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Enhancement
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Resolution: Fixed
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P5
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None
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b09
Currently, the Java implementation of numberOfTrailingZeros() is derived from HD:
public static int numberOfTrailingZeros(long i) {
// HD, Figure 5-14
int x, y;
if (i == 0) return 64;
int n = 63;
y = (int)i; if (y != 0) { n = n -32; x = y; } else x = (int)(i>>>32);
y = x <<16; if (y != 0) { n = n -16; x = y; }
y = x << 8; if (y != 0) { n = n - 8; x = y; }
y = x << 4; if (y != 0) { n = n - 4; x = y; }
y = x << 2; if (y != 0) { n = n - 2; x = y; }
return n - ((x << 1) >>> 31);
}
It can be simplified through re-use of numberOfLeadingZeros() as
public static int numberOfTrailingZeros_02(int i) {
if (i == 0) return 32;
return 31 - numberOfLeadingZeros(i & -i);
}
This will reduce bytecode size, but also this shows better performance with both c1 and c2 compilers.
Long.numberOfTrailingZeros() can also be simplified via delegating to Integer's version.
This variant also shows better performance.
public static int numberOfTrailingZeros(long i) {
// HD, Figure 5-14
int x, y;
if (i == 0) return 64;
int n = 63;
y = (int)i; if (y != 0) { n = n -32; x = y; } else x = (int)(i>>>32);
y = x <<16; if (y != 0) { n = n -16; x = y; }
y = x << 8; if (y != 0) { n = n - 8; x = y; }
y = x << 4; if (y != 0) { n = n - 4; x = y; }
y = x << 2; if (y != 0) { n = n - 2; x = y; }
return n - ((x << 1) >>> 31);
}
It can be simplified through re-use of numberOfLeadingZeros() as
public static int numberOfTrailingZeros_02(int i) {
if (i == 0) return 32;
return 31 - numberOfLeadingZeros(i & -i);
}
This will reduce bytecode size, but also this shows better performance with both c1 and c2 compilers.
Long.numberOfTrailingZeros() can also be simplified via delegating to Integer's version.
This variant also shows better performance.
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