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eval: Allow specifying the variable id.

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Reviewed-by: Nicolas George <nicolas.george@normalesup.org>
Signed-off-by: Michael Niedermayer <michaelni@gmx.at>
This commit is contained in:
Michael Niedermayer 2012-02-22 17:21:04 +01:00
parent 02670762d2
commit 923092697a
3 changed files with 13 additions and 8 deletions
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5
doc/eval.texi
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@ -107,9 +107,10 @@ the evaluation of @var{y}, return 0 otherwise.
Evaluate @var{x}, and if the result is zero return the result of the
evaluation of @var{y}, return 0 otherwise.
@item taylor(expr, x)
@item taylor(expr, x) taylor(expr, x, id)
Evaluate a taylor series at x.
expr represents the LD(0)-th derivates of f(x) at 0.
expr represents the LD(id)-th derivates of f(x) at 0. If id is not specified
then 0 is assumed.
note, when you have the derivatives at y instead of 0
taylor(expr, x-y) can be used
When the series does not converge the results are undefined.

12
libavutil/eval.c
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@ -184,18 +184,19 @@ static double eval_expr(Parser *p, AVExpr *e)
case e_taylor: {
double t = 1, d = 0, v;
double x = eval_expr(p, e->param[1]);
int id = e->param[2] ? av_clip(eval_expr(p, e->param[2]), 0, VARS-1) : 0;
int i;
double var0 = p->var[0];
double var0 = p->var[id];
for(i=0; i<1000; i++) {
double ld = d;
p->var[0] = i;
p->var[id] = i;
v = eval_expr(p, e->param[0]);
d += t*v;
if(ld==d && v)
break;
t *= x / (i+1);
}
p->var[0] = var0;
p->var[id] = var0;
return d;
}
default: {
@ -523,6 +524,9 @@ static int verify_expr(AVExpr *e)
case e_not:
case e_random:
return verify_expr(e->param[0]) && !e->param[2];
case e_taylor:
return verify_expr(e->param[0]) && verify_expr(e->param[1])
&& (!e->param[2] || verify_expr(e->param[2]));
default: return verify_expr(e->param[0]) && verify_expr(e->param[1]) && !e->param[2];
}
}
@ -722,7 +726,7 @@ int main(int argc, char **argv)
"ifnot(0, 23)",
"ifnot(1, NaN) + if(0, 1)",
"taylor(1, 1)",
"taylor(eq(mod(ld(0),4),1)-eq(mod(ld(0),4),3), PI/2)",
"taylor(eq(mod(ld(1),4),1)-eq(mod(ld(1),4),3), PI/2, 1)",
NULL
};

4
tests/ref/fate/eval
View File

@ -172,8 +172,8 @@ Evaluating 'ifnot(1, NaN) + if(0, 1)'
Evaluating 'taylor(1, 1)'
'taylor(1, 1)' -> 2.718282
Evaluating 'taylor(eq(mod(ld(0),4),1)-eq(mod(ld(0),4),3), PI/2)'
'taylor(eq(mod(ld(0),4),1)-eq(mod(ld(0),4),3), PI/2)' -> 1.000000
Evaluating 'taylor(eq(mod(ld(1),4),1)-eq(mod(ld(1),4),3), PI/2, 1)'
'taylor(eq(mod(ld(1),4),1)-eq(mod(ld(1),4),3), PI/2, 1)' -> 1.000000
12.700000 == 12.7
0.931323 == 0.931322575