C :: Dividing Two Integers
May 14, 2014
i am relatively new to C programming so i run into problems on daily basis. But this time i have something i just cant figuer out and i was hoping you could point me towards the right track. I am trying to divide two integers.DevValue by KpTotal. for some reason my micro controller allways crashes.Y is a variable of a distance measuring sensor. i have a 4x4 keypad to enter a three digit number (e.i 123) so Kp1 = 1 Kp2 = 2 Kp3 =3.
Code:
int kp1, kp2, kp3, kpTotal = 0;
char txt[6] = ""
int keypadPort at PORTD;
sbit LCD_RS at RB4_bit;
sbit LCD_EN at RB5_bit;
}
[code]....
i think it has something to do with the format of the value. i read that the micro controller crash when dividing by zero.
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Dec 2, 2014
So I have a simple calculator that does a few operations (+ - * / %) Pretty basic stuff
I declared int x, y for the numbers, char operation, and float result.
the code is based on switch(operation)
The program is running alright, but when I divide 8/7 it returns 1 as the result, I tried changing the x and y to float but that won't work because of the case '%'
I also tried making local float variables in the case'/' but it won't compile "E2126 Case bypasses initialization of a local variable"
How can I make the division work and return a float value?
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Apr 4, 2015
I wrote the following code to divide 100 customers into three clusters but it kept on hanging during execution. I used while loop. Attached is the text file to use. and the code below :
#include <iostream>
#include <iomanip>
#include <fstream>
#include <string>
#include <stdlib.h>
#include <cstdlib>
#include <ctime>
#include <stdio.h>
#include <math.h>
[Code] ....
Attached File(s)
c201.txt (7.13K)
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Jun 21, 2013
i made a calculator that you have to give terms and operators one by one but now i want to improve it. the user now should be able to type something like 5*78+325/sin(3*pi)-1 and the program should be able to calculate it. but i don't know how to use the signs '*','+','-' and '/' as delimiters and turn the input string into a string array. i mean the string "5*78+325/sin(3*pi)-1" will become a string array like "5,*,78,+,325,/,sin(3*pi),-,1,"
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Mar 2, 2013
So i have this -
time_t t4 = time(NULL);
//and then some stuff
time_t t5 = time(NULL);
iElapsedYah = t5-t4;
fElapsedYah2 = iElapsedYah/iNumTimes;
printf("It took %f seconds", fElapsedYah2);
but it always prints 0.0000000, and i dont know why even though individually
iElapsedYah = not zero and iNumTimes = not zero
I am using windows 7 and m complier is Dev C++
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May 5, 2014
I need to read a float number and show the rest of his division by an integer, but i'm having the following error message:
Quote
error: invalid operands of types 'float' and 'int' to binary 'operator%'
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Mar 12, 2014
I'm supposed to store the value of a countrys population. Then gather out the percentage that countrys population holds when compared with the global population.
Anyway here's the code:
Code:
#include <iostream>
long swe_pop = 9644864;
int main ()
[Code] .....
The result I'm getting is 0%.
I was under the impression that long (or long long) integers could hold high values. And that I could then divide these and answer with a float type value. Giving space for the decimals.
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Jan 9, 2014
Im using the remquo function in the cmath library as follows:
int quotient;
double a = remquo (10.3, 4.5, "ient);
This gives the correct remainder (a = 1.3) and quotient (quotient = 2).
Infact about 50% of the answers are right when I play around, however, trying something like:
int quotient;
double a = remquo (2.25, 1.5, "ient);
yields an incorrect quotient of 2 and remainder of 0.
I do think this has something to do with float arithmetic. I recall tinkering with the float number 0.500 and that the CPU actually saves it as 0.50000000000000231. However if my suspicion of float arithmetic as the suspect is correct, I do not understand why a tenth decimal would make such a drastic difference as changing the quotient result.
