I can't seem to make this printf line work
printf("THE NUMBER PI
To six decimal places, pi is equal to \%f" ,3.141593 " which is roughly \%i", 3 "
");
I'm not sure what the correct syntax is to start and end a new write.
main()
{
printf("%lf",5/2);
}
this code is giving runtime error ....why ?
Code:
#include<stdio.h>
#include<conio.h>
void main()
{
float i;
i=0.7;
[Code] ....
If i do run the above program in turbo C/C++ complier, it outputs "h". But,if i change the code as i=0.6 and if (i<0.6), it outputs "w". Even if i change it to i=0.8 and if(i<0.8), then also it outputs "w".
Code:
#include<stdio.h>
#include<conio.h>
float square(float);
void main() { clrscr();
float a,b;
printf("ENter a Number");
scanf("%f",&a);
[Code] ....
In the above program, I am calculating the square of float number. But sometimes the number is entered as NAN and sometimes Output is NAN. What is NAN? I am entering floating point number, then y NAN is entered?
SEE the Image attached for the OUTPUT.
if we have a decimal number like c=3.46
And i want to set two number, a and b
now a= static_type<int>(c); so a=3;
and i want b= 46
which is the two numbers after the decimal how can I do that ? how can I set b = 46 ?
P.S: i do not know what c equals to. now it's two number after the floating point but it might be more or less
I am having trouble understanding the mantissa of a floating point number. I have divided up the floating point number into the sign bit, the exponent and the mantissa, I have found the exponent, but I am not sure what to do with the mantissa? From what I have gathered so far i divide the mantissa by ten until I get a number between 1 and 10. after that i convert the number into a decimal with everything after the decimal point (or radix) being a fractional number. But when I do that on paper I dont get my intended number. How do i put the exponent and mantissa together to make a decimal from my floating point?
ex. input is 00111010000111111111011000001000
sign is 0
exponent is 01110100 which is 64+32+16+4-127=-11
mantissa is 00111111111011000001000 which would be 1.11111111011000001
When i convert that i get 1.99756622314 i dont know what to do with the -11 exponent and the answer i want is 6.1e-4
I'm trying to convert 4 hex register into floating point value using IEEE 754 floating point format. My device will reply 4 register value. The problem is that it always reply for example 0x10 as 10 when i use getc() hence using char variable to store it is not ideal.
Code:
union {
char c[4];
float f;
} conv;
Write a function named "sum" that takes as its arguments the following:
(1) an array of floating point values.
(2) an integer that tells how many floating point values are in the array.
The function should return as its value the sum of the floating point values. For example, if the array that's passed to the function looks like this:
0 1 2 3 4
5.8|2.6|9.0|3.4|7.1
The function should return the value as 27.9 as its value...I know I need to declare and intialize the array in the main function. I don't how to write the sum function.
I'm currently trying to learn about floating point representation in depth, so I played around a bit. While doing so, I stumbled on some strange behaviour; I can't really work out what's happening...
#include <iostream>
#include <cmath>
using namespace std;
int main(){
float minVal = pow(2,-149); // set to smallest float possible
float nextCheck = ((float)((minVal/2.0f))); // divide by two
[Code] ....
Essentially what's happening is:
- I set minVal to be the smallest float that can be represented using single precision
- Dividing by 2 should yield 0 -- we're at the minimum
- Indeed, isZero2 does return true, but isZero returns false.
What's going on -- I would have thought them to be identical? Is the compiler trying to be clever, saying that dividing any number cannot possibly yield zero?
Suppose I wanted to check if a given floating point value is within a certain range. What would be Your recommended approach when considering simplicity, speed, portability, etc.?
View 7 Replies View RelatedFunction is not returning a decimal point value. Here is my function
int meanValueFunction(vector<int> arrayValues){
int sum = 0;
sum = sumFunction(arrayValues);
float meanValue = sum/arrayValues.size();
cout<< meanValue << endl;
return meanValue;
I want result in decimal point i.e 27.2 for the values (2 4 20 10 100) but it returns 27 instead.
I need to implement a C++ program that asks the user for four floating-point numbers. The program should then calculate the average using two different functions, one value returning and one void. The program should output the average of the four numbers. For this program I need to use float instead of int for the types of variables. Below is a proto-type code that I am able to use to do this program.
#include <iostream>
using namespace std;
int sum(int,int);
[Code].....
I'm displaying a table of floating point numbers with setprecision(5). If the number is "1.25" it will display "1.2500" which is what I want. However, if the number is "0.25" it will display "0.25000"
How can I make numbers with a base number of zero display properly?
And it is not running successfully... abnormal termination
View 3 Replies View RelatedCan we change the floating point number format from scientific format to below example format ?
