C code for calculating simple interest
#include <stdio.h>
int main()
{
float principle, rate, years;
float simpleInterest = (principle * rate * years)/100;
printf("Enter the principle");
scanf("%f", &principle);
printf("Enter the rate");
scanf("%f", &rate);
printf("Enter the years");
scanf("%f", &years);
printf("The simple interest is %f", simpleInterest);
}
C variables are not mathematical variables: they do not express a relationship:
int y = x + 1; // this does NOT mean that `y` is always equal to `x + 1`
Instead, statements are executed from top to down, and when a variable name comes up in the code, the C language will read its contents[1]. Therefore:
int x = 5;
// now x == 5
int y = x + 1;
// now x == 5 and y == 6
x = 8;
// now x == 8 and y == 6
That means that your code should be:
#include <stdio.h>
int main(void) // NOTE: use `int main(void)` instead of `int main()`
{
float principle, rate, years;
printf("Enter the principle");
scanf("%f", &principle);
printf("Enter the rate");
scanf("%f", &rate);
printf("Enter the years");
scanf("%f", &years);
float simpleInterest = (principle * rate * years) / 100.0f;
printf("The simple interest is %f\n", simpleInterest);
return 0;
}
NOTES:
[1] = not always true, because of short circuiting when using operators like && and ||, and because operators like sizeof do not evaluate the expressions. But this is beyond the scope of my answer.
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The output always comes: 6.35 (I think it's a garbage value)
the code is
#include<stdio.h>
#include<math.h>
float formula(float,float);
int main()
{
float l=0,a=0;
printf("enter the length of the pendulum(l)\n");
scanf("%f",&l);
printf("Enter the angle of displacemnt(a)\n");
scanf("%f",&a);
printf("the length is %0.2f\n",l);
printf("the angle of displacemnt is %0.2f\n",a);
printf("the period of pendulum is %0.2f",formula(l,a));
return 0;
}
float formula(float l, float a)
{
float P=0,ran=0,g = 9.8;
ran = (l/g) * (1 + ((1/4)*(pow((float)(sin((double)a/2)),2))));
P = 2 * M_PI * ((float)sqrt((double)ran));
return P;
}
I don't know what is happening 😐
You can simplify this implementation by using doubles or floats exclusively. The 1/4 integer division yields 0. Here's an example:
#include <stdio.h>
#include <math.h>
float formula(float l, float a)
{
float ran = l / 9.8f * (1.0f + powf(sinf(a/2.0f), 2.0f) / 4.0f);
return 2.0f * (float) M_PI * sqrtf(ran);
}
int main(void)
{
float l;
float a;
printf("Enter the length of the pendulum(l): ");
scanf("%f", &l);
printf("Enter the angle of displacemnt(a): ");
scanf("%f", &a);
printf("The length is %0.2f\n", l);
printf("The angle of displacemnt is %0.2f\n", a);
printf("The period of pendulum is %0.2f\n", formula(l,a));
return 0;
}
I have written a program to calculate compound interest.
Here is the code:
#include <stdio.h>
#include <math.h>
int main(void) {
float value, rate, years,r;
int column = 0, tmp;
printf("Enter money values: ");
scanf("%f",&value);
printf("Enter a interest rate: ");
scanf("%f",&rate);
printf("Enter number of years: ");
scanf("%f",&years);
printf("\nYears ");
tmp = rate + 4;
r = rate;
for (int a = rate; a <= tmp; a++) {
printf(" %d ", a);
column++;
}
for (int b = 1; b <= column; b++) {
printf("\n %d",b);
for (int i = 1; i<= column; i++) {
printf(" %.2f ", (float) pow ( (value)*(1.0+((r/100.0)/(1.0))) , (1.0*b))
r++;
}
r = rate;
printf("\n");
}
// I = P*R*T
// P= AMOUNT (value)
// R=RATE (r)
// T=YEARS (b)
return 0;
}
It asks the user for a value (money), interest rate, number of years and displays the interest rate like so:
Enter money values: 100
Enter a interest rate: 6
Enter number of years: 5
Years 6 7 8 9 10
1 106.00 107.00 108.00 109.00 110.00
2 11236.00 11449.00 11664.00 11881.00 12100.00
3 1191016.00 1225043.00 1259712.00 1295029.00 1331000.00
4 126247696.00 131079600.00 136048896.00 141158160.00 146410000.00
5 13382255616.00 14025517056.00 14693280768.00 15386240000.00 16105100288.00
But my problem is the floating point calculation.
As you may be able to tell the numbers in the output above and very long and have many trailing digits
which i am very confused about.
For example in the 2nd row the first output is 11236.00,
this is wrong since it should be outputting 112.36 but for some reason the decimal has moved
forward two spaces. Why is this? and how could i fix this problem and print the correct solution
with the decimal in the correct place.
You have the value inside the pow. So when you square for two years, you are squaring the amount. Move the (value)* to output the pow call.
