strint with Arrays?
Author |
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Flikerator
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Posted: Sat Mar 05, 2005 5:25 pm Post subject: strint with Arrays? |
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I want to convert my array of string to an array of int. Can I do that?
code: |
var file : int
var numstring : array 1 .. 6 of string
var nums : array 1 .. 6 of int
open : file, "DATA1.txt", get
for i : 1 .. 6
get : file, numstring (i)
end for
for i : 1 .. 6
end for
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Convert where the second for statement is, I put it in there incase its needed. You can take it out if its not needed. |
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Sponsor Sponsor
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Cervantes
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Posted: Sat Mar 05, 2005 6:06 pm Post subject: (No subject) |
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Sure you can do that.
code: |
var numstring : array 1 .. 6 of string
var nums : array 1 .. 6 of int
for i : 1 .. 6
numstring (i) := intstr (Rand.Int (-100, 100))
end for
for i : 1 .. 6
nums (i) := strint (numstring (i))
end for
for i : 1 .. 6
put nums (i)
end for
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However, let's improve on that a bit, using the strintok() function.
code: |
var numstring : array 1 .. 30 of string
var nums : array 1 .. 30 of int
for i : 1 .. 30
numstring (i) := chr (Rand.Int (40, 60))
end for
for i : 1 .. 30
if strintok (numstring (i)) then
nums (i) := strint (numstring (i))
else
nums (i) := -1
end if
end for
for i : 1 .. 30
put nums (i)
end for
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This just prevents the program from crashing. |
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Flikerator
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Posted: Sat Mar 05, 2005 6:14 pm Post subject: (No subject) |
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code: | var file : int
var numstring : array 1 .. 6 of string
var nums : array 1 .. 6 of int
open : file, "DATA1.txt", get
for i : 1 .. 6
get : file, numstring (i)
end for
for i : 1 .. 6
nums (i) := strint (numstring (i))
end for |
Overflow in result of strint
Because the value is "12345678987654321"
Is there anything I can do? |
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Cervantes
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Posted: Sat Mar 05, 2005 6:37 pm Post subject: (No subject) |
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Use real variables, instead of integers.
That, or you could break the number up into different parts. But that's just too much extra work. |
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