Siemens Interview Question for Software Engineer in Tests






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

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

- Anonymous 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 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 November 06, 2009 | Flag Reply
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way to go

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

- Anonymous 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 March 27, 2010 | Flag Reply
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beautiful!!!

- Anonymous June 18, 2011 | Flag


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