Hi All,
I hope I am posting a C question instead of a C++ this time. I have
the following program:
#include<stdlib .h>
#include<stdio. h>
#include<math.h >
#include<time.h >
int main(int argc,char** argv){
int i;
srand(time(0));
for(i = 0; i < 20; i++){
int x = 1 + (int) (20.0 * rand() / (RAND_MAX + 1.0));
int y = 1 + (int) (20 * rand() / (RAND_MAX + 1.0));
int z = 1 + (int) (20.0 * rand() / (RAND_MAX + 1 ));
printf("%d %d %d\n",x,y,z);
}
}
and it prints:
12 1 -18
4 1 -17
14 1 -12
17 1 -13
8 1 0
9 1 -2
.... etc
it looks like x is always between 1 and 20. y is always 1 and z is
always between negative and 0. i don't understand why. i guess i
understand why x is always positive because the whole expression is
evaluated in long(?) so there won't be any overflow, right? But is y
and z behave so differently? I though C promote the expression if one
of them is long instead of int, right? Can someone explain what
happened? Thanks! 6 1505
"pembed2003 " <pe********@yah oo.com> wrote in message int z = 1 + (int) (20.0 * rand() / (RAND_MAX + 1 )); it looks like x is always between 1 and 20. y is always 1 and z is always between negative and 0. i don't understand why.
The expression rand() / (RAND_MAX + 1) will be evaluated first as an integer
expression. RAND_MAX on your machine must equal INT_MAX, so RAND_MAX + 1 is
the lowest possible negative integer.
In article <c5**********@n ewsg1.svr.pol.c o.uk>,
"Malcolm" <ma*****@55bank .freeserve.co.u k> wrote: "pembed2003 " <pe********@yah oo.com> wrote in message int z = 1 + (int) (20.0 * rand() / (RAND_MAX + 1 )); it looks like x is always between 1 and 20. y is always 1 and z is always between negative and 0. i don't understand why.
The expression rand() / (RAND_MAX + 1) will be evaluated first as an integer expression. RAND_MAX on your machine must equal INT_MAX, so RAND_MAX + 1 is the lowest possible negative integer.
You should really know that this is not true.
20.0 * rand() / (RAND_MAX + 1) is evaluated as follows:
20.0 => Value 20, type double.
rand () => Random value, type int, from 0 to RAND_MAX
(RAND_MAX + 1) => Type int. If RAND_MAX == INT_MAX then -> overflow,
undefined behavior; many implementations will produce INT_MIN.
20.0 * rand () => rand () is converted to double, result is of type
double, from 0 to 20 * RAND_MAX.
20.0 * rand () / (RAND_MAX + 1) => RAND_MAX + 1 is converted to
double. If there was no overflow in calculating RAND_MAX + 1 then the
result is of type double, >= 0.0 and < 20.0. If there is an overflow,
then undefined behavior, but most likely the result of (RAND_MAX + 1) is
INT_MIN, which is often - (INT_MAX + 1) = - (RAND_MAX + 1), so the
result is <= 0 and > -20.
Christian Bau <ch***********@ cbau.freeserve. co.uk> wrote in message news:<ch******* *************** ***********@slb-newsm1.svr.pol. co.uk>... In article <c5**********@n ewsg1.svr.pol.c o.uk>, "Malcolm" <ma*****@55bank .freeserve.co.u k> wrote:
"pembed2003 " <pe********@yah oo.com> wrote in message int z = 1 + (int) (20.0 * rand() / (RAND_MAX + 1 )); it looks like x is always between 1 and 20. y is always 1 and z is always between negative and 0. i don't understand why.
The expression rand() / (RAND_MAX + 1) will be evaluated first as an integer expression. RAND_MAX on your machine must equal INT_MAX, so RAND_MAX + 1 is the lowest possible negative integer.
You should really know that this is not true.
