NaN problem

[beism]

Hi all,

I have a nasty problem with floating point arithmetic.
Let E be a double variable which is assigned a value using an expression like:
E=A+B*((i<j) ? f(C1)/(1-f(C2)) : 0);
where A,B,C1,C2 are double variables, f is a function returning a double and i,j are integers.

The problem I have is that the value assigned to E is 'nan' but when I use
printf(&quot;E=%f\n&quot;,A+B*f(C1)/(1-f(C2)));
the actual value of E (which is in the range [-5,5]) appears correctly! It seems that there is a problem during the assignment.

Any help will be welcome,

Mike

 

[xwb]

Instead of using the triadic operator, try an if statement.

if (i < j)
E = A + B * f(C1)/(1-f(C2));
else
E = A;

if this still causes problems, work out the divisor first. Some compilers are just plain weird. QuickC used to give these problems all the time.

 

[sskumar]

I guess U have to use
E=A+B*((i<j) ? (f(C1)/(1-f(C2))) : 0);
instead of
E=A+B*((i<j) ? f(C1)/(1-f(C2)) : 0);

please see that if works....