I have a shell script which does the encryption of a file where i am passing the file name as a command line argument,but later on the script waits on the screen to enter Y or N
what is the command i should be using on the shell script
#!/bin/bash -x
outfilename=file.out
echo... (8 Replies)
The following program takes two command line arguments.
I want the second argument (fileCount) to be stored/printed as a int value. I tried my best to typecast the char to int (check the printf statement at last) but is not working...the output is some junk value.
This program is in its... (3 Replies)
Hi,
I am facing a problem to pass command line arguments that looks like
<script name> aa bb "cc" dd "ee"
I want to pass all 5 elements include the " (brackets). when I print the @ARGV the " disappear. I hope I explain myself
Regards,
Ziv (4 Replies)
Hi,
I had written a shell script to pass command line argument to variable in a function.
Here is my code:
main
if ; then
.$1
echo $1
get_input_file
else
echo "input file $1 is not available"
fi
get_input_file()
{
FILE = "$1"
echo $FILE
} (10 Replies)
I have one working awk command line. Which taking data from the “J1202523.TXT” file and generating the “brazil.dat” file. PFB code.
awk '{ DUNS = substr($0,0,9);if ( substr($0,14,3) == "089" ) print DUNS }' J1202523.TXT > Brazil.dat
But now I want to pass two parameter as a command line argument... (4 Replies)
Hi,
In directory "inoutfiles", I have folders fold0001, fold0002 and so on. Every folder has corresponding file file0001.txt, file0002.txt and so on. I want to perform a certain action on multiple files in one go. The cpp file is in the same directory as "inoutfiles".
This is my code :
... (0 Replies)
Hello All,
I have a Bash Script and an Expect script that together will SSH to another server and
do some stuff there... From within the Bash Script I process the Command Line Arguments,
which are Required Args and Optional Args.
When I call the Expect script from the Bash Script, I pass... (4 Replies)
Hi,
I am calling a Perl script in my shell script. When Perl script is executed it asks for a answer to be entered by user from terminal. How can i pass that value from my shell script ??
I know I can change perl script to default the answer but i dont have access to do that so only option i... (5 Replies)
Hi,
I am trying to pass a variable as an argument to another script. While substitution of variable, I am facing a problem.
varaiable "a" value should be -b "FPT MAIN".
When we pass "a" to another script, we are expecing it to get substitue as ./test.sh -b "FPT MAIN". But, it is getting... (9 Replies)
Discussion started by: Manasa Pradeep
9 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)