如何使用PHP清除共享内存？

Question

I want to ask about why after I use shmop_delete($shmid) and shmop_close($shmid) there is nothing happen? the first string is still attaced.

This is my php code: $key = 864; $mode = 'c'; $permission = 0644; $size = 1024;

$shmid = shmop_open($key, $mode, $permission, 1024);

$string = "Hello World !!!";

shmop_write($shmid, $string, 0);

$size = shmop_size($shmid);
echo shmop_read($shmid, 0, $size);

shmop_delete($shmid);
shmop_close($shmid);

First Result: Hello World !!!

Then I change $string = "TEST".

After that I run the code again with result : TESTo World !!! (Unwanted Result). The result I want is : TEST

why the 'o World !!!' is still attached although I already did shmop_delete and shmop_close? And How I can erase it from Shared Memory? Thx...

Answer 1

When you first write "Hello World" it's writing the length of those 11 chars. When "Test" is written, it only writes the length of "Test" into the shared memory. That is why the rest of the string is still attached. One potential solution is to take the length written like so:

$shm_bytes_written = shmop_write($shmid, $string, 0);

and pad the next string to that length. Or I guess you could write the entire size of the shared block, but that's probably not what you want.

Edit:

The code should be something like this (not tested nor compiled). This is for ASCII. If you're using UTF-8, you'll need to handle getting the string length a bit differently. Hopefully you get the idea

$shmid = shmop_open($key, $mode, $permission, 1024);

$string = "Hello World !!!";

$s1bytes = strlen( $string );

$shm_bytes_written = shmop_write($shmid, $string, 0);

$string2 = "Test";

$padLen = strlen( $string ) - strlen( $string2 );

$s2Pad = str_pad( "Test", $padLen );

$shm_bytes_written = shmop_write($shmid, $s2Pad, 0);

$size = shmop_size($shmid);
echo shmop_read($shmid, 0, $size);

shmop_delete($shmid);
shmop_close($shmid);