Calculate the constant e to 14+ decimal places using integer maths.
Hi guys...
I am loving this integer maths thing.
64 bit systems are certainly easier than 32 bit, but hey, I don't intend to leave out my fav' platform.
Using one of the 'Brothers' methods, URL inside the code.
Result in 64 bit mode, my usual platform.
These 2 Users Gave Thanks to wisecracker For This Post:
I have a perl script that reports the avg time of a application call and the total number of calls. This works fine, however I would like to trim the number of decimal places reported from 12 to like 3 and I don't know how.
Any suggestions? Here is what I use to get the avg time...
for $eRef (... (2 Replies)
i have a script in which awk prints "($2-1700)/10000"
and the answer is -0.07,but i want the answer in 4 decimal places.
that is -0.0700.
How can i sue awk to get my results in four decimal places (4 Replies)
i need to multiplay a number with 1.00.. so that the output should contain two decimal places at end..
for example...
236 * 1.00 = 236.00
245.8 * 1.00 = 245.80
but when i perform multiplication it shows output as.
236
245.8
can anyone help me to get the actual output of... (11 Replies)
Hi All
I've made a few scripts which using GDAL extract the value of a pixel within a given raster. The purpose is to work out the combine value of every pixel. I thought there may have been an easier way to do this but alas!
The code below extracts the pixel value at position X Y. The... (3 Replies)
I am trying to perform arithmetric, for example, to increment the value of variable $a (say 3) by 0.05 but when I tried the following expression
let a=a+0.05
or a=$((a+0.05))
both returned
3.0499999999999998
I want to keep 2 decimal places so it returns 3.05 instead. (6 Replies)
Please help me in rounding up value upto 2 decimal palces using sed command
#!/usr/bin/bash
a=15.42
b=13.33
c=`echo $a*$b |bc -l`
echo $c
above code is is giving output "205.5486" but i want the output as "205.55"
Thank you... (15 Replies)
I have input file like below,
201424|9999|OSS|622010|RGT|00378228764
201424|8888|OM|587079|RGT|00284329675
201424|7777|OM|587076|RGT|00128671024
201424|6666|OM|581528|RGT|00113552084
Output should be like below, should add decimal (.) from last 4 digits.
... (2 Replies)
I used the below script to Sum up a field in a file based on some unique values. But the problem is when it is summing up the units, it is truncating to 2 decimals and not 6 decimals as in the input file (Input file has the units with up to 6 Decimals – Sample data below, when the units in the 2... (4 Replies)
Discussion started by: brlsubbu
4 Replies
LEARN ABOUT PHP
datefmt_set_lenient
DATEFMT_SET_LENIENT(3) 1 DATEFMT_SET_LENIENT(3)IntlDateFormatter::setLenient - Set the leniency of the parser
Object oriented style
SYNOPSIS
public bool IntlDateFormatter::setLenient (bool $lenient)
DESCRIPTION
Procedural style
bool datefmt_set_lenient (IntlDateFormatter $fmt, bool $lenient)
Define if the parser is strict or lenient in interpreting inputs that do not match the pattern exactly. Enabling lenient parsing allows
the parser to accept otherwise flawed date or time patterns, parsing as much as possible to obtain a value. Extra space, unrecognized
tokens, or invalid values ("February 30th") are not accepted.
PARAMETERS
o $fmt
- The formatter resource
o $lenient
- Sets whether the parser is lenient or not, default is TRUE (lenient).
RETURN VALUES
Returns TRUE on success or FALSE on failure.
EXAMPLES
Example #1
datefmt_set_lenient(3) example
<?php
$fmt = datefmt_create(
'en_US',
IntlDateFormatter::FULL,
IntlDateFormatter::FULL,
'America/Los_Angeles',
IntlDateFormatter::GREGORIAN,
'dd/MM/yyyy'
);
echo 'lenient of the formatter is : ';
if ($fmt->isLenient()) {
echo 'TRUE';
} else {
echo 'FALSE';
}
datefmt_parse($fmt, '35/13/1971');
echo "
Trying to do parse('35/13/1971').
Result is : " . datefmt_parse($fmt, '35/13/1971');
if (intl_get_error_code() != 0) {
echo "
Error_msg is : " . intl_get_error_message();
echo "
Error_code is : " . intl_get_error_code();
}
datefmt_set_lenient($fmt, false);
echo "
Now lenient of the formatter is : ";
if ($fmt->isLenient()) {
echo 'TRUE';
} else {
echo 'FALSE';
}
datefmt_parse($fmt, '35/13/1971');
echo "
Trying to do parse('35/13/1971').
Result is : " . datefmt_parse($fmt, '35/13/1971');
if (intl_get_error_code() != 0) {
echo "
Error_msg is : ".intl_get_error_message();
echo "
Error_code is : ".intl_get_error_code();
}
?>
Example #2
OO example
<?php
$fmt = new IntlDateFormatter(
'en_US',
IntlDateFormatter::FULL,
IntlDateFormatter::FULL,
'America/Los_Angeles',
IntlDateFormatter::GREGORIAN,
'dd/MM/yyyy'
);
echo 'lenient of the formatter is : ';
if ($fmt->isLenient()) {
echo 'TRUE';
} else {
echo 'FALSE';
}
$fmt->parse('35/13/1971');
echo "
Trying to do parse('35/13/1971').
Result is : " . $fmt->parse('35/13/1971');
if (intl_get_error_code() != 0) {
echo "
Error_msg is : " . intl_get_error_message();
echo "
Error_code is : " . intl_get_error_code();
}
$fmt->setLenient(FALSE);
echo "
Now lenient of the formatter is : ";
if ($fmt->isLenient()) {
echo 'TRUE';
} else {
echo 'FALSE';
}
$fmt->parse('35/13/1971');
echo "
Trying to do parse('35/13/1971').
Result is : " . $fmt->parse('35/13/1971');
if (intl_get_error_code() != 0) {
echo "
Error_msg is : " . intl_get_error_message();
echo "
Error_code is : " . intl_get_error_code();
}
?>
The above example will output:
lenient of the formatter is : TRUE
Trying to do parse('35/13/1971').
Result is : 66038400
Now lenient of the formatter is : FALSE
Trying to do parse('35/13/1971').
Result is :
Error_msg is : Date parsing failed: U_PARSE_ERROR
Error_code is : 9
SEE ALSO datefmt_is_lenient(3), datefmt_create(3).
PHP Documentation Group DATEFMT_SET_LENIENT(3)