20.0 * rand() / (RAND_MAX + 1) is evaluated as follows:
20.0 => Value 20, type double. rand () => Random value, type int, from 0 to RAND_MAX (RAND_MAX + 1) => Type int. If RAND_MAX == INT_MAX then -> overflow, undefined behavior; many implementations will produce INT_MIN.
20.0 * rand () => rand () is converted to double, result is of type double, from 0 to 20 * RAND_MAX. 20.0 * rand () / (RAND_MAX + 1) => RAND_MAX + 1 is converted to double. If there was no overflow in calculating RAND_MAX + 1 then the result is of type double, >= 0.0 and < 20.0. If there is an overflow, then undefined behavior, but most likely the result of (RAND_MAX + 1) is INT_MIN, which is often - (INT_MAX + 1) = - (RAND_MAX + 1), so the result is <= 0 and > -20.
Thanks for the explaination. Here is how I look at it:
(20.0 * rand() / (RAND_MAX + 1 ))
If I put parenthesis:
(((20.0) * (rand())) / ((RAND_MAX) + (1)))
so essentially:
(((double) * (int)) / ((int) + (int)))
double * int gives:
((double) / ((int) + (int)))
now since the second expression overflow (in my machine RAND_MAX ==
MAX_INT), we have:
((doubld) / ((MAX_INT) + (1))) = ((double) / (overflow))
which results in a nagetive number. make sense. But I can't understand
why:
(20 * rand() / (RAND_MAX + 1.0))
always result in a 1?
(((20) * (rand())) / ((RAND_MAX) + (1.0)))
so it's:
(((int) * (int)) / ((int) + (double)))
and then:
(((int) * (int)) / (double))
and then from here, I can't understand why it's always a 1. I can
understand that an int * int might overflow but will C promote it to a
double because we have a double just on the right side of /
Thanks!
On Fri, 15 Apr 2004, pembed2003 wrote: Christian Bau <ch***********@ cbau.freeserve. co.uk> wrote 20.0 * rand() / (RAND_MAX + 1) is evaluated as follows:
20.0 => Value 20, type double. rand () => Random value, type int, from 0 to RAND_MAX (RAND_MAX + 1) => Type int. If RAND_MAX == INT_MAX then -> overflow, undefined behavior; many implementations will produce INT_MIN.
20.0 * rand () => rand () is converted to double, result is of type double, from 0 to 20 * RAND_MAX. 20.0 * rand () / (RAND_MAX + 1) => RAND_MAX + 1 is converted to double. If there was no overflow in calculating RAND_MAX + 1 then the result is of type double, >= 0.0 and < 20.0. If there is an overflow, then undefined behavior, but most likely the result of (RAND_MAX + 1) is INT_MIN, which is often - (INT_MAX + 1) = - (RAND_MAX + 1), so the result is <= 0 and > -20.
[Why does]
(20 * rand() / (RAND_MAX + 1.0))
always result in a 1?
It doesn't, of course. It results in a floating-point 'double'
value, on your machine probably between -1.0 and 1.0 (even though of
course technically it could do whatever it liked).
Consider:
20 --> type 'int', value 20
rand() --> type 'int', value 0..RAND_MAX
20 * rand() --> type 'int', value 0..20*RAND_MAX
Now, at this point we realize that RAND_MAX is probably equal to
INT_MAX (as it is on most modern systems); so 20*RAND_MAX overflows
and we have a case of undefined behavior. Let's assume that 'int'
values overflow by "wrapping around," so that the value of 20*rand()
is still a random number between INT_MIN and INT_MAX.
20 * rand() --> type 'int', value INT_MIN..INT_MA X
RAND_MAX + 1.0 --> type 'double', value RAND_MAX+1
whole expression --> type 'double', value...
The value of the whole expression, assuming what we assumed above about
overflow on your machine, is a random number between
(INT_MIN/(RAND_MAX+1.)) and (INT_MAX/(RAND_MAX+1.)) . And since RAND_MAX
is assumed to be INT_MAX, and we can also assume that on your machine
INT_MIN is just -INT_MAX-1, our resulting value is a 'double' value
between -1.0 and INT_MAX/(INT_MAX+1).
