convert to GB


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# 1  
convert to GB

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:

Code:
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:

Code:
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:

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

Regards

Last edited by Franklin52; 03-03-2008 at 02:20 PM.. Reason: linguistic correction
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