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Fill integer array
#3
Thanks both work great!
Just so I understand syntax on first one:

the integer pointer (*) variable p initially points to the address (&) of the first argument e0.

I am a little unsure what the [i] after the p means - is it incrementing the pointer to the address in memory by 1 to pull in the additional arguments.

I see from the help menu


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A single character can be accessed using [] operator:

;

str s = "Cat"
int char = s[1]
;now char is 97 ('a' character code)
s[0] = 66
;now s is "Bat" (66 is 'B' character code)
s[2] = 'r'
;now s is "Bar"


but why is the i-th character in the address of e0 equivalent to the next argument's value?

Thanks,
S


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