convert to GB

Shell Programming and Scripting
1

hello,

my objective is to calculate the swap size

-bash-3.00# swap -l
swapfile dev swaplo blocks free
/dev/md/dsk/d1 85,1 8 33559776 16966160

so size=blocks*512/(1024*1024*1024)
since blocks are 512-blocks

any idea how to get it?
so far I was able to get the 4th value:

swap -l |grep -v blocks| awk '{print $4}'
33559776

how to get the size in GB?
thanks

2

actually I want also to get the 4th column of the 2nd line in the file:

more /opt/SUNWexplo/output/explorer.zeus/disks/swap-l.out
swapfile dev swaplo blocks free
/dev/md/dsk/d1 85,1 8 33559776 16966152

I want 33559776

how to get it?
how to convert it to GB? (*512/(1024)^3

thanks

3

Let awk do the job for you:

awk 'NR==2{print $4*512/(1024*1024*1024)}' /opt/SUNWexplo/output/explorer.zeus/disks/swap-l.out

Regards

4

thanks indeed

1] how to concact GB to the result?

2] how to have an output such as:

Swap Space: xxxxxx GB

3] Any way to round this number? 16.0025 to 16?

thanks

5

hello again

let me start with this:
echo "Swap Space in GB:"
awk 'NR==2{print $4*512/(1024*1024*1024)}' /opt/SUNWexplo/output/explorer.zeus/disks/swap-l.out

gives:
Swap Space in GB:
16.0025

I want to get
Swap Space in GB: 16.0025

i.e concatenate the 2 outputs in one line.

how to do that?

many thanks

6

It is ok:

v_swap1="Swap Space in GB:"
v_swap2=`awk 'NR==2{print $4*512/(1024*1024*1024)}' /opt/SUNWexplo/output/explorer.zeus/disks/swap-l.out`
echo $v_swap1 $v_swap2

any idea how to round 16.0025 to 16??
thanks

7

If the printf statement rounding is "unbiased" (system dependent) which means it doesn't round a trailing 0.5 up you can just add 0.5 to the result:

awk '{printf("Swap Space: %d\n", $4*512/(1024*1024*1024)+0.5)}' file

To verify weather printf is rounding biased or not you can try this:

awk 'BEGIN{printf("%d\n", 15.9)}'

Regards