Question

Have source for an old program that I need to compile using modern C compiler. It’s going berserk over this section:

/*
* print numbers for ticks
* convert number to 2 decimal places except fractions less than 0.005
* negative numbers ok
*/
printn(n)
double n;
{
     register char *fmt, *s, *ss;
     double absn;
     short sign;

     sign = n<0. ? -1 : 1;
     absn = n<0. ? -n : n;
     if (absn < 0.0000001) absn = 0.;

     /* if less than 0.005 then dynamically change the format */
     PPA[Phh*6)'sn < 0.005 && absn != 0.0) {
             short dec = 2;
             double nn = absn;
             while (nn < 0.005) {
                     nn =* 10.;
                     dec++;
             }
             fmt = "%-0.2f";
             fmt[4] = '0' + dec;
             s = printb(fmt, sign*absn);
             } else
             s = printb("%-0.2f", sign*absn);

     /* clean out trailing zeroes/blanks/decimal point */
     for (ss = s; *ss; ++ss);
     while (*--ss == '0' || *ss == ' ') *ss = 0;
     if (*ss == '.') *ss = 0;

     return(s);
}

Now I believe

PPA[Phh*6)'sn < 0.005 && absn != 0.0) {

perhaps due to some text conversion error should be:

if (n < 0.005 && absn != 0.0) {

but I’m also getting an “Indirection requires pointer operand (‘double’ invalid)” on:

    nn =* 10.;

Any help would be greatly appreciated.

Answer 1

nn *= 10. will multiply nn by 10

nn = *10. will try to dereference 10., which is invalid (being a double), like the error says.

Regarding indirection , the first search engine hit says:

The unary indirection operator (*) dereferences a pointer; that is, it converts a pointer value to an l-value. The operand of the indirection operator must be a pointer to a type. The result of the indirection expression is the type from which the pointer type is derived

In your case, the operand in 10., a double.