GO Classes 2023 | Weekly Quiz 8 | Question: 8

Answer 1

Answer: C

Each variable of “type” must be stored at a location divisible by “sizeof(type)”.

Ints must be stored at locations 0, 4, 400, 256, 1024 etc.

chars can be stored anywhere since sizeof(char) = 1B and 1|anything.

To solve these types of questions first make a structure as wide as its largest sized element, then grow it in either upwards or downwards direction without altering the order of elements of structure.

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int d1; char c1; int d2;

int d1: size 4B: location 0 – 3
char c1: size 1B: location 4 – 4	3B padded: location 5 – 7
int d2: size 4B: location 8 – 11

Width: 4B. Total size = 12B

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int d1; char c1; int d2; char c2; short s;

int d1: size 4B: location 0 – 3
char c1: size 1B: location 4 – 4		3B padded: location 5 – 7
int d2: size 4B: location 8 – 11
char c2: size 1B: location 12 – 12	1B padded: location 13 – 13	short s: size 2B: location 14 – 15

Width: 4B. Total size = 16B

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int d1; int d2; char c1; char c2; short s;

int d1: size 4B: location 0 – 3
int d2: size 4B: location 4 – 7
char c1: size 1B: location 8 – 8	char c2: size 1B: location 9 –   9	shot s: size 2B: location 10 – 11

Width: 4B. Total size = 12B

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char c1; int d1; short s; int d2; char c2;

char c1: size 1B: location 0 – 0	3 bytes padded: location 1 – 3
int d1: size 4B: location 4 – 7
short s: size 2B: location 8 – 9	2 bytes padded: location 10 -11
int d2: size 4B: location 12 – 15
char c2: size 1B: location 16 – 16	3 bytes padded: location 17  – 19

Width: 4B. Total size = 20B

Comments

Nice explanation…

Just one minor correction needed in structure foo2.

Here, short s will not store at location 13-14, instead it will be stored at location 14-15 and 1B padded location will be 13-13.

 

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Yes @ASNR1010 my bad, I have updated the answer. :)

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beautiful presentation.

The main point of this question is : 

Each variable of “type” must be stored at a location divisible by “sizeof(type)”.