NVIDIA Interview Question for Software Engineer / Developers






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2
of 4 vote

sizeof is neither a macro nor a function. Its a unary operator like ! or ~. Its implemented by(in) compiler.

- Tulley March 15, 2011 | Flag Reply
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1
of 1 vote

It's a compile time operator. Reason is during runtime byte code would not have type info of the variable(unless u have rtti.). So runtime function cannot handle this.

- Gayathri February 18, 2012 | Flag Reply
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0
of 0 vote

yes, it is an operator

- Anonymous March 16, 2011 | Flag Reply
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0
of 0 vote

operator!! man how can u not know this :D

- greed March 16, 2011 | Flag Reply
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0
of 0 vote

sizeof is a unary operator not a macro
reference-dennis ritchie

- neal caffrey January 29, 2012 | Flag Reply
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0
of 0 vote

It's a compile time operator. Reason is during runtime byte code would not have type info of the variable(unless u have rtti.). So runtime function cannot handle this.

- Gayathri February 18, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

It's a compile time operator. Reason is during runtime byte code would not have type info of the variable(unless u have rtti.). So runtime function cannot handle this.

- Gayathri February 18, 2012 | Flag Reply
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0
of 0 vote

of course, its an operator.
In almost all cases, sizeof is evaluated based on static type information (at compile-time, basically).

One exception (the only one, I think) is in the case of C99's variable-length arrays (VLAs).

- Aashish June 24, 2012 | Flag Reply
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-1
of 3 vote

Ekta Kapoor's rascala fan commenting 3 times. Mind it...

- Rajnikanth March 12, 2012 | Flag Reply


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