Yes.

You should most definitely always have swap enabled, except if there is a very compelling, forbidding reason (like, no disk at all, or only network disk present). Should you have a swap on the order of the often recommended ridiculous sizes (such as, twice the amount of RAM)? Well, no.

The reason is that swap is not only useful when your applications consume more memory than there is physical RAM (actually, in that case, swap is not very useful at all because it seriously impacts performance). The main incentive for swap nowadays is not to magically turn 16GiB of RAM into 32 GiB, but to make more efficient use of the installed, available RAM.

On a modern computer, RAM does not go unused. Unused RAM is something that you could just as well not have bought and saved the money instead. Therefore, anything you load or anything that is otherwise memory-mapped, anything that could possibly be reused by anyone any time later (limited by security constraints) is being cached. Very soon after the machine has booted, all physical RAM will have been used for something.

Whenever you ask for a new memory page from the operating system, the memory manager has to make an educated decision:

1. Purge a page from the buffer cache
2. Purge a page from a mapping (effectively the same as #1, on most systems)
3. Move a page that has not been accessed for a long time -- preferably never -- to swap (this could in fact even happen proactively, not necessarily at the very last moment)
4. Kill your process, or kill a random process (OOM)
5. Kernel panic

Options #4 and #5 are very undesirable and will only happen if the operating system has absolutely no other choice. Options #1 and #2 mean that you throw something away that you will possibly be needing soon again. This negatively impacts performance.

Option #3 means you move something that you (probably) don't need any time soon onto slow storage. That's fine because now something that you do need can use the fast RAM.

By removing option #3, you have effectively limited the operating system to doing either #1 or #2. Reloading a page from disk is the same as reloading it from swap, except having to reload from swap is usually less likely (due to making proper paging decisions).

In other words, by disabling swap you gain nothing, but you limit the operation system's number of useful options in dealing with a memory request. Which might not be, but very possibly may be a disadvantage (and will never be an advantage).

[EDIT]

The careful reader of the mmap manpage, specifically the description of MAP_NORESERVE, will notice another good reason why swap is somewhat of a necessity even on a system with "enough" physical memory:

"When swap space is not reserved one might get SIGSEGV upon a write if no physical memory is available."

-- Wait a moment, what does that mean?

If you map a file, you can access the file's contents directly as if the file was somehow, by magic, in your program's address space. For read-only access, the operating system needs in principle no more than a single page of physical memory which it can repopulate with different data every time you access a different virtual page (for efficiency reasons, that's of course not what is done, but in principle you could access terabytes worth of data with a single page of physical memory). Now what if you also write to a file mapping? In this case, the operating system must have a physical page -- or swap space -- ready for every page written to. There's no other way to keep the data around until the dirty pages writeback process has done its work (which can be several seconds). For this reason, the OS reserves (but doesn't necessarily ever commit) swap space, so in case you are writing to a mapping while there happens to be no physical page unused (that's a quite possible, and normal condition), you're guaranteed that it will still work.

Now what if there is no swap? It means that no swap can be reserved (duh!), and this means that as soon as there are no free physical pages left, and you're writing to a page, you are getting a pleasant surprise in the form of your process receiving a segmentation fault, and probably being killed.

[/EDIT]

However, the traditional recommendation of making swap twice the size of RAM is nonsensical. Although disk space is cheap, it does not make sense to assign that much swap. Wasting something that is cheap is still wasteful, and you absolutely don't want to be continually swapping in and out working sets several hundreds of megabytes (or larger) in size.

There is no single "correct" swap size (there are as many "correct" sizes as there are users and opinions). I usually assign a fixed 512MiB, regardless of RAM size, which works very well for me. The reasoning behind that is that 512MiB is something that you can always afford nowadays, even on a small disk. On the other hand, adding several gigabytes of swap is none better. You are not going to use them, except if something is going seriously wrong.

Even on a SSD, swap is orders of magnitude slower than RAM (due to bus bandwidth and latency), and while it is very acceptable to move something to swap that probably won't be needed again (i.e. you most likely won't be swapping it in again, so your pool of available pages is effectively enlarged for free), if you really need considerable amounts of swap (that is, you have an application that uses e.g. a 50GiB dataset), you're pretty much lost.

Once your computer starts swapping in and out gigabytes worth of pages, everything goes to a crawl. So, for most people (including me) this is not an option, and having that much swap therefore makes no sense.

I recently had the same problem, and created this to clean the swap out quickly:

#Single-proc function to core-dump swapped ranges > 1M to /dev/null
unswap(){ (awk -F'[ \t-]+' '/^[a-f0-9]*-[a-f0-9]* /{recent="0x"$1" 0x"$2}/Swap:/&&$2>1000{print recent}' /proc/$1/smaps | while read astart aend; do gdb --batch --pid $1 -ex "dump memory /dev/null $astart $aend" &>/dev/null; done&)2>/dev/null;};

#Loop to run unswap on the top 20 swap-consuming processes
grep VmSwap /proc/*/status 2>/dev/null | sort -nk2 | tail -n20 | cut -d/ -f3 | while read line; do unswap $line; done;

#Observe the number of core dumps currently running, along with free swap, over time. 
echo "Dumps Free(m)"; rcount=10; while [[ $rcount -gt 0 ]]; do rcount=$(ps fauxww | grep "dump memory" | grep -v grep | wc -l); echo "$rcount        $(free -m | awk '/Swap/{print $4}')"; sleep 1; done 

If this doesn't work in your specific environment, the process is as follows:

1. Get a list of the processes consuming the most swap.

In my case, I'm checking /proc/$pid/status, with the swap use in the VmSwap row.

# grep VmSwap /proc/*/status 2>/dev/null | sort -nk2 | tail -n5
/proc/22457/status:VmSwap:      3780 kB
/proc/22684/status:VmSwap:      4260 kB
/proc/7408/status:VmSwap:       4396 kB
/proc/31992/status:VmSwap:      9176 kB
/proc/2967/status:VmSwap:      60840 kB

2. For the high-swap procs, I find the memory address ranges which are contributing the most swap use.

I get this from /proc/$pid/smaps:

7f2fd1bc4000-7f2fd1d24000 rw-p 00000000 00:00 0 <<< Address range
Size:               1408 kB
Rss:                 900 kB
Pss:                 900 kB
Shared_Clean:          0 kB
Shared_Dirty:          0 kB
Private_Clean:         0 kB
Private_Dirty:       900 kB
Referenced:            4 kB
Anonymous:           900 kB
AnonHugePages:         0 kB
Swap:                508 kB    << Swap used
KernelPageSize:        4 kB
MMUPageSize:           4 kB

3. I use gdb to do core dumps of those address ranges straight to /dev/null

This forces the system to access that memory and pull it out of swap. Going to /dev/null avoids unnecessary IO.

gdb --batch --pid $pid -ex "dump memory /dev/null $astart $aend"

The process I mentioned originally omits any memory regions using less than 1M of swap in order to get the majority of the swap with considerably fewer core dumps, but this is not mandatory.