#include <stdio.h>
#include <math.h>
#define M_PI 3.14
int main(void)
{
// Declaring Variables
float time, p1, p2, p3, voltage, p3_num, p3_dem, epsilon;
int index, type;
// Asking user for time and index values into equation
printf("Please enter a time value: ");
scanf("%f", &time);
printf("Please enter an index value: ");
scanf("%d", &index);
// Calculate value for given time, part a, b, or c
printf("Do you want to run type a, b, or c?: ");
scanf("%d", &type);
char again = 'Y';
while (again == 'Y')
{
if (type == 'a') {
int going = 'Y';
while (going == 'Y')
{
// Different parts of formula to ensure accuracy
printf("Please enter an index value: ");
scanf("%d", &index);
p1 = ((3 / 2) - 12 / pow((M_PI), 2));
p2 = (1 / pow((double)(2 * index - 1), 2));
p3_num = ((2 * index - 1) * M_PI * time);
p3_dem = 3;
p3 = cos(p3_num / p3_dem);
voltage = p1 * p2 * p3;
printf("time = %f", time);
printf("index = %i", index);
printf("voltage = %f", voltage);
printf("Again? (Y/N)");
scanf("%c", &going);
}
}
}
}
There is more underneath, but this is the main bulk. What I am confused about is that none of this running, none. I start at the very beginning and comment out everything else, not even my print statement is running. What is wrong?
You should divide your program up into smaller function so you can test them individually.
For example here would be a function that reads in one integer.
int GetInteger(const char * msg)
{
//read up on c style formatting here is a source http://en.cppreference.com/w/c/io/fscanf
int i = 0;
printf("%s", msg);
scanf("%i", &i);
printf("Program confirms: %i\n", i);
return i;
}
This will help you down the road. Read up on SOLID principles and Test Driven Development.
I'm trying to write a code that calculates monthly pay for a project.
This is the formula I was given:
(Rate + Rate/((1+Rate)^Months)-1) * Principle
Rate according to this formula is Rate/1200 so as an example if the rate is 7% it would be 7/1200 which is 0.00583333333. I'm trying to get the exact number 0.00583333333 into my program but then I get the error "illegal use of floating point".
Here's my code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
float r;
int m, y;
int p;
//int mp;
printf("Enter Rate: ");
scanf("%d", &r);
r = r%1200;
printf("Enter number of years: ");
scanf("%d", &y);
m = y*12;
printf("%.10lf\n",r);
printf("%d",m);
return 0;
}
How do I get 0.00583333333 to be a part of my calculation in the program?
try to change scanf("%d", &r); by scanf("%f", &r); and r = r%1200 by r = r/1200
I am confused on how to complete this for loop. The mission is to read input in unix. For the input if the radius is >0 it should prompt the user each time and then if <=0 it should terminate. I am going from centimeters to square inches. My current configuration requires 2 inputs (1 prompted, 1 not) before giving output to the console. Cheers.
#include <stdio.h>
#define PI 3.14159
main()
{
float r, a;
int y = 9999999;
for(int i =0; i <y; i++){
printf("Enter the circle's radius (in centimeters): ");
scanf ("%f", &r);
if(r>0){
r=r;
a = PI * r * r *2.54;
printf("Its area is %3.2f square inches.\n", a);
} else {}
}
}
Your code flow is the following:
for (infinite condition) {
scan input
if (input > 0) {
do things
}
else {
do nothing
}
}
So there's no way to exit out of the loop, that's why the break statement exists, to force quitting an iterative block of code:
while (true) {
scanf ("%f", &r);
if (r > 0) {
// do whatever;
}
else
break;
}
The break will stop the cycle when executed, just going out of the loop.
You may want to try using a while loop instead so that the question is continually prompted until the user inputs a value =>0. see if below helps (also your conversion factor was not quite right);
#include <stdio.h>
#define PI 3.14159
void main()
{
float r, a;
printf("Enter the cirle's radius (in centimeters):");
scanf("%f",&r);
while (r>0)
{
a=PI*r*r*0.155; // conversion from sqcm to sqin is ~0.155
printf("Its area is %3.2f square inches \n", a);
printf("Enter the cirle's radius (in centimeters):");
scanf("%f",&r);
}
}
r=1.0f;
// break if no. of cases exhausted or r is negative or zero
for(int i =0; i < y && r > 0; i++)
{
printf("Enter the circle's radius (in centimeters): ");
if( scanf ("%f", &r) == 1) // Always check for successful scanf
{
a = PI * r * r/2.54/2.54; //This is correct formula
printf("Its area is %3.2f square inches.\n", a);
}
}
Consider a while loop instead:
#include <stdio.h>
#define PI 3.14159
main(){
float r, a;
int continueBool = 1;
while(continueBool == 1){
printf("Enter the circle's radius (in centimeters): ");
scanf ("%f", &r);
if(r>0){
a = PI * r * r *2.54;
//the above formula may be wrong, so consider trying:
//a = PI * r * r/2.54/2.54;
printf("Its area is %3.2f square inches.\n", a);
}
else{
continueBool = 0;
}
}
}
The break statement can be dangerous if you are new to C programming, so I recommend not using it until you get a better understanding of C and break. If you do want to use break, then this could be your solution:
#include <stdio.h>
#define PI 3.14159
main(){
float r, a;
while(1){
printf("Enter the circle's radius (in centimeters): ");
scanf ("%f", &r);
if(r<=0){
break;
}
a = PI * r * r *2.54;
//the above formula may be wrong, so consider trying:
//a = PI * r * r/2.54/2.54;
printf("Its area is %3.2f square inches.\n", a);
}
}
Use this:
for(int i =0; i < y; i++)
{
printf("Enter the circle's radius (in centimeters): ");
scanf ("%f", &r);
if(r > 0)
{
a = PI * r * r *2.54;
printf("Its area is %3.2f square inches.\n", a);
}
else
{
break;
}
}