So my question relates to double, I am trying to get an input from the user in decimal point for any value and its exponent also in decimal point to display the result after calculation in another function where the variables will pass values as double and I have used double the output as well but the end result is 1.00000 even though I have used the output specifier as %lf%.
#include <stdio.h>
double pwra (double, double);
int main()
{
double number, power, xx;
printf("Enter Number: ");
scanf("%lf", &number);
printf("Enter Number: ");
scanf("%lf", &power);
xx=pwra (number,power);
printf("Result: %lf", xx);
return 0;
}
double pwra (double num, double pwr)
{
int count;
int result = 1;
for(count=1;count<=pwr;count++)
{
result = result*num;
}
return result;
}
You have used the wrong type for result in the pwrs() function.
Change:
int result = 1;
to:
double result = 1.0;
Note that this type of simple mistake is easily identified if you learn to use your debugger. Further reading: How to debug small programs.
Also note that pwr should be an int, not a double, since your function only works with integer exponents.
Related
Do you know how a number can be rounded up and printed with a precision that is not fixed by some natural number, but by some variable? If the user needs to enter how many decimal places to round, how to solve it?
#include <stdio.h>
int main()
{
int r;
double var = 37.66666;
scanf("%d", &r);
printf("%.2f", var);
return 0;
}
Here's simple approach that would work in your case:
You just need to put * before f, and that's it.
#include <stdio.h>
int main()
{
int r;
double var = 37.66666;
scanf("%d", &r);
printf("%.*f",r, var);
return 0;
}
You can put a * in place of the precision, in which case you can specify an int as the precision.
printf("%.*f", r, var);
I wrote the following code in c to evaluate exponent of a number without using the math library
#include <stdio.h>
float powr(float,int);
int main(){
float a;
int b;
printf("Enter base and exponent a^b: ");
scanf("%.2f %d",&a,&b);
float p=powr(a,b);
printf("%.2f",p);
return 0;
}
float powr(float x,int y){
float r=1;
for(int i=1;i<=y;i++){
r=r*x;
}
return(r);
}
but no matter what base and exponent I input, the ouput always comes out to be 1.00. I can't find any mistake in this program and I tried running the powr function algorithm inside main() in a seperate program and it works.
In scanf() it onlys accepts field-width format, but no precision, see here.
You can just input the value, like this:
scanf("%f %d",&a,&b);
Also, you should always check the return value of scanf(). Something like this:
numOfItems = scanf("%.2f %d",&a,&b);
if(numOfItems != 2) // uh-oh
{
printf("Error while input!");
}
There are two floating point numbers 23.54 and 33.22 have to make a program to add them with just left side integral value like 23+33=56.
Here's the code that I tried:
int sum;
sum=(int)num1+(int)num2;
printf("%d",sum);
or
printf("%d",(int)num1+(int)num2);
The datatype of num1 is float and I'm using (int) to typecast to integer type.
Since we typecast the datatype this is called explicit typecasting!
Another option would be to define a function that explicitly names the types used, for example:
static int int_sum(int a, int b) {
return a + b;
}
which could be used as:
#include <stdio.h>
int main()
{
float a, b;
if (scanf("%f%f", &a, &b) == 2) {
printf("%d\n", int_sum(a, b));
}
return 0;
}
note that it's good practice to check that scanf successfully parsed the values it was supposed to before using them. You might instead want to return something else, or maybe print an error message, if the user didn't supply valid input.
#include <stdio.h>
int main(){
float sum;
float num1;
float num2;
printf("first number:");
scanf("%f",&num1);
printf("second number:");
scanf("%f",&num2);
sum=(int)num1+int(num2);
printf("sum is %.2f",sum);
}
We have been learning about recursion vs iteration in C this week and we were required to make a program that recursively determines the value of the nth term of a geometric sequence defined by the terms a, ar, ar^2, ... ar^(n-q).
For the most part, I think I have it figured out, as it seems to display the correct values per run, but it doesn't manage to break the recursion when the tested value reaches zero. Also, if possible to get a better explanation of recursion, and some examples of when recursion would be preferred over iteration as I'm still struggling with the concept.
// 2/20/2018
//Lab 6 Solution for Page 369 PE 4 B
//including libraries to be used
#include <stdio.h>
#include <math.h>
int main() {
//Function prototype
double goAnswer(int *, double, double, double, double *, int);
//Declaring variables
int nValue = 0;
double ratio = 0;
double firstTerm = 0;
double answer = 0;
double addedAnswer = 0;
int count = 1;
//Setting up to ask for each value
printf("Please enter in the value of n: ");
scanf("%d", &nValue);
printf("Please enter in the ratio you'd like to use: ");
scanf("%lf", &ratio);
printf("Please enter in the first term to use: ");
scanf("%lf", &firstTerm);
addedAnswer = goAnswer(&nValue, ratio, firstTerm, answer, &addedAnswer,
count);
//Printing out the value of the first nth terms
printf("The value of all terms added together is: %lf\n", addedAnswer);
return 0;
}
//function header
double goAnswer(int *nValue, double ratio, double firstTerm, double answer,
double *addedAnswer, int count) {
if (nValue == 0){
return 0;
}
else{ //This part calculates the answer, prints the value to the screen,
adds the answer to a running sum, decreases the nValue by one and calls the
function again with the lower nValue
answer = firstTerm * pow(ratio, count);
printf("The value of term %d is: %lf\n", count, answer);
printf("This is the nValue: %d \n", *nValue);
*addedAnswer += answer;
nValue -= 1;
return (goAnswer(nValue, ratio, firstTerm, answer, addedAnswer,
(count + 1)));
}
}
I am novice to C. I am trying to write a program to calculate the Hypotenuse, but when I run it I get garbage values and I don't know how to handle it.
double Hypotenuse(double base, double perpendicular)
{
double hyp = sqrt((base*base) + (perpendicular*perpendicular));
return hyp;
}
int _tmain(void)
{
double b, p;
printf("Enter the lenght of base\n");
scanf_s("%f",&b);
printf("Enter the lenght of perpendicular\n");
scanf_s("%f",&p);
printf("The hypotenuse of the triangle is %3.3f",Hypotenuse(b,p));
return 0;
}
problem is in your scanf statements.
As a thumb rule you should always use "%lf" for doubles.
This works perfectly:
double Hypotenuse(double base, double perpendicular)
{
double hyp = sqrt((base*base) + (perpendicular*perpendicular));
return hyp;
}
int _tmain(void)
{
double b, p;
printf("Enter the lenght of base\n");
scanf_s("%lf",&b);
printf("Enter the lenght of perpendicular\n");
scanf_s("%lf",&p);
printf("The hypotenuse of the triangle is %3.3lf",Hypotenuse(b,p));
return 0;
}
So, the best way to see what is going on in your code or where u get garbage, after all scanf print the value
ex:
double a;
scanf("%f",&a);
printf("a=%f",a); //OR
printf("a=%3.3f",a);
and you can easy see if the problem is in your scanf or somewhere else..