What would be the best way to remove the decimal point from a float number? For instance if I have a float number .2546 I would like to be able to remove the dot and use just the number (int) 2546 for some calculations.
Maybe, convert to string than remove the first character from the string than convert the string back to an int?
I was going through the exercises in one C programming ebook.There is this question which asks me to print a float variable in fixed decimal notation, field size of 6, right-justified, no digits after decimal point.I got printf("%6f", x );
x = float variable.
But the above code prints numbers after the decimal point, so I changed it to %d. But %d doesn't work with float variables..
I have a simple input output problem using float point numbers and after the first input the program skips the other cin functions is there something that I did wrong? It compiles fine also.
Code: #include <iostream> #include <float.h> using namespace std; int main() { int x; int y; int z;
I am entering numbers to float ... I want program to find out, which first number is not from specific interval. How to do it ? Example: Enter input : 5 10 20 30 50 46 . 30 is invalid. Here is the code :
Code:
while(scanf("%f",&input)!=EOF || input==0) { sum=input+sum; if (getchar() == '
How do you convert a number float in a range of -10.0f to 17.0f to a eqivalent number in the range of 0.0f to 1.0f?The code does not work well. floaty is the float to change.
//change range to 0..1 diamond[x][y] = (floaty - minY) / (maxY - minY);
I am unable to find why my code is going into infinite loop as below. This works perfectly fine if I keep entering just the integer values but if I enter a float number it ends up in an infinite loop
int main() { int x; while(1){ cin>>x; cout <<x; } return 0; }
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?
So, I'm supposed to do : Create a function with unlimited number of arguments, which forms a dynamic string based on the following form (%d, %s, %f, %lf, %c), with the following prototype:
char*create(char*form, ...);
The function is supposed to have the following output:
create("Peter is %d years old and is in %s-%c class.",7,"second",'A'); -> Peter is 7 years old and is in 7-A class. create("His GPA is %lf.",4.96); -> His GPA is 4.96. create("His favourite subject is math!"); -> His favourite subject is math!
The part with %d and %s string was not that hard, but now I'm supposed to convert %f and %lf to string, I've tried using sprintf but I've had no luck so far, another problem is the fact that I've gotta use lists to complete the task. I've been trying to convert float to string for the past 2 hours, but I'm drawing a blank now.
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
How to return the value after the decimal point. For example:if two integer numbers are 3, 4 then (3+4)/2 is 3.5 the 5 after the decimal point is to be returned. if suppose it is 34.456 i have to return 456.
I have one code that use MPI broadcast and I want to change it into Asynchronous Point to Point communication. I am newbie in Parallel programming. Looking for implementation of one simple same program in broadcast and P2P ?
determining if a value entered in loop is an odd or even number. Also, the value can't be int because it may be a decimal value(therefore i cant use the if(x%==0).
I need to replace the if(value%2 ==0) else num_even++ statement in my code with something else that will work with float to determine odd vs even.
#include <iostream> using namespace std; int main() { int num_values; float sum_values = 0; int num_neg_values = 0; int num_pos_values = 0;
N=1st number M=2nd number(accrding to my prof we will name it N and M ctr= increment of factorial of N, ctr2= increment of factorial of (N-M)
the problem is when i got the factorial of N / factorial of (N-M) i need to get the last non zero digit. so i use mod if it has zero in it. but mod can be only used with an int value. and when i change it to int value, the value of fact1 which is a float change
I am looking for a math/big num library, that allows me to convert 32/64/80 bot float numbers to string and vice versa.
Precision & accuracy is of importance here, and since this is an IEEE standard, i have high hopes that there are libraries for this out there, which would save me the hassle of trying to implement this myself...
I am working from my "ansi c" book by steven lawlor, page 73 program 2. write a program that accepts two numbers from the keyboard and prints the following information.
variables first second execution First number ? 7 Second number ? 2 the second goes into the first 3 times with a remainder of 1. the quotient is 3.5.
Code: #include <stdio.h> main() { int first, second; scanf(" %1i %1i", &first, &second); printf("First number ? %1i
[Code] ....
//I included this as I had some error message come up // before, not sure if this is correct though? } it shows what is expected but I cant get the 3.5.
I have tried %f and variations of width/precision but still not luck. Also, when I click on the application and put in the variables I press enter, the program executes and disappears so I cant see the result. how do I get the program to stay up until I want to get rid of it?
I just checking but confused with float. in that code same size int, and same type double are working but float showing nothing in printf..why?? i'm using GCC compiler int 32bit win7 os