Question

unsigned long long int a[4][10] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,37,38,39,40};
printf("%lu",a); //5120 this is given

printf("%lu",a+2); //5280 since a is 80 bytes, and 2*80 = 160, and 5120 + 160 = 5280

printf("%lu",a[2]); //also 5280

printf("%lu",*(a+2)+9); //answer is 5352

I don't understand how the answer is 5352

@photonics

Answer 1

took me a while but i figured it out

Answer 2

essentially the issue at play is that you're doing pointer arithmetic (something we never learned), but you have an array which is a pointer of pointers (in a sense) and you've dereferenced the pointer once

Answer 3

printf("%lu",*(a+2)+9); //answer is 5352
so *(a+2) would yield address 5280, but as an unsigned long long is 8 bytes having already dereferenced your pointer you get 5280+8*9=5352

Answer 4

admittedly I probably would not get this question correct on an exam, but hey, i already made it through the exams that would ask :D

Answer 5

and if I dereference that entire thing

printf("*lu*, *(*a+2)+9);

that would be *(5352) which is the content at memory location 5352.
right?

Answer 6

Huh, it was 30 on the exam.

Answer 7

doing math is hard

Answer 8

okay well yeah I get the idea at least.

Answer 9

i mean you get the idea

Answer 10

the issue is that on an exam you have to do it exactly and not just the idea

Answer 11

but that's not my problem *shrug*

Answer 12

Have you done R-values and L-values before?

Answer 13

yeah but i don't remember them

Answer 14

but hey i never even learned pointer arithmetic sooo

Answer 15

okay xd. I'll keep that in mind.

Answer 16

I actually didn't know that an array in C was an array of pointers.

Answer 17

it's not, but you can treat it as such

Answer 18

it's really a single pointer wherein data is assumed to be stored in row major order