FROM TO
==========================
2.06374E-03 ---> 206370-8
-4.30311E-01 ---> -.430310
-4.28146E-04 ---> -42815-8
==========================
I have written this regex to match a floating point literal:
(^[[:space:]]*)(([0-9]+.?[0-9]*([eE][+-]?[0-9]+)?)|"
"(.[0-9]+([eE][+-]?[0-9]+)?))([fFdD]?[[:space:]]*)$
and when I match it with string like "123e" or "e2" it works while it shouldn't and I can't find the reason why.
I have the problem of trying to find the smallest natural number that makes two consecutive terms in single precision floating point notation in the riemann zeta function equal. So basically the riemann function of 2 is given by:
sum of 1/(k^2) from k=1 until infinity, so : 1/(1^2) + 1/(2^2) + 1/(3^2) + ...... until infinity.
Now the question asks to stop at the smallest natural number n, at which the sum 1/1^2 + 1/2^2 + ......+ 1/(n^2) is equal to the sum 1/1^2 + 1/2^2 + ..... + 1/((n+1)^2) in single precision floating point notation.
Now well the obvious way to look for n would be on this way:
float i = 1.0;
float n = 1/(i);
float n1 = 1/(i+1.0);
while ( n != n1){
i += 1.0;
n = 1/i;
n1 = 1/(i+1.0);}
But first of all this is obviously completely inefficient and I dont think it will yield a valid answer for any float variable, i.e. I dont think the sum until 1/n^2 and 1/(n+1)^2 will ever differ. I tried it out with the largest possible value of a variable of type float and the values were still seen as unequal in C++. How to do this? Will C++ find a value for n for which the condition holds? Is the compiler or my hardware important for this, i.e. would I get different results on a different pc?
I heard that the speed of floating point multiplication is much faster than division. Is it still the case today?
View 14 Replies View RelatedI have a program which generates lots of data points in 2D (x and y co-ordinates). Perhaps 1,000,000+ of these points. These are floating point numbers. The values all fall within a specific range -R/2 and +R/2.
I want to impose an 'imaginary' grid. Such that I can collect the number of each of these points falling within a specific grid location. The actual 'grid' size is variable.
Should I try and collate this data from my C++ program and then post-process it in some other s/w like matlab?
The question was to evaluate postfix expression (floating point numbers). I had been able to implement stack data structure using one way singly linked list linked list but I am not been able to extract the original input by the user expressions like
ex:
1. 252.124 3453.7 * 46.3 346.2 23.6 ^/$
2.45.23 87.045 * 6.5 ^$
etc,($ELIMETER)
How to take such inputs from the user for proper evaluation . Previously I tried to extract separate digits from integer and decimal fields and computed numbers. The method is very lengthy. Any optimised way for taking such input!
I have to code a program to calculate magnetization and strain. I have everything working except for the "INPUT LOAD"(bold lines). It returns as an illegal use of floating point.
# include <stdio.h>
# include <stdlib.h>
# include <conio.h>
# include <math.h>
# include <string.h>
# define KTOL 500000
# define Num_pos 100
[Code].....
I am trying to make sure no floating point numbers can be input at any time in my roulette program. My code is too long to put in one post so I shall piece it out
#include <iostream>
#include <iomanip>
#include <random>
#include <cstdlib>
#include <ctime>
using namespace std;
// structure of arrays to pass to functions
[Code] .....
If I have the following code:
long lSecondsSum = 8039;
double dNumDays = lSecondsSum / (24 * 3600);
I expect to get 0.093044 but for some reason I am getting dNumDays = 0.0000000000.
However, if I write the code as follows:
long lSecondsSum = 8039;
double dNumDays = lSecondsSum/24;
dNumDays = dNumDays/3600;
then I get correct dNumDays = 0.092777777777777778.
Also, how do I avoid all these floating point errors.
This code snippet just won't compile in a conforming compiler:
Code:
template<> struct series<0.0, 0, 0>
Because of the floating point template parameter.... If this "specialization" is useful, how come MS would have discarded it?
I wrote a pong game and it keeps crashing, it only happens on a score and says floating point exception.
Code:
//The headers
#include"SDL/SDL.h"
#include"SDL/SDL_image.h"
#include"SDL/SDL_ttf.h"
#include<string>
#include<stdio.h>
#include<math.h>
[Code] ....
I wrote a program that use a struct to represent an athlete. The program allocates memory for an array of 5 Athletes, and, soon after I enter the fourth data (height) for the first athlete, I get the message "runtime Error R6002 - floating point support not loaded". The program doesn't reach the line where __LINE__ would be printed.
Code:
#include <stdio.h>
#include <stdlib.h>
typedef struct athlete {
char name[ 100 ];
char sport[ 100 ];
double height;
[Code] .......
