Write a program that uses two pointer variables to read two double numbers and display the absolute value of their sum?
this is my code and I don not know where it gets wrong:
int main(void)
{
double *p1,*p2, val1,val2;
p1 = &val1;
p2 = &val2,
printf("Enter two number: ");
scanf("%f %f", p1,p2);
if(*p1+*p2 >= 0)
printf("%f\n", *p1+*p2);
else
printf("%f\n", -(*p1+*p2));
return 0;
}
if you have to scan doubles you use "lf" and to print them as well. It was your only mistake. "f" it is just for floats.
int main(void)
{
double *p1,*p2, val1,val2;
p1 = &val1;
p2 = &val2,
printf("Enter two number: ");
scanf("%lf %lf", p1,p2);
if(*p1+*p2 >= 0)
printf("%lf\n", *p1+*p2);
else
printf("%lf\n", -(*p1+*p2));
return 0;
}
http://www.cplusplus.com/reference/cstdio/scanf/
Please consult this website or a similar one for errors or warnings.
%f is used for float values, and c compilers often issue warnings or stop compilation when type conversions occur without specifying them.
%lf is used for doubles.
Related
I've tried to build a simple calculator for physics force experiments.
//
// main.c
#include <stdio.h>
int main() {
char name[20];
int month,date;
int difference_percentage1, difference_percentage2, difference_percentage3;
double force1, force2, force3;
scanf("%s", name);
scanf("%d %d", &month, &date);
scanf("%lf %lf %lf", &force1, &force2, &force3);
double Favg=(force1 + force2 + force3)/3.000;
puts(name);
difference_percentage1=100*(force1-Favg)/Favg;
difference_percentage2=100*(force2-Favg)/Favg;
difference_percentage3=100*(force3-Favg)/Favg;
printf("%d %d %d\n", difference_percentage1, difference_percentage2, difference_percentage3);
getchar();
return 0;
}
The calculate doesn't match what I've typed for scanf().
The o's of the % mark for the %f look a bit larger to me than for the %d, so my guess is that you used the wrong % for the %f format specifier which isn't recognized by the compiler; it interprets the double value as int.
Edit: one more reason to post text instead of a screenshot of the code! (o;
Edit2: Yup, your % mark in %f is actually a "EF BC 85" in hex but should be "25"
I have problem with following fragment of code:
int main()
{
int n = 3;
complex_t t[3];
for(int i = 0; i < n; i++)
{
double x, y;
printf("Enter complex number: ");
scanf("%f %f", &x, &y);
t[i].re = x;
t[i].im