fsolve: check if evalf() return value is actually a number.
authorAlexei Sheplyakov <alexei.sheplyakov@gmail.com>
Wed, 18 Aug 2010 21:07:13 +0000 (00:07 +0300)
committerJens Vollinga <jensv@nikhef.nl>
Sat, 21 Aug 2010 17:57:21 +0000 (19:57 +0200)
Fixes the segfault triggered by

fsolve((1/(sqrt(2*Pi)))*integral(t,0,x,exp(-1/2*t^2))==0.5,x,0,100)

In general, ex_to is unsafe, and should be used only after proper checks.
evalf() may return non-numeric expression for various reasons (bad
convergence, floating point over- or underflow, out of memory, etc).
So let's add missing checks.

Thanks to Ernst Moritz Hahn for a bug report.

ginac/inifcns.cpp

index 981e7b2..f366e77 100644 (file)
@@ -1007,8 +1007,16 @@ fsolve(const ex& f_in, const symbol& x, const numeric& x1, const numeric& x2)
        do {
                xxprev = xx[side];
                fxprev = fx[side];
-               xx[side] += ex_to<numeric>(ff.subs(x==xx[side]).evalf());
-               fx[side] = ex_to<numeric>(f.subs(x==xx[side]).evalf());
+               ex dx_ = ff.subs(x == xx[side]).evalf();
+               if (!is_a<numeric>(dx_))
+                       throw std::runtime_error("fsolve(): function derivative does not evaluate numerically");
+               xx[side] += ex_to<numeric>(dx_);
+
+               ex f_x = f.subs(x == xx[side]).evalf();
+               if (!is_a<numeric>(f_x))
+                       throw std::runtime_error("fsolve(): function does not evaluate numerically");
+               fx[side] = ex_to<numeric>(f_x);
+
                if ((side==0 && xx[0]<xxprev) || (side==1 && xx[1]>xxprev) || xx[0]>xx[1]) {
                        // Oops, Newton-Raphson method shot out of the interval.
                        // Restore, and try again with the other side instead!
@@ -1017,8 +1025,16 @@ fsolve(const ex& f_in, const symbol& x, const numeric& x1, const numeric& x2)
                        side = !side;
                        xxprev = xx[side];
                        fxprev = fx[side];
-                       xx[side] += ex_to<numeric>(ff.subs(x==xx[side]).evalf());
-                       fx[side] = ex_to<numeric>(f.subs(x==xx[side]).evalf());
+
+                       ex dx_ = ff.subs(x == xx[side]).evalf();
+                       if (!is_a<numeric>(dx_))
+                               throw std::runtime_error("fsolve(): function derivative does not evaluate numerically [2]");
+                       xx[side] += ex_to<numeric>(dx_);
+
+                       ex f_x = f.subs(x==xx[side]).evalf();
+                       if (!is_a<numeric>(f_x))
+                               throw std::runtime_error("fsolve(): function does not evaluate numerically [2]");
+                       fx[side] = ex_to<numeric>(f_x);
                }
                if ((fx[side]<0 && fx[!side]<0) || (fx[side]>0 && fx[!side]>0)) {
                        // Oops, the root isn't bracketed any more.
@@ -1042,7 +1058,10 @@ fsolve(const ex& f_in, const symbol& x, const numeric& x1, const numeric& x2)
                        static const double secant_weight = 0.984375;  // == 63/64 < 1
                        numeric xxmid = (1-secant_weight)*0.5*(xx[0]+xx[1])
                            + secant_weight*(xx[0]+fx[0]*(xx[0]-xx[1])/(fx[1]-fx[0]));
-                       numeric fxmid = ex_to<numeric>(f.subs(x==xxmid).evalf());
+                       ex fxmid_ = f.subs(x == xxmid).evalf();
+                       if (!is_a<numeric>(fxmid_))
+                               throw std::runtime_error("fsolve(): function does not evaluate numerically [3]");
+                       numeric fxmid = ex_to<numeric>(fxmid_);
                        if (fxmid.is_zero()) {
                                // Luck strikes...
                                return xxmid;