what I do I cannot get a division to work:
Code:
//END RANGE INPUT
long double End;
printf("
Please enter the start of the range (Lower Bound):
");
[Code]...
No matter what I input for the values of 'Start', 'End' and 'Interval', the value of 'SizeL' always seems to be -2.
I have the following details
double x= 1.5
double y= -1.5
int m= 20
int n= 4
my question is 5 * x - n / 5 at which what would n / 5 equal to, I think its zero since its integer division? or would the 5 be considered a real number?
I want to find the remainder of the division between a and b, but without using the reminder operator a%b.I thought to subtract b from a as long as a>b, that will give the remainder, but I don't know how to write it in code.
View 11 Replies View RelatedHow do i get this with decimal part?:
for(i=1;i<=4;i++){
printf("%d
",m);
s = s + m/i
m = m + 2;
}
m/i?
I have to build a program that calculates the remainder of the expression
(2^10)!/((2^10-500)! * 500!)
when divided by 10^9+7
where x^y = x*x*x*x...*x (y times)
and x! = x*(x-1)...*1
How can I do that? I know how to calculate the remainder of x! and the remainder of y!, but I do not know how t calculate the remainder of x!/y!. I can´t even store this in a variable because x! is very large.
Question about instead of using the division operator to display the output of user"s input....
View 4 Replies View RelatedAt one point in my C++, non-CLR program, the following code:
Code:
unsigned int size = 3;
float maxX = (float)(int(size-1))/2.0f;
std::cout << maxX;
outputs 107. Is it something about a conversion from unsigned int to float?
I heard that the speed of floating point multiplication is much faster than division. Is it still the case today?
View 14 Replies View Relatedwhy are the arithmetic operations like division(p/q) and multiplication(p*q) invalid on pointers?.here p and q both are pointers .
View 6 Replies View RelatedSo I am using Visual Studio 2012 Professional, this is C++ code. I am just trying to get the remainder from a very simple division. Nothing difficult, heres the code:
double getProbability(){
int rd = random();
int max = numeric_limits<int>::max();
double result = rd % max;
cout << "Probability: " << result << "
";
return result;
}
When I look at the values in debug I get:
max 2147483647
rd 1804289383
result 1804289383.0000000
That is completely wrong. The answer should be 0.840188. What is going on here?
random() just returns a number from a vector that was prepopulated with "random" integers. Not really random, but that isn't all that important. What is important is why on earth is a % operation returning such a huge number. I assigned the values to variables so I could look at them in the debugger. I know I am going to probably get a thousand different ways that I could do this "better" but again, that isn't what I am looking for. I would just like to know why the % operation is doing what it is doing?
C++ only allow addition and subtraction operation with pointer .why multiplication and division is not allowed? Then how to perform multiplication and division with pointer
View 3 Replies View RelatedThis code works very oddly.
Code:
#include <algorithm>
#include <cstdlib>
#include <iostream>
[Code].....
For class I need to write a program that inputs a file (the dividend), performs binary division on the file (using 0x12 as the divisor), and outputs the remainder(checksum).
I have researched binary division algorithms and I get the general gist, but I'm still unsure where to start. How would I store the dividend and divisor? As two arrays of bits?
Then, I need to figure out how to perform shifts and XORs on the the binary numbers. Maybe I should use bitwise operations?
Say I wanted to overload the modulus operator to return the remainder of a division between two floating point numbers. Why isn't a custom double operator%(double, double) allowed even though that function isn't available in the standard anyway?
View 5 Replies View RelatedI was required to write a program that takes a baseball players statistics and displays there averages. I was required to make 3 function in the file to perform this tasks. my problem I am having a division problem in the SLG function. My compiler does not require the system ("PAUSE"); command.
OUTPUT
The player's batting average is: 0.347
The player's on-base percentage is: 0.375
The player's slugging percentage is:
(test)AB = 101
(test)Tot Base = 58
0.000
Code:
/* Batting Average Program
file: batavg1CPP.cpp
Glossary of abbreviations:
BA = batting average
PA = plate appearances
H = hits
BB = bases on balls (walks)
[Code] ....
I've just started learning from "Jumping into C++". Great book. Unfortunately, I've also encountered my first snag. Practice problem number 3 from chapter 4 tells me to make a small calculator that takes one of the four arithmetic operations and its two arguments as input and give the result as output. Here's what my newbie mind came up with:
Code:
#include <iostream>
#include <string>
using namespace std;
int main()
[Code] .....
The thing that I don't get is why doesn't the function "getline" work. It's worked in previous programs. The program seems to work if I simply replace getline with a simple "cin". I could easily use that as a cheap fix but I am interested in knowing why "getline" refuses to work anymore....