Thanks!
You're welcome.
-Arthur
Christian Bau <ch***********@ cbau.freeserve. co.uk> wrote in message news:<ch******* *************** ***********@slb-newsm1.svr.pol. co.uk>... In article <c5**********@n ewsg1.svr.pol.c o.uk>, "Malcolm" <ma*****@55bank .freeserve.co.u k> wrote:
"pembed2003 " <pe********@yah oo.com> wrote in message int z = 1 + (int) (20.0 * rand() / (RAND_MAX + 1 )); it looks like x is always between 1 and 20. y is always 1 and z is always between negative and 0. i don't understand why.
The expression rand() / (RAND_MAX + 1) will be evaluated first as an integer expression. RAND_MAX on your machine must equal INT_MAX, so RAND_MAX + 1 is the lowest possible negative integer.
You should really know that this is not true.
20.0 * rand() / (RAND_MAX + 1) is evaluated as follows:
20.0 => Value 20, type double. rand () => Random value, type int, from 0 to RAND_MAX (RAND_MAX + 1) => Type int. If RAND_MAX == INT_MAX then -> overflow, undefined behavior; many implementations will produce INT_MIN.
20.0 * rand () => rand () is converted to double, result is of type double, from 0 to 20 * RAND_MAX. 20.0 * rand () / (RAND_MAX + 1) => RAND_MAX + 1 is converted to double. If there was no overflow in calculating RAND_MAX + 1 then the result is of type double, >= 0.0 and < 20.0. If there is an overflow, then undefined behavior, but most likely the result of (RAND_MAX + 1) is INT_MIN, which is often - (INT_MAX + 1) = - (RAND_MAX + 1), so the result is <= 0 and > -20.
Thanks for the explaination. Here is how I look at it:
(20.0 * rand() / (RAND_MAX + 1 ))
If I put parenthesis:
(((20.0) * (rand())) / ((RAND_MAX) + (1)))
so essentially:
(((double) * (int)) / ((int) + (int)))
double * int gives:
((double) / ((int) + (int)))
now since the second expression overflow (in my machine RAND_MAX ==
MAX_INT), we have:
((doubld) / ((MAX_INT) + (1))) = ((double) / (overflow))
which results in a nagetive number. make sense. But I can't understand
why:
(20 * rand() / (RAND_MAX + 1.0))
always result in a 1?
(((20) * (rand())) / ((RAND_MAX) + (1.0)))
so it's:
(((int) * (int)) / ((int) + (double)))
and then:
(((int) * (int)) / (double))
and then from here, I can't understand why it's always a 1. I can
understand that an int * int might overflow but will C promote it to a
double because we have a double just on the right side of /
Thanks!
On Fri, 15 Apr 2004, pembed2003 wrote: Christian Bau <ch***********@ cbau.freeserve. co.uk> wrote 20.0 * rand() / (RAND_MAX + 1) is evaluated as follows:
20.0 => Value 20, type double. rand () => Random value, type int, from 0 to RAND_MAX (RAND_MAX + 1) => Type int. If RAND_MAX == INT_MAX then -> overflow, undefined behavior; many implementations will produce INT_MIN.
20.0 * rand () => rand () is converted to double, result is of type double, from 0 to 20 * RAND_MAX. 20.0 * rand () / (RAND_MAX + 1) => RAND_MAX + 1 is converted to double. If there was no overflow in calculating RAND_MAX + 1 then the result is of type double, >= 0.0 and < 20.0. If there is an overflow, then undefined behavior, but most likely the result of (RAND_MAX + 1) is INT_MIN, which is often - (INT_MAX + 1) = - (RAND_MAX + 1), so the result is <= 0 and > -20.
[Why does]
(20 * rand() / (RAND_MAX + 1.0))
always result in a 1?
It doesn't, of course. It results in a floating-point 'double'
value, on your machine probably between -1.0 and 1.0 (even though of
course technically it could do whatever it liked).
Consider:
20 --> type 'int', value 20
rand() --> type 'int', value 0..RAND_MAX
20 * rand() --> type 'int', value 0..20*RAND_MAX
Now, at this point we realize that RAND_MAX is probably equal to
INT_MAX (as it is on most modern systems); so 20*RAND_MAX overflows
and we have a case of undefined behavior. Let's assume that 'int'
values overflow by "wrapping around," so that the value of 20*rand()
is still a random number between INT_MIN and INT_MAX.
20 * rand() --> type 'int', value INT_MIN..INT_MA X
RAND_MAX + 1.0 --> type 'double', value RAND_MAX+1
whole expression --> type 'double', value...
The value of the whole expression, assuming what we assumed above about
overflow on your machine, is a random number between
(INT_MIN/(RAND_MAX+1.)) and (INT_MAX/(RAND_MAX+1.)) . And since RAND_MAX
is assumed to be INT_MAX, and we can also assume that on your machine
INT_MIN is just -INT_MAX-1, our resulting value is a 'double' value
between -1.0 and INT_MAX/(INT_MAX+1).
Thanks!
You're welcome.
-Arthur This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
by: Niels Dekker (no reply address) |
last post by:
Is it possible for a standard compliant C++ compiler to have
( sizeof(short) < sizeof(int) )
and
( sizeof(short) == sizeof((short)0 + (short)0) )
?
Regards,
Niels Dekker
http://www.xs4all.nl/~nd/dekkerware
|
by: BigMan |
last post by:
Can someone cite the rules for type promotion in C++?
And, in particular, what is the type of the result of adding 2 values
of type char?
|
by: Andy |
last post by:
Hi...
i'm trying to understand the concept of function name overloading in
c++.
to understand the resolving system it's important to understand the
diffrent
levels of typecasting (exact match, promotion, conversion, not
possible)
that is, what i don't really understand. so i tried do make a little
|
by: Carsten Hansen |
last post by:
Suppose I'm using an implementation where an int is 16 bits.
In the program below, what function is called in the first case,
and what is called in the second case?
Also, if there is a difference between C89 and C99, I would
like to know.
I have tried with different compilers, and I see some differences.
Before I file a bug report with my C vendor, I would like to
know what the correct behavior is.
struct S
|
by: TTroy |
last post by:
Hello, I'm relatively new to C and have gone through more than 4 books
on it. None mentioned anything about integral promotion, arithmetic
conversion, value preserving and unsigned preserving. And K&R2
mentions "signed extension" everywhere.
Reading some old clc posts, I've beginning to realize that these books
are over-generalizing the topic. I am just wondering what the
difference between the following pairs of terms are:
1)...
| |
by: pachanga |
last post by:
Last week, I got a promotion to a higher position with new
responsibilities and new supervisor. My supervisor ask me about a new
position opening, and I took the position, but I totally forgot to ask
how much is my raise. A week has past and no one has mention it to me.
Next week I start training. What should I do? I feel kinda weird, one
thing is I do want to know now. And, the other is should I wait till I
go to the new position? What is...
|
by: Jon Paul Jones |
last post by:
For some time now, I have been looking for a build system for ASP.NET that
will merge the file based ease of deployment of classic ASP with the type
saftey and prebuilt nature of ASP.NET with codebehind. The team that I work
on manages several web sites and we have many different projects (not VS
projects) going on in each site simultaneously. Each of these projects may
have different deployment schedules. In a classis ASP world this was...
|
by: gen_tricomi |
last post by:
Python 2.4.2 (#67, Sep 28 2005, 12:41:11)
on win32
Type "copyright", "credits" or "license()" for more information.
****************************************************************
Personal firewall software may warn about the connection IDLE
makes to its subprocess using this computer's internal loopback
interface. This connection is not visible on any external
interface and no data is sent to or received from the Internet....
|
by: Frederick Gotham |
last post by:
I set about trying to find a portable way to set the value of UCHAR_MAX. At
first, I thought the following would work:
#define UCHAR_MAX ~( (unsigned char)0 )
However, it didn't work for me. Could someone please explain to me what's
going on? I would have thought that the following happens:
(1) The literal, 0, whose type is int, gets converted to an unsigned char.
|
by: sarathy |
last post by:
Hi,
What is integer promotion? How is it different from
arithmetic conversion?
Regards,
Sarathy
|
by: marktang |
last post by:
ONU (Optical Network Unit) is one of the key components for providing high-speed Internet services. Its primary function is to act as an endpoint device located at the user's premises. However, people are often confused as to whether an ONU can Work As a Router. In this blog post, we’ll explore What is ONU, What Is Router, ONU & Router’s main usage, and What is the difference between ONU and Router. Let’s take a closer look !
Part I. Meaning of...
| |
by: Hystou |
last post by:
Most computers default to English, but sometimes we require a different language, especially when relocating. Forgot to request a specific language before your computer shipped? No problem! You can effortlessly switch the default language on Windows 10 without reinstalling. I'll walk you through it.
First, let's disable language synchronization. With a Microsoft account, language settings sync across devices. To prevent any complications,...
|
by: jinu1996 |
last post by:
In today's digital age, having a compelling online presence is paramount for businesses aiming to thrive in a competitive landscape. At the heart of this digital strategy lies an intricately woven tapestry of website design and digital marketing. It's not merely about having a website; it's about crafting an immersive digital experience that captivates audiences and drives business growth.
The Art of Business Website Design
Your website is...
|
by: Hystou |
last post by:
Overview:
Windows 11 and 10 have less user interface control over operating system update behaviour than previous versions of Windows. In Windows 11 and 10, there is no way to turn off the Windows Update option using the Control Panel or Settings app; it automatically checks for updates and installs any it finds, whether you like it or not. For most users, this new feature is actually very convenient. If you want to control the update process,...
|
by: isladogs |
last post by:
The next Access Europe User Group meeting will be on Wednesday 1 May 2024 starting at 18:00 UK time (6PM UTC+1) and finishing by 19:30 (7.30PM).
In this session, we are pleased to welcome a new presenter, Adolph Dupré who will be discussing some powerful techniques for using class modules.
He will explain when you may want to use classes instead of User Defined Types (UDT). For example, to manage the data in unbound forms.
Adolph will...
|
by: conductexam |
last post by:
I have .net C# application in which I am extracting data from word file and save it in database particularly. To store word all data as it is I am converting the whole word file firstly in HTML and then checking html paragraph one by one.
At the time of converting from word file to html my equations which are in the word document file was convert into image.
Globals.ThisAddIn.Application.ActiveDocument.Select();...
|
by: TSSRALBI |
last post by:
Hello
I'm a network technician in training and I need your help.
I am currently learning how to create and manage the different types of VPNs and I have a question about LAN-to-LAN VPNs.
The last exercise I practiced was to create a LAN-to-LAN VPN between two Pfsense firewalls, by using IPSEC protocols.
I succeeded, with both firewalls in the same network. But I'm wondering if it's possible to do the same thing, with 2 Pfsense firewalls...
| |
by: 6302768590 |
last post by:
Hai team
i want code for transfer the data from one system to another through IP address by using C# our system has to for every 5mins then we have to update the data what the data is updated we have to send another system
|
by: bsmnconsultancy |
last post by:
In today's digital era, a well-designed website is crucial for businesses looking to succeed. Whether you're a small business owner or a large corporation in Toronto, having a strong online presence can significantly impact your brand's success. BSMN Consultancy, a leader in Website Development in Toronto offers valuable insights into creating effective websites that not only look great but also perform exceptionally well. In this comprehensive...
| |