exactlyOneBitIsSet(i) provides a very fast way to determine whether an integral type has exactly one bit set.
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exactlyOneBitIsSet(i) provides a very fast way to determine whether an integral type has exactly one bit set.
For unsigned and positive signed values, this is equivalent to the value being a power of two. Note that negative powers of two are not represented with a single bit set – negate it first if you want to use this routine to determine if a signed value is a negative power of two.
- See also
- atMostOneBitIsSet()
bool SimTK::exactlyOneBitIsSet |
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unsigned char |
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bool SimTK::exactlyOneBitIsSet |
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unsigned short |
v | ) |
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bool SimTK::exactlyOneBitIsSet |
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unsigned int |
v | ) |
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bool SimTK::exactlyOneBitIsSet |
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unsigned long |
v | ) |
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bool SimTK::exactlyOneBitIsSet |
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unsigned long long |
v | ) |
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bool SimTK::exactlyOneBitIsSet |
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signed char |
v | ) |
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inline |
bool SimTK::exactlyOneBitIsSet |
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char |
v | ) |
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inline |
bool SimTK::exactlyOneBitIsSet |
( |
short |
v | ) |
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inline |
bool SimTK::exactlyOneBitIsSet |
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int |
v | ) |
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inline |
bool SimTK::exactlyOneBitIsSet |
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long |
v | ) |
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inline |
bool SimTK::exactlyOneBitIsSet |
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long long |
v | ) |
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inline |