Siemens Interview Question Software Engineer in Tests

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    Write a function to determine whether the binary representation of a specified 32-bit integer is a palindrome. For example, the 32-bit integer 0xFCA8153F is a palindrome, but 0xFCA88ACF is not. Show how you would test this function.

    - SReddy on November 02, 2009 Report Duplicate | Flag
    Siemens Software Engineer in Test Coding



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You could do this bit by bit or nibble by nibble...or even byte by byte (look up table)

- EveryoneLovesMicrosoft on November 03, 2009 | Flag Reply
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This can be done by checking the starting nibbles and end nibbles and the sum of both should be 15(F). This should be true for all the nibbles till you cover all the nibbles.

- Anonymous on November 03, 2009 | Flag Reply
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F = 1111- F
E = 1110- 7
D = 1101- B
C = 1100- 3
B = 1011- D
A = 1010- 5
9 = 1001- 9
8 = 1000- 1
7 = 0111- E
6 = 0110- 6
5 = 0101- A
4 = 0100- 2
3 = 0011- D
2 = 0010- 4
1 = 0001- 8
0 = 0000- 0

- Anonymous on November 04, 2009 | Flag
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F = 1111- F
E = 1110- 7
D = 1101- B
C = 1100- 3
B = 1011- D
A = 1010- 5
9 = 1001- 9
8 = 1000- 1
7 = 0111- E
6 = 0110- 6
5 = 0101- A
4 = 0100- 2
3 = 0011- D
2 = 0010- 4
1 = 0001- 8
0 = 0000- 0

- Anonymous on November 04, 2009 | Flag Reply
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F = 1111- F
E = 1110- 7
D = 1101- B
C = 1100- 3
B = 1011- D
A = 1010- 5
9 = 1001- 9
8 = 1000- 1
7 = 0111- E
6 = 0110- 6
5 = 0101- A
4 = 0100- 2
3 = 0011- D
2 = 0010- 4
1 = 0001- 8
0 = 0000- 0

- Anonymous on November 04, 2009 | Flag Reply
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They do not have to be summed up as 15.

- Anonymous on November 04, 2009 | Flag Reply
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#define IS_PALINDROME (num) !(num ^ ((num & (1 << 0 ) ) << 31   \
| (num & (1 << 1 ) ) << 29      \
| (num & (1 << 2 ) ) << 27     \
| (num & (1 << 3 ) ) << 25     \
| (num & (1 << 4 ) ) << 23     \
| (num & (1 << 5 ) ) << 21     \ 
| (num & (1 << 6 ) ) << 19     \
| (num & (1 << 7 ) ) << 17     \
| (num & (1 << 8 ) ) << 15     \
| (num & (1 << 9 ) ) << 13     \
| (num & (1 << 10 ) ) << 11    \
| (num & (1 << 11 ) ) << 09    \
| (num & (1 << 12 ) ) << 07    \
| (num & (1 << 13 ) ) << 05    \
| (num & (1 << 14 ) ) << 03    \
| (num & (1 << 15 ) ) << 01    \
| (num & (1 << 16 ) ) >> 1     \
| (num & (1 << 17 ) ) >> 3     \
| (num & (1 << 18 ) ) >> 5     \
| (num & (1 << 19 ) ) >> 7     \
| (num & (1 << 20 ) ) >> 9     \
| (num & (1 << 21 ) ) >> 11    \
| (num & (1 << 22 ) ) >> 13    \
| (num & (1 << 23 ) ) >> 15    \
| (num & (1 << 24 ) ) >> 17    \
| (num & (1 << 25 ) ) >> 19    \
| (num & (1 << 26 ) ) >> 21    \
| (num & (1 << 27 ) ) >> 23    \
| (num & (1 << 28 ) ) >> 25    \
| (num & (1 << 29 ) ) >> 27    \
| (num & (1 << 30 ) ) >> 29    \
| (num & (1 << 31 ) ) >> 31))

- Ashutosh on November 05, 2009 | Flag Reply
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int reverse(unsigned int num)
{
        unsigned int rev=0;
        int s = sizeof(num) * 8;
        while(num)
        {
                rev <<= 1;
                rev |= num & 1;
                num>>=1;
                --s;
        }
        rev <<= s;
        return rev;
}
void checkPalindrome(unsigned int number)
{
        if(reverse(number)==number)
                printf("Palindrome\n");
        else
                printf("Not Palindrome\n");
}

- GK on November 06, 2009 | Flag Reply
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way to go

- Anonymous on November 06, 2009 | Flag
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reverse function doesn't work for 0001, 0010, 0100 which is less than 8.

- Anonymous on February 25, 2011 | Flag
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void palindrome_32bit_no()
{
unsigned long int x=0xFCA88ACF,A,B;
int i=0,flag=0;

for(i=0;i<16;i++)
{
A=(x & (unsigned long)pow(2.0, 32-i-1) );
B=(x & (unsigned long)pow(2.0, i) );
if((A && B)|| (!A && !B))
flag=1;
else
{
flag=0;
cout<<"\nThe no is not palindrome";
break;
}
}
if(flag==1)
cout<<"\nPalindrome";

}

- Hemant on March 27, 2010 | Flag Reply
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beautiful!!!

- Anonymous on June 18, 2011 | Flag


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