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NVIDIA Interview Question for Software Engineer / Developers


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1
of 1 vote

#define setbit(i,j,k) i=(k==0)?((1<<j)&i):((1<<j)|i)

- rampurearun September 16, 2012 | Flag Reply
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0
of 0 vote

#define MACRO(i,j,k) i = i&(((((i>>j)<<1)|k)<<(j-1))|(~(1<<j)))

- Tulley February 26, 2011 | Flag Reply
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0
of 0 vote

#define MACRO(i, j, k) i = ((((i >> j) & 1) ^ k) << j) ^ i

- drivehappy February 27, 2011 | Flag Reply
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0
of 0 votes

This is almost the best solution. Small correction as !k should be included.
#define MACRO(i, j, k) i = ((((i >> j) & 1) ^ !k) << j) ^ i

- gavinashg March 19, 2011 | Flag
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0
of 0 votes

When jth bit is 0 and k=0 then i think your algo is not working...
ex- i=0x0000001A j=3rd bit k=0
check it

- Qurious October 25, 2011 | Flag
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0
of 0 votes

When jth bit is 0 and k=0 then i think your algo is not working...
ex- i=0x0000001A j=3rd bit k=0
check it

- Qurious October 25, 2011 | Flag
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0
of 0 vote

doesn't the question say hex number? won't it effect our code?

if not this will work - #define setbit(i,j,k) i=i^(k<<j)

- sai February 27, 2011 | Flag Reply
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0
of 0 votes

sorry, its
#define setbit(i,j,k) i=(i&(~(1<<j)))|(k<<j)

- sai February 27, 2011 | Flag
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0
of 0 vote

Is it correct :-

i = i | ( (1 << j) & (k << j));

- sandygupta February 28, 2011 | Flag Reply
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0
of 0 votes

i^(k<<j-1)

- rajat ahuja February 28, 2011 | Flag
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0
of 0 vote

@sandy i guess no. because if i has 1 in the jth bit, k as 0, ( (1 << j) & (k << j)); of your code returns 0 but the '|' wont be able to set it. see the #define MACRO(i,j,k) i = ((i | (1 << j))&(k << j)) that codebuddy defined above.

- sai March 02, 2011 | Flag Reply
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0
of 0 vote

first set the j'th bit to 0, then add j'th bit to it.

#define MACRO(i,j,k) i = (i & (~(1<<j)) + (k<<j))

- pc March 05, 2011 | Flag Reply
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0
of 0 votes

To optimize:
#define MACRO(i,j,k) i = (i & (~(1<<j)) | (k<<j))

I realized, it is same as sai's macro.

- pc March 05, 2011 | Flag
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0
of 0 vote

#define SET(i,j,k) (i^(((i>>j&1)^k)<<j))

- simransraj July 31, 2014 | Flag Reply
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0
of 0 votes

Can some one explain the logic

- anon July 31, 2014 | Flag
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0
of 0 vote

#define REPLACE_BIT(i, j, k) ((k == 1) ? ((i) | (1 << j)) : ((i) & ~(1 << j)))

- Anonymous September 18, 2014 | Flag Reply
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-1
of 1 vote

#define MACRO(i,j,k) i=i&((a<<j)&k))

- Anonymous February 26, 2011 | Flag Reply
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0
of 0 votes

#define MACRO(i,j,k) i=i&((1<<j)&k))

- Anonymous February 26, 2011 | Flag
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0
of 0 votes

if k =0 then i will become 0 always.

- Tulley February 26, 2011 | Flag
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-1
of 3 vote

#define MACRO(i,j,k) i = ((i | (1 << j))&(k << j))

- codebuddy February 26, 2011 | Flag Reply


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