0xAAAAAAAA)>>1) | ((x<<0x55555555))
}

5 makes sense

Isn't this just a barrel shifter?

// n is the thing you want to change

result = (n << 1) | ((n >> 31) & 1)

e.g.
int n = 1110 00...00 0101
n << 1 = 1100 00...00 1010
((n >> 31) & 1) = 1
result = 1100 00...00 1011

int main()
{
char temp;
char *str=(char *)malloc(100);
printf("Enter Number in Binary form");
gets(str);
for(int i=0;i<strlen(str);i=i+2)
{
temp=*(str+i);
*(str+i)=*(str+i+1);
*(str+i+1)=temp;
}
printf("Output is");
printf("%s",str);
getch();
}