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Microsoft Interview Question


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0
of 0 vote

#define sizeof(var) (((int)&(var)+1)-((int)&var))

- mail2vcp@gmail.com September 29, 2008 | Flag Reply
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0
of 0 votes

sizeof is an operator, most probably cannot be implemented in user code. just try your code with sizeof(char), sizeof('c') etc...

- greg October 09, 2008 | Flag
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1
of 0 votes

ooh don't be so pessimistic...
Since we are not restricted from using RTTI,
can write something like:
works equally well for both types and variables

#define SIZEOF_type(T) \
((char *)((typeof(T) *)0 + 1))

- asm December 01, 2008 | Flag
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0
of 0 votes

The macro does not work.

- russoue February 05, 2010 | Flag
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0
of 0 vote

Yeah. sizeof is actually a macro.

- Anonymous November 30, 2009 | Flag Reply
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0
of 0 vote

in c i think its macro

#define sizeof(A) (char*((A*)1+1)- (char*)(A*)1);

is impletend in C++ by template function ??

- ridercoder October 21, 2010 | Flag Reply
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0
of 0 vote

sizeof is unary operator;

- vijaymukilan July 17, 2013 | Flag Reply


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