Hallo everyone
I might just be being dumb, but I am using the BASH shell and cannot get the following script to work:
x=0
while
do
echo $x
x=´echo "$x + 1" | bc´
done
Can anybody help me out. I am just get a repeating output saying:
bc: command not found
0 + 1: command not... (5 Replies)
I'm just trying to make a script that runs in command line to echo each line in a text file. Everything i found on google is telling me to do it like this but when I run it it just echos removethese.txt and thats it. Anyone know what im doing wrong?
for i in removethese.txt; do echo $i; done
... (4 Replies)
i have the following process running in background:
when i give "ps -lef"
------------------------------------------------------------------------
user2
user1
user1
user3
user1
user4
user5
user4
user3
user4
user2
user1
user1
user3
user1
user4 (3 Replies)
i have a script called file2
#!/bin/ksh
i=0
while
do
echo $i >> result.txt
i=`expr $i + 1`
done
echo "***********************" >> result
------------------------------------------------------------------- (10 Replies)
I have what I believe is a simple programming question. I have a text file that looks like:
mol 1 G:\stereo01.hin
block text
...
...
...
endmol 1
However, I would like a file that repeats this entire block of text several times over. The lines of text in the middle remain the same for each... (2 Replies)
Hi expert,
I'm using csh
Code:
#!/bin/csh
set x = 0
set number = `awk '{array=$0} END {print array;}'`
i want to use for loop to store data to $number repeatly
untill x = 23
How to use c shell for loop? (2 Replies)
I need to chmod a bunch of files with a specific extension in one directory.
If I understand correctly first I would run ls command like this
ls -R | grep .mp3 > /tmp/list
once I have the output file I should be able to run a loop to chmod all the files in the list created.
This is where... (5 Replies)
Hello forum memebers.
can you correct the simple while program.
#! /bin/ksh
count=10
while
do
echo $count
count='expr$count-1'
done
I think it will print 10 to 1 numbers but it running for indefinite times. (2 Replies)
Does any body can help me with a loop in this example?
if
then
if
then
runner=$(grep "$1" "$2")
runne=$(grep "$1" "$3")
run=$(grep "$1" "$4")
fi
fi
#
# Message on screen... (3 Replies)
Dear experts,
I am writing a bash script. At some point of the program I need to have 'for' loop. For simplicity I tried with some other simple code. The format of the loop is given below.
k=51
m=55
for j in {$k..$m};do
w=$(($j+2))
z=$(($j+9))
echo "$w, $z"
done
But my... (4 Replies)
Discussion started by: vjramana
4 Replies
LEARN ABOUT CENTOS
integer
integer(3pm) Perl Programmers Reference Guide integer(3pm)NAME
integer - Perl pragma to use integer arithmetic instead of floating point
SYNOPSIS
use integer;
$x = 10/3;
# $x is now 3, not 3.33333333333333333
DESCRIPTION
This tells the compiler to use integer operations from here to the end of the enclosing BLOCK. On many machines, this doesn't matter a
great deal for most computations, but on those without floating point hardware, it can make a big difference in performance.
Note that this only affects how most of the arithmetic and relational operators handle their operands and results, and not how all numbers
everywhere are treated. Specifically, "use integer;" has the effect that before computing the results of the arithmetic operators (+, -,
*, /, %, +=, -=, *=, /=, %=, and unary minus), the comparison operators (<, <=, >, >=, ==, !=, <=>), and the bitwise operators (|, &, ^,
<<, >>, |=, &=, ^=, <<=, >>=), the operands have their fractional portions truncated (or floored), and the result will have its fractional
portion truncated as well. In addition, the range of operands and results is restricted to that of familiar two's complement integers,
i.e., -(2**31) .. (2**31-1) on 32-bit architectures, and -(2**63) .. (2**63-1) on 64-bit architectures. For example, this code
use integer;
$x = 5.8;
$y = 2.5;
$z = 2.7;
$a = 2**31 - 1; # Largest positive integer on 32-bit machines
$, = ", ";
print $x, -$x, $x + $y, $x - $y, $x / $y, $x * $y, $y == $z, $a, $a + 1;
will print: 5.8, -5, 7, 3, 2, 10, 1, 2147483647, -2147483648
Note that $x is still printed as having its true non-integer value of 5.8 since it wasn't operated on. And note too the wrap-around from
the largest positive integer to the largest negative one. Also, arguments passed to functions and the values returned by them are not
affected by "use integer;". E.g.,
srand(1.5);
$, = ", ";
print sin(.5), cos(.5), atan2(1,2), sqrt(2), rand(10);
will give the same result with or without "use integer;" The power operator "**" is also not affected, so that 2 ** .5 is always the
square root of 2. Now, it so happens that the pre- and post- increment and decrement operators, ++ and --, are not affected by "use
integer;" either. Some may rightly consider this to be a bug -- but at least it's a long-standing one.
Finally, "use integer;" also has an additional affect on the bitwise operators. Normally, the operands and results are treated as unsigned
integers, but with "use integer;" the operands and results are signed. This means, among other things, that ~0 is -1, and -2 & -5 is -6.
Internally, native integer arithmetic (as provided by your C compiler) is used. This means that Perl's own semantics for arithmetic
operations may not be preserved. One common source of trouble is the modulus of negative numbers, which Perl does one way, but your
hardware may do another.
% perl -le 'print (4 % -3)'
-2
% perl -Minteger -le 'print (4 % -3)'
1
See "Pragmatic Modules" in perlmodlib, "Integer Arithmetic" in perlop
perl v5.16.3 2013-02-26 integer(3pm)