05-05-2015
As in the Perl example, just use integer arithmetic and you'll be fine. Separate dollars from cents at the end of your script.
This User Gave Thanks to MaroonedAlien For This Post:
10 More Discussions You Might Find Interesting
1. Programming
Alright, umm i cant get this to work.
im looking at some example and a book i have.
when i try to compile my program i get an error message.
ld: 0711-317 ERROR: Undefined symbol: .sqrt
ld: 0711-345 Use the -bloadmap or -bnoquiet option to obtain more information. I did #include<math.h> after my... (2 Replies)
Discussion started by: primal
2 Replies
2. Programming
Hi, I got an easy problem for you but really difficult for me 'cause I am pretty new to this field
I got header file <math.h> included in my .c file , then I write the code as below:
k = sqrt(i); /* both variables k and i are int */
then I cc temp.c
it says like this
undefined... (4 Replies)
Discussion started by: blf0
4 Replies
3. Programming
I want to calculate secant method using C language
That is a program---->
#include<stdio.h>
#include<math.h>
#include<stdlib.h>
main()
{
double fx(double x);
double x0,x1,x2,f0,f1,f2,err;
int n,i;
printf("\n\n f(x) =x*x*x-5*x-7");
printf("\n\nEnter an interval in"
... (4 Replies)
Discussion started by: cdfd123
4 Replies
4. Programming
Hey all,
How do I link the math library in a gnu make makefile? I have tried using -lm with the CFLAGS varibale - flags like -Wall and -ggdb work, but -lm does not. I am running gcc - 4.1.2 on a linux machine. (2 Replies)
Discussion started by: kermit
2 Replies
5. UNIX for Dummies Questions & Answers
I have 2 variables
a=2
b=1
i want to add a and b
how do i do this in unix using just the echo command and by assigning it to a different variable like c? (13 Replies)
Discussion started by: khestoi
13 Replies
6. Shell Programming and Scripting
$ x=1
$ y=1.5
$ z=$((x*y))
bash: 1.5: syntax error: invalid arithmetic operator (error token is ".5")
What's wrong? (2 Replies)
Discussion started by: rockbike
2 Replies
7. Shell Programming and Scripting
I am trying to do some math, so that I can compare the average of six numbers to a variable.
Here is what it looks like (note that when I divide really big numbers, it isn't a real number):
$ tail -n 6 named.stats | awk -F\, '{print$1}'
1141804
1140566
1139429
1134210
1084682
895045... (3 Replies)
Discussion started by: brianjb
3 Replies
8. Shell Programming and Scripting
Hi all,
I am new to PERL scripts, and i have made my first script which i am posting here.
This math tool performs all basic arithmatic functions.
#!/usr/bin/perl
print "\t----------Welcome to Maths Tool-----------\n";
do
{
print "Enter your choice :";
print... (2 Replies)
Discussion started by: PranavEcstasy
2 Replies
9. Programming
I have
int miles, yards;
float kilometers;
float kilometers2;
miles = 26;
yards = 385;
kilometers = 1.609 * (miles + yards / 1760.0);
where int/float remains a float. How ever if I change it to
kilometers = 1.609 * (miles + yards / 1760);
... (7 Replies)
Discussion started by: Fingerz
7 Replies
10. UNIX for Dummies Questions & Answers
I am struggling with scripting this challenge a friend and I have.
You have file1 and its contents is a single number
you have file 2 and its contents are a different number
you want to add file1 to file2 and have the output be put into file3 (3 Replies)
Discussion started by: minkyboodle
3 Replies
LEARN ABOUT PLAN9
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 v5.12.1 2010-04-26 integer(3pm)