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
took me a while but i figured it out
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
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
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
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?
Huh, it was 30 on the exam.
doing math is hard
okay well yeah I get the idea at least.
i mean you get the idea
the issue is that on an exam you have to do it exactly and not just the idea
but that's not my problem *shrug*
Have you done R-values and L-values before?
yeah but i don't remember them
but hey i never even learned pointer arithmetic sooo
okay xd. I'll keep that in mind.
I actually didn't know that an array in C was an array of pointers.
it's not, but you can treat it as such
it's really a single pointer wherein data is assumed to be stored in row major order
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