[EMAIL PROTECTED] wrote:
top says convert is using 14496 bytes of virtual memory
That makes sense if the pixels are cached to disk. If you use the limits we
recommended its possible your system is denying the memory request. We call
malloc() and malloc can fail to honor a request. If you want to see
what's going on behind the scenes, try this command:
convert -debug cache -limit area 8192 -limit memory 8192 \
bigassimage.jpg bigassimage.png
[EMAIL PROTECTED]:~/tmp$ convert -verbose -type bilevel -monochrome
-compress group4 -debug cache gray0008.jpg gray0008.tif
21:23:13 0:01 0.000u 6.2.3 Cache convert[8265]:
cache.c/DestroyCacheInfo/1279/Cache
destroy
21:23:13 0:01 0.000u 6.2.3 Cache convert[8265]: cache.c/OpenCache/2623/Cache
open gray0008.jpg[0] (/tmp/magick-XX37jHkW[3], disk, 1.93gb)
Looks like it's going straight to disk.
The lines afterward all look like:
21:23:13 0:01 0.000u 6.2.3 Cache convert[8265]:
cache.c/WriteCachePixels/3901/Cache
16801x1+0+0
So does malloc try to acquire enough memory for the entire process?
Like 1.93gb. If so, why does it have to be all or nothing? Can't it
use RAM and a swap file?
Thanks for your help.
Randall
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