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Languguage OS 2
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Languguage OS II Version 10-94 (Knowledge Media)(1994).ISO
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libg_261.zip
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libg_261
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libg++
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tests
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tBitString.exp
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1992-05-06
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2KB
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79 lines
a =
b = 1000000001
c = 10101010101010101010
d = 00110011001100110011
e = 11110000111100001111
f = b = 1000000001
g = ~e = 00001111000011110000
h = d = 00110011001100110011
bits in e:
0 1 2 3 8 9 10 11 16 17 18 19
clear bits in g (reverse order):
19 18 17 16 11 10 9 8 3 2 1 0
~c = 01010101010101010101
c & d = 00100010001000100010
c | d = 10111011101110111011
c - d = 10001000100010001000
c ^ d = 10011001100110011001
c + d = 1010101010101010101000110011001100110011
c <<2 = 0010101010101010101010
c >>2 = 101010101010101010
f &= c = 1000000000
f |= d = 10110011001100110011
f -= e = 10000000000000000000
f ^= c = 00101010101010101010
f += b = 001010101010101010101000000001
f <<=5 = 00000001010101010101010101000000001
f >>=10= 0101010101010101000000001
l = 101010101010101010100011001100110011001110101010101010101010
BitPattern pat = 0011XXXX0011XXXX0011
pat.pattern = 00110011001100110011
pat.mask = 11110000111100001111
l.index(pat) = 20
l.index(pat,-1)= 20
l.before(pat) = 10101010101010101010
l.at(pat) = 00110011001100110011
l.after(pat) = 10101010101010101010
b.set(0) :1000000001
b.set(65):
100000000100000000000000000000000000000000000000000000000000000001
b.clear(2):
100000000100000000000000000000000000000000000000000000000000000001
b.invert(11,20):
100000000100111111111000000000000000000000000000000000000000000001
b.set(21,30):
100000000100111111111111111111100000000000000000000000000000000001
b.clear(31, 40):
100000000100111111111111111111100000000000000000000000000000000001
b.set(0) :10110011001100110011
b.set(65):
101100110011001100110000000000000000000000000000000000000000000001
b.clear(2):
100100110011001100110000000000000000000000000000000000000000000001
b.invert(11,20):
100100110010110011001000000000000000000000000000000000000000000001
b.set(21,30):
100100110010110011001111111111100000000000000000000000000000000001
b.clear(31, 40):
100100110010110011001111111111100000000000000000000000000000000001
k = 0101
c.before(k) = 1
c.at(k) = 0101
c.after(k) = 010101010101010
c.after(k)=k :101010101
c.before(k)=k:010101010101
reverse(k) = 1010
k.left_trim(0) : 101
k.right_trim(1) : 10
k = 0110
c.before(k) = 0
c.at(k) = 0110
c.after(k) = 011001100110011
c.after(k)=k :001100110
c.before(k)=k:011001100110
reverse(k) = 0110
k.left_trim(0) : 110
k.right_trim(1) : 110
End of test.