awk, floating point and rounding


 
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# 1  
Old 01-18-2011
awk, floating point and rounding

I had a person bring an interesting problem to me that appears to involve some sort of rounding inside awk. I've verified this with awk and nawk on Solaris as well as with gawk 3.1.5 on a Linux box.

The original code fragment he brought me was thus:

Code:
for (index=0; index < 1; index=index+.1)
                printf "Index=%f\n",index

What was odd was this would print an index of 1 even though the script should have found "index < 1" false when index==1.

I wrote a little follow on script to highlight this further

Code:
BEGIN   {
        for (i=5; i < 7; i+=1) {
                printf "Pass %s...\n",i
                for (j=0.0; j+0.0 < (.1*i); j=j+0.1) {
                        printf "j=%f\n",j;
                }
        }
exit
}'

I would expect the output of this to print 5 lines, then 6 lines but instead we see 5 lines then 7 lines:

Code:
Pass 5...
j=0.000000
j=0.100000
j=0.200000
j=0.300000
j=0.400000
Pass 6...
j=0.000000
j=0.100000
j=0.200000
j=0.300000
j=0.400000
j=0.500000
j=0.600000

The fact that the change occurs around .5 makes me think we're seeing some sort of integer rounding in action but I don't know how to force a variable to always be interpreted as a float to avoid this -- as best as I can tell awk doesn't support any for of typecasting. In the example you'll see that I always use a floating point value when doing math on the variables but that doesn't seem to affect things one way or another.

Any thoughts appreciated!
# 2  
Old 01-18-2011
Hi

This works for me:


Code:
BEGIN   {
        for (i=5; i <=7; i+=1) {
                printf "Pass %s...\n",i
                for (j=0.0; (float)j < (.1*i); j=j+0.1) {
                        printf "j=%f\n",j;
                }
        }
exit
}


Guru.
# 3  
Old 01-18-2011
Indeed it does!

I looked for information on typecasting in the awk/nawk/gawk man pages but found none but apparently it works and it supported even in the ancient 'awk' that Solaris has by default.

Thanks for the pointer!
# 4  
Old 01-18-2011
There is nothing special about "(float)j". Awk interprets it as a concatenation of two strings. First the uninitialized variable "float" is treated as an empty string, then the variable j is converted to string using the default "%.6g" format.
Code:
awk 'BEGIN{for(x=0.1;x<1;x+=.1)printf("%.20f\n",x)}'
0.10000000000000000555
0.20000000000000001110
0.30000000000000004441
0.40000000000000002220
0.50000000000000000000
0.59999999999999997780
0.69999999999999995559
0.79999999999999993339
0.89999999999999991118
0.99999999999999988898

Here you can see that the last number is very close to 1. When it is converted to string using the default "%.6g", it will be 1 exactly.
# 5  
Old 01-19-2011
The "error" is due to the default rounding.
Code:
BEGIN   {
        for (i=5; i < 7; i+=1) {
                printf "Pass %s...\n",i
                for (j = 0.0; j < (0.1*i); j = j + 0.1) {
                        printf "j=%.20f %.20f\n", j, .1*i;
                }
        }
exit
}

Code:
Pass 5...
j=0.00000000000000000000 0.50000000000000000000
j=0.10000000000000000555 0.50000000000000000000
j=0.20000000000000001110 0.50000000000000000000
j=0.30000000000000004441 0.50000000000000000000
j=0.40000000000000002220 0.50000000000000000000
Pass 6...
j=0.00000000000000000000 0.60000000000000008882
j=0.10000000000000000555 0.60000000000000008882
j=0.20000000000000001110 0.60000000000000008882
j=0.30000000000000004441 0.60000000000000008882
j=0.40000000000000002220 0.60000000000000008882
j=0.50000000000000000000 0.60000000000000008882
j=0.59999999999999997780 0.60000000000000008882

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