public int lisphashCode(int n) {
int r = 19937;
for (int i = 0; i < vec.length; i++) {
LispObject b = vec[i];
if (b == null) r = 54321 * r;
else if (!b.atom) r = 169 * r + ((Cons) b).lisphashCode(b, n - 10);
else r = 0x8040201 * r + b.lisphashCode();
}
return r;
}
public boolean leq(LispObject a) throws Exception {
return (value <= a.doubleValue());
}
public boolean ge(LispObject a) throws Exception {
return (value > a.doubleValue());
}
public boolean neqn(LispObject a) throws Exception {
return (value != a.doubleValue());
}
public LispObject min(LispObject a) throws Exception {
return (value <= a.doubleValue() ? this : a);
}
public LispObject expt(LispObject a) throws Exception {
// it is possible that I should delect cases where a is an integer
// and raise to a power using some alternative scheme, like repeated
// multiplication.
return new LispFloat(Math.pow(value, a.doubleValue()));
}
public LispObject remainder(LispObject a) throws Exception {
return new LispFloat(value % a.doubleValue());
}
public LispObject divide(LispObject a) throws Exception {
return new LispFloat(value / a.doubleValue());
}
public LispObject multiply(LispObject a) throws Exception {
return new LispFloat(value * a.doubleValue());
}
public LispObject subtract(LispObject a) throws Exception {
return new LispFloat(value - a.doubleValue());
}
public LispObject add(LispObject a) throws Exception {
return new LispFloat(value + a.doubleValue());
}
public void print(int fg) throws ResourceException {
Jlisp.print("#CALL" + (nargs & 0xf) + "as" + ((nargs >> 4) & 0xf) + "<");
body.print(fg);
Jlisp.print(">");
}
