I want to pass an array in my function, And my function will be changing the elements of the array in the fuction, but it should not affect the values in my array variable of main function (1 Reply)
Hi all..
Does anyone know have an example of passing the contents of a ksharray to oracle?
basically I am looking to loop through the contents of a file and store each line into a bash ksh. Once i have this I can then pass the array into an oracle procedure that accepts an array as an... (1 Reply)
Please excuse my ineptitude for a bit as I've been spoiled for the past few months with only writing perl code instead of C.
So ok, I've been thinking about some code to change the crc32 values that are held within central directory headers of zip files.
Because I'm lazy I decided to just... (3 Replies)
Hi experts,
I am here again with another Issue.
I need to pass the array as parameter / argument to another script.
I tried it as follows . ( I got this idea from the link )
$ cat test1.sh
#! /usr/bin/ksh
set -a arr1
echo "...In Test1...."
arr1="APPS_DEV"
arr1="TEST_DEV"
echo... (16 Replies)
Hi,
Please guide to pass an array as a arg to a script...
for example,
I have a script small.sh to find the small no of given arg as below...
#! /bin/sh
# this script is for finding the small number
set -A arr_no_updates
small=$1
i=1
for arr in $@
do
if (3 Replies)
Hi all,
Hereby wish to have your advise for below:
Main concept is
I intend to get current directory of my script file.
This script file will be copied to /etc/init.d.
A string in this copy will be replaced with current directory value.
Below is original script file:
... (6 Replies)
Hi, all
suppose I have following myfile (delimited by tab)
aa bb
cc dd
ee ffand I have following awk command:
awk 'BEGIN{FS="\t"}{AwkArrayVar_1=$1;AwkArrayVar_2=$2};END{for(i=0; i<NR; i++) print i, AwkArrayVar_1, AwkArrayVar_2,}' myfileMy question is: how can I assign the awk array... (7 Replies)
How do i pass an array from test4.sh to a function in another shell script test5.sh, basically i am sourcing the test5.sh in test4.sh and printing the contents, but not working below are my trial scripts, please help, thank you.
#!/bin/bash
# /usr/local/dw/archive/test5.sh
print_array()
{... (5 Replies)
Hi All
I have multiple arrays like below.
set -A val1 1 2 4 5
set -A val2 a b c d
.
.
.
Now i would like to pass the individual arrays one by one to a function and display/ do some action.
Note : I am using ksh
Can you please advise any solution...
Thanks in advance. (7 Replies)
Hi,
I am creating filesystem for block device, but I want to pass array value one by one acording to block device count.
$tmp1 = block device count 3
$blockdevice =
So I want to first pass sdb1 alone in loop, how to take only block device seprately from $blockdevice array. (1 Reply)
Discussion started by: stew
1 Replies
LEARN ABOUT PHP
debug_zval_dump
DEBUG_ZVAL_DUMP(3) 1 DEBUG_ZVAL_DUMP(3)debug_zval_dump - Dumps a string representation of an internal zend value to outputSYNOPSIS
void debug_zval_dump (mixed $variable, [mixed $...])
DESCRIPTION
Dumps a string representation of an internal zend value to output.
PARAMETERS
o $variable
- The variable being evaluated.
RETURN VALUES
No value is returned.
EXAMPLES
Example #1
debug_zval_dump(3) example
<?php
$var1 = 'Hello World';
$var2 = '';
$var2 =& $var1;
debug_zval_dump(&$var1);
?>
The above example will output:
&string(11) "Hello World" refcount(3)
Note
Beware the refcount
The refcount value returned by this function is non-obvious in certain circumstances. For example, a developer might expect the
above example to indicate a refcount of 2. The third reference is created when actually calling debug_zval_dump(3).
This behavior is further compounded when a variable is not passed to debug_zval_dump(3) by reference. To illustrate, consider a
slightly modified version of the above example:
Example #2
<?php
$var1 = 'Hello World';
$var2 = '';
$var2 =& $var1;
debug_zval_dump($var1); // not passed by reference, this time
?>
The above example will output:
string(11) "Hello World" refcount(1)
Why refcount(1)? Because a copy of $var1 is being made, when the function is called.
This function becomes even more confusing when a variable with a refcount of 1 is passed (by copy/value):
Example #3
<?php
$var1 = 'Hello World';
debug_zval_dump($var1);
?>
The above example will output:
string(11) "Hello World" refcount(2)
A refcount of 2, here, is extremely non-obvious. Especially considering the above examples. So what's happening?
When a variable has a single reference (as did $var1 before it was used as an argument to debug_zval_dump(3)), PHP's engine opti-
mizes the manner in which it is passed to a function. Internally, PHP treats $var1 like a reference (in that the refcount is
increased for the scope of this function), with the caveat that if the passed reference happens to be written to, a copy is made,
but only at the moment of writing. This is known as "copy on write."
So, if debug_zval_dump(3) happened to write to its sole parameter (and it doesn't), then a copy would be made. Until then, the
parameter remains a reference, causing the refcount to be incremented to 2 for the scope of the function call.
SEE ALSO var_dump(3), debug_backtrace(3), References Explained, References Explained (by Derick Rethans).
PHP Documentation Group DEBUG_ZVAL_DUMP(3)