How size of structure can be calculated? What is byte alignment used while allocating memory for structure members.

svyengalwar
Profile Answers by svyengalwar Questions by svyengalwar
Apr 14th, 2007
15
2000

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Mathi

Apr 18th, 2007

The size of the structutre will be the total size of fields holded by that structure.For Eg
struct emp
{
int empid;
char name[20];
float bpay;
}e;
sizeof (e) = sizeof(empid)+sizeof(name[20])+sizeof(bpay)=2+20+4=26.

J

Nov 22nd, 2007

Size for the above example would be 28

4 + 20 + 4

gokul_tn
Profile Answers by gokul_tn

Feb 4th, 2008

I suppose it is 4+20+4 = 28

prabhjots
Profile Answers by prabhjots

Feb 6th, 2008

hey first of all i got this as 26 (running on windows)..
and the other thing you are asking about memory aligment is dat there are restrictions by some compilers(or by machines) that every int (or float ) should start from an adress which should be divisible by 2 or 4(bcoz of some performance issues). i have heard dat it is not true for intel processors(but not sure)., so on those machines the size of structure can be different if it has used some "padding bytes or alignement bytes"
hope this clears some of your doubts