I'm trying to write what I expected to be a reasonably simple awk routine to generate a list of numbers as part of a larger bash shell script (since I'll need this done for many different files). For example: echo | awk '{ x = 0.0; y = 0.0; do { printf("%.1f\n",x) if ( x < 3.0 ) { y = 0.3 } else if ( x < 6.0 ) { y = 1.0 } else if ( x < 15.0 ) { y = 3.0 } else if ( x < 25.0 ) { y = 5.0 } else { y = 10.0 } x = x + y; } while ( x < 55.0 ) }' I expected this to produce this list of numbers: 0 0.3 0.6 ... 2.4 2.7 3.0 4.0 5.0 6.0 9.0 12 15 20 25 35 45 55 what I get is: 0 0.3 0.6 ... 2.7 3.0 3.3 4.3 5.3 ... when x = 3 in the loop (printed x after each "if" conditional) the comparison seems to fail. i.e. ( x < 3 ) returns true, even when x = 3, and y remains equal to 0.3 for one iteration longer than it should ( as if I used "<=" instead of "<" ). I manage to get the right results if I initialize x = 0.0001 and use printf("%.1f\n",x). Fine, but why is this necessary? (Its terribly unsatisfying). I found post 8067 (http://www.computing.net/unix/wwwboard/forum/8067 .html) which suggests it might be because awk treats the value of x as a string, but I tried adding 0 to x ( x += 0; or x = x + 0 ) and the results are the same. I've tried every variation I can think of and searched a number of forums (but my question is so general I'm not sure I know how to search for it). I think I even has a case where ( x < 3 ) did not work inside the "do" loop but ( x < 55 ) did work as the while condition.... but I haven't done that again. Can anyone explain why my " x < 3 " is acting like " x <= 3 "? I would greatly appreciate your help (and I might be able to move on and get more work done). Thank you!

try to print out your variables at each stage and see what's wrong with the logic.

$ echo | awk '{ > x = 0.0; y = 0.0; > do { > printf("%.1f\t",x) > if (x < 3) {y = 0.3; print "(x<3) x=" x "; y=" y } > else if (x < 6) {y = 1; print "(x<6) x=" x "; y=" y } > else if (x < 15) {y = 3; print "(x<15) x=" x "; y=" y } > x += y; > } while ( x <= 15 ) > }' 0.0 (x < 3) x = 0; y = 0.3 0.3 (x < 3) x = 0.3; y = 0.3 0.6 (x < 3) x = 0.6; y = 0.3 0.9 (x < 3) x = 0.9; y = 0.3 1.2 (x < 3) x = 1.2; y = 0.3 1.5 (x < 3) x = 1.5; y = 0.3 1.8 (x < 3) x = 1.8; y = 0.3 2.1 (x < 3) x = 2.1; y = 0.3 2.4 (x < 3) x = 2.4; y = 0.3 2.7 (x < 3) x = 2.7; y = 0.3 3.0 (x < 3) x = 3; y = 0.3 <-- ( 3 < 3 ) --> TRUE?!? 3.3 (x < 6) x = 3.3; y = 1 4.3 (x < 6) x = 4.3; y = 1 5.3 (x < 6) x = 5.3; y = 1 6.3 (x < 15) x = 6.3; y = 3 9.3 (x < 15) x = 9.3; y = 3 12.3 (x < 15) x = 12.3; y = 3 $ When x = 3.0, the statement "if (x < 3)" evaluates true(?) and sets y = 0.3 ... It looks like its acting as "<=" instead of "<". Should I be initializing the variables differently to force awk to see that 3.0 == 3.0 and is not less than 3.0? I tried using inserting "x += 0" right before "if (x < 3.0)..." and this didn't work either. Thank you

floating point numbers are approximations only. so 3.0 may not be exactly 3. it may be 3.00001 for example. you might want think of a workaround for your particular problem.

Thank you for the sanity check. initializing x = 0.00001 is a rough workaround. but (x < 3) does work sometimes. maybe just when dealing with integers? perhaps bc or perl would work better for anything more complicated than a list of numbers. Thanks again.

I observed the same thing.  I put in checks for x in the range of 0.3 to 3.3. The 0.9 did not test exact, nor anything past 2.4. x is now exactly 0.3 0.300000 x is now exactly 0.6 0.600000 0.900000 x is now exactly 1.2 1.200000 x is now exactly 1.5 1.500000 x is now exactly 1.8 1.800000 x is now exactly 2.1 2.100000 x is now exactly 2.4 2.400000 2.700000 3.000000 3.300000 My workaround for this is to avoid the decimal points.  I just multiplied everything by 10, and changed the printf to print x/10.  But it also works to do: xfinal=x/10 printf xfinal (and 30/10 did yield an exactly 3.0) echo | awk '{ x = 0; y = 0; do { printf("%.1f\n",x/10) if ( x < 30 ) { y = 3 } else if ( x < 60 ) { y = 10 } else if ( x < 150 ) { y = 30 } else if ( x < 250 ) { y = 50 } else { y = 100 } x = x + y; } while ( x < 550 ) }' 0.0 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4 2.7 3.0 4.0 5.0 6.0 9.0 12.0 15.0 20.0 25.0 35.0 45.0

Thank you, I've used your 10x trick before to keep track of decimals as integers in bash and avoid using awk or bc. Its strange that the exact test fails for x = 0.9 and x > 2.4 ( and 30/10 == 3 ). $ a=0; b=0.3; $ echo $a $b | awk '{ for (x=0; x<=12; ++x) { printf("a =%18.15f b =%18.15f a + b =%23.20f\n",$1,$2,$1 + $2) $1 = $1 + $2; } }' ... ## I don't know how to format the output nicely for a ## post, so here's an excerpt: ... a = 0.000000000000000 \ b = 0.300000000000000 \ a + b = 0.29999999999999998890 a = 0.300000000000000 \ b = 0.300000000000000 \ a + b = 0.59999999999999997780 ... a = 1.200000000000000 \ b = 0.300000000000000 \ a + b = 1.50000000000000000000 a = 1.500000000000000 \ b = 0.300000000000000 \ a + b = 1.80000000000000004441 Right off the bat, a = 0.0 and a+b = 0.2999..., but the next a is 0.3000. the only exception is when a = 1.20000... and a+b = 1.500000... (maybe I just need more arbitrary decimal places). The "exactly" test fails at 0.9 and above 2.4 again. hmmm... something in the subtleties of floating point values I guess... (I read somewhere that the decimal 0.1 can't be represented exactly as a floating point number in base 2). I'll have to stick with your 10x suggestion and get back to work. Thank you!