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Apr 21, 2014
I'm a beginner at c++ and I need to write a program that reads a set of integers and then finds and prints the sum of the even and odd integers. The program cannot tell the user how many integers to enter. I need to have separate totals for the even and odd numbers. what would I need to use so that I can read whatever number of values the user inputs and get the sum of even and odd?
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Apr 7, 2013
I'm writing a program to calculate a final grade by adding 4 numbers minus the lowest grade and dividing by 3. My knowledge in c is not extensive I thought that a simple assigment operator would do the job but I'm getting a strange large numbers in the execution.
Code:
#include <stdio.h>
#include <stdlib.h>
main(){
int eg, g1, g2, g3, g4, fg, s1, s2, sg;
[Code] ....
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Oct 5, 2014
Write a program, sumit.c, that reads an integer value n from the user and then sums the integers between 0 and n. Your program should be able to handle both positive and negative values of n and should write the sum to the output file sumitout.txt. Note that your sum should not actually include 0, since that doesn't affect the sum. Your program should open the file sumitout.txt for appending so that the file can record a series of runs.
For marking purposes run your program twice with values n = −10 and n = 5.
Here is a sample run:
Enter n: -4 The sum of integers from -1 to -4 is -1
#include<stdio.h>
int main(void) {
//declare your variables
int n, i, term, total = 0;
//promt user to enter and read a value of n
printf("Enter n: ");
scanf("%d", &n);
//calculate the sum
[Code] ....
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Mar 12, 2014
How to write a simple function that will take 3 ints and find the sum of the higher 2? This is what i got so far:
int findsum(int a,int b,int c)// will find the highest int and return it to our main program {
int max,max2;// this sets our local variable max
// next we will find the larger of our first 2 variables
if( a>=b)
{ max=a;
[Code] ....
How to get the second highest number and add it to max
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Apr 18, 2014
I am working on an assignment where I have to subtract two very large integers. How can I write it's code provided that I have both numbers in a character array with each index holding a fraction of the original number. Like if I have a number 123456789 in an array then
arr[0]='1';
arr[1]='2';
arr[2]='3';
arr[3]='4';
and so on. n
nNw if i have to subtract this number from an other number, how can I do that?
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Nov 8, 2014
Code:
#include <iostream>
#include <string>
using namespace std;
int main()
[Code] ......
How do I link the strings to the integers?
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Oct 17, 2014
I'm trying to make a C++ program that generate 6 random numbers ( from 100,000 to 999,999 ), okay, so I wrote a few lines..
Code:
srand(time(0));
for (int i = 0; i < 5; i++)
{
std::cout << 100000 + (rand() % 999999) << std::endl;
}
The problem is, that it generates numbers like this:
117,207
123,303
131,083
... etc etc..
They're all starts with 100K, i want them to be an actual random..
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Dec 3, 2013
I am doing a program and i need to read integers that are in a string and i dont know if i am doing it correctly.
Code:
unsigned int j=0, h1;
char num[100] , str1[100], str2[100], a[100], b[100];
sscanf(str1, "%[^;] ; %[^;] ; %d ", a, b, &h1); the string is : "0048915433190;24/10/2013;13h 7m 47s;13h 8m 59s;0;1;"
And what i want to read with sscanf is the hours, minutes and seconds.
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Nov 12, 2013
I am trying to create a program that reads my file filled with random words, it then compares the words after they are put into a 2d array and sees if there is any matching words.. unfortunately the count is not working for me (in function2 and function3) and I am not sure why..
Code:
#include<stdio.h>
#include<string.h>
char function1(char words_array[][17]);
int function2(char words_array[][17]);
void function3(int pairs, char words_array[][17]);
int main( void )
{ char words_array[20][17];
int x = 0;
[Code]...
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Oct 15, 2014
How i can concatenate two integers into one integer. I have code that concatenate two integers but if the 2nd integer is zero it won't work. How can i modify it so that it can cater the case of y=0 too.
Code:
int concatenate(int x, int y) {
int pow = 10;
while(y >= pow)
pow *= 10;
return x * pow + y;
}
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Dec 2, 2014
I compiled a simple program using a function to add two integers, in order to understand the difference of calling by value vs by reference.
