does anyone know what the TASK ONE assignment was that year? It had something to do with Vikings....or Task two? it would make it easier to follow along!
@snazelle The task was writing a program that took in a number and printed out how a viking would've pronounced it. Here's the test program used for demonstration: htAtp://cgi.cse.unsw.edu.Aau/~hs1917/10s1/bjorn/index.cgi (remove upper case A's)
For any who are curious, at 37:00 when he guesses that the 2 empty bytes are due to divisibility by 4, he is actually correct. The problem is he forgot it was allocating from the bottom of his graph. If you look at it you will see that there are 6 bytes stored in 2 4-byte chunks, leaving a two byte gap at the end between the byte variables and the int. This means the final address of each 4-byte chunk is divisible by 4 (the final empty byte is 48, which divides evenly).
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Thanks for this! I mean, really. Thank you!!
katheryncruz24 3 months ago
He talks really fast like a computer. But I like the way he teaches. It interests me.
jhamien920 3 months ago
does anyone know what the TASK ONE assignment was that year? It had something to do with Vikings....or Task two? it would make it easier to follow along!
thanks!
snazelle 10 months ago
@snazelle The task was writing a program that took in a number and printed out how a viking would've pronounced it. Here's the test program used for demonstration: htAtp://cgi.cse.unsw.edu.Aau/~hs1917/10s1/bjorn/index.cgi (remove upper case A's)
navidsonpuzzle 4 months ago
I lie to the compiler.
ASeventhSign 10 months ago
The rest of the COMP1917 lectures aren't easy to find. Help?
J90JAM 11 months ago
39:06 hilarious!
Desrathedemon 11 months ago
For any who are curious, at 37:00 when he guesses that the 2 empty bytes are due to divisibility by 4, he is actually correct. The problem is he forgot it was allocating from the bottom of his graph. If you look at it you will see that there are 6 bytes stored in 2 4-byte chunks, leaving a two byte gap at the end between the byte variables and the int. This means the final address of each 4-byte chunk is divisible by 4 (the final empty byte is 48, which divides evenly).
mrmadmaxalot 1 year ago 2
Parolu Esperanton!
thisusernameistaken2 1 year ago
Thanks for the lectures, I love the stories Richard tells :)
dpilankar 2 years ago 5
"Don't piss on the sign" LOL
3dgaras 2 years ago
15:00 nice wallpaper
intindse 2 years ago
But we, New Zealanders, can say six!
TheJohanLiebert 2 years ago
Any paraody of Richard's lecture style would have to involve a hand on the forehead.
SauronsEye 2 years ago 11
Ow wow, this guy is so good with metaphores. I absolutely loved the way he explained memorymaps using his movie collection shelving.
Excellent!
devjock 3 years ago 5
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i dont think u gotta do it like
pointer+1; if its a byte pointer
could u do like
pointer+=6 or whatever pointer=pointer+6;
sspoke 3 years ago