I am writing a simple console-based tic-tac-toe game. I am trying to write a function to check whether someone has won the game.
The board data is saved in an array called board: Code: int board[3][3] with each element corresponding either to an empty spot, an X, or an O, using the numbers 0, 1, and 2, respectively. For clarity:
Code:
#define EMPTY 0
#define X 1
#define O 2 Here is my function: Code: int check_state(int board[3][3]) {
int winner = 0;
[Code].....
I just dont see what the issue is here. I have stared at this thing forever. Im trying to make a calendar from scratch so I can be prepared for my second test on Friday.
Code:
#include<stdio.h>
#include<stdlib.h>
int main(void)
{
int i, n, s;
[Code]....
Assuming that we have the string :
Code:
***HELLO&SIR. & and * is the delimiters.
My question is how strtok will work in the beginning?
According to this to determine the beginning and the end of a token, the function first scans from the starting location for the first character not contained in delimiters (which becomes the beginning of the token). And then scans starting from this beginning of the token for the first character contained in delimiters, which becomes the end of the token. The scan also stops if the terminating null character is found.
This end of the token is automatically replaced by a null-character, and the beginning of the token is returned by the function. strtok will replace the first * with '' and the next call will start from the second * puts '' there and again the same with the third * ?
I am unable to get the string seaching function to work. it always says "Nothing found" I am stumped.
Code:
#include <stdio.h>
#include <string.h>
char tracks[][80] = {
"I left my Heart at Harvard Medical School",
"Newark, Newark, You suck balls",
"Dancing with a dork",
"From Here to maternity",
"The Girl from Iwo Jima",
[Code]....
how the -bounds annotations work, and I came up with the following test program.The alloc_char function ensures maxRead/ maxSet(str) == len. This is, in turn, ensured by malloc and memset (see the comments; I found this in /usr/share/splint/lib/standard.h), but SPLint doesn't recognize that.However, SPLint correctly sees that requires maxRead(str) >= i is fulfilled in main.
Code:
/*
* splint -strict
*/
#include
#include
#include
#include
}
[code]....
(Of course the main function is contrived and I could use puts instead, but that's not the point. The point is to make it see that i is not out of bounds.)I have struggled with -bounds for a very long time. Earlier I gave up and used -likelybounds instead. But now I have written a library which is annotated for splint -strict with -likelybounds, and I want people to be able to use it with the full -bounds checks. I can't get those ensures and requires annotations for maxRead/Set to work.
I have a program that spawns a number of threads. Each thread generates a table of output. During the time that it is creating this output, I don't want the other threads to be outputting, so I've attempted to create a mutex semaphore to allow only one thread at time to ouput.
Code:
sem_t mutex;
void* node(void *thread_arg) {
// set a mutex lock to allow the output for this next step to complete for this thread
sem_wait(&mutex);
// generate table with various fprintf statements
}
[code]....
This approach isn't working.
puts(pd) does not give any output in the below code. Also, the first puts(ps) does give an output, but the second puts(ps) after "while" does not give any output.
Code:
#include <stdio.h>
int main()
{
char source[50]="Alice has apples.";
char destination[60];
char *pd, *ps;
}
[code].....
I have an issue getting boost libraries to work with QT IDE. Here are relevant lines from my .pro file...
INCLUDEPATH += c:/users/bob/desktop/boost_1_53_0/
LIBS += -LC:c:/users/bob/desktop/boost_1_53_0/stage/lib64
After adding #include <boost/filesystem.hpp> I get error...
C:UsersobDesktopqtpuntitled5mainwindow.cpp:3: error: C1083: Cannot open include file: 'boost/filesystem.hpp': No such file or directory
The IDE does not highlight the line. The error only appears on compilation.
I want to access the body of the Add() of a list in c# to see how it works, but it only just gives me the declaration.
[DebuggerTypeProxy(typeof (Mscorlib_CollectionDebugView<>))]
[DebuggerDisplay("Count = {Count}")]
[Serializable]
public class List<T> : IList<T>, ICollection<T>, IList, ICollection, IReadOnlyList<T>, IReadOnlyCollection<T>, IEnumerable<T>, IEnumerable {
public void Add(T item); // thats all. I tried go to declaration but still gives me this line of code. This is from metadata in Visual studio.
}
How this thing work. It just a declaration not a definition yet its still doing something. How is that possible.
#include <iostream>
#include <string>
using namespace std ;
int main() {
string bored ;
do {
cout << " program" <<endl ;
[Code] .....
I made this as a simple do/while program, and if i run it, the second do/while statement will keep on going forever, without the [cin >> bored;] line working?