Code:
#include <stdio.h>
int add(int a, int b, int sum);
int main()
{
int a, b, sum;
printf("Please enter two integers
");
scanf("%d%d", &a, &b);
[Code]...
I used these two (one, two) pages I found in the net while studying the subject. So here are my questions...
(a) Do I understand correctly that in the above code all a, b and sum are passed by value?
(b) If I wanted only sum to be passed by reference, how the above code would change?
(c) If I wanted all (a, b, sum) to be passed by reference, how the above code would change?
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Jan 22, 2013
"Write a program that prompts the user for an integer number from the keyboard, and store it in a variable num. After each number is typed in, add num to a total.
Do this six times, each time adding num to a total. (Be sure to reuse the same variable named num for each of the six numbers entered.) When all six numbers have been entered from the keyboard, calculate an average. Display the total and average calculated. "
Here is what I have so far:
Code:
#include<stdio.h>
int main() {
int num, total1, total2, total3, total4, total5, total6, avg;
printf("Enter first number:");
scanf("%2d",&num);
[Code] .....
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Apr 5, 2013
from k&r book (4.2 - functions retuning non integers)
(doesnt compile - 1 error)
Code:
#include <stdio.h>
#define MAXLINE 100
/* rudimentary calculator */
main()
{
double sum, atof(char []);
char line[MAXLINE];
int getline(char line[], int max);
sum = 0;
while (getline(line, MAXLINE) > 0)
printf(" %g
", sum += atof(line));
return 0;
}
Code:
int getline(char line[], int max); is no different from:
Code:
getline(char line[], int max); ?
so, the int (return type) is optional when using a function. i assume return type is casted double? why the (char [])? just weird cuz i never seen it before
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Mar 25, 2013
I am trying to make a game where you have a secret code that is coded with colors like ROYG (red,orange,yellow,green) and I am having trouble when it tells you when you have a right color in the right spot or a right color in the wrong spot when you guess a color. How can I change my code under the function int comparearray where it will compare pointers to pointers and not integers and give me the correct number of "almost" and "correct".
Code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define ROWS 100
#define COLS 4
}
[code]....
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Feb 25, 2014
How do I do the operation of two integers that gives you the results?
I'm supposed to write a program that:
Asks the user for an integer
Asks the user for one of '+', '-', '*', or '/' (as a character)
Asks the user for another integer
Performs the given operation on the two integers, and prints out the result of
Please enter an integer: 4
Please enter an operation (+, - , *, /): *
Please enter another integer: 5
4 * 5 = 20
Code:
#include <iostream>
using namespace std;
int main() {
int x;
int c;
int d;
char e;
[Code] ....
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Apr 5, 2013
I have a file where the first column shows letters, second column shows numbers.
How could I make it to print everything exactly how it is in the file – a vector/array with strings and integers?
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Oct 2, 2014
I took a month's break from programming, and got rusty. Here's the problem:
The character 'b' is char('a'+1), 'c' is char('a'+2), etc. Use a loop to write out a table of characters with their corresponding integer values:
a (tab) 97
b (tab) 98
...
z (tab) 122
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Jan 26, 2014
I'm confused about the actual value of 8 bytes for unsigned integers.
The below code suggests the value is 13217906525252912201:
#include <stdio.h>
#include <inttypes.h>
typedef uint64_t byte_int_t;
int main(void){
byte_int_t t;
printf("%" PRIu64 "
", t);
}
./runprogram
13217906525252912201
However, when I use a calculator, I get a different value: 2^64= 1.8446744e+19
So I was wondering is this really 8 bytes? So I try below test and it produces 8, as expected:
#include <stdio.h>
#include <inttypes.h>
typedef uint64_t byte_int_t;
int main(void) {
byte_int_t t;
printf("%u
", sizeof(t));
return 0;
}
So why does C and my calculator provide two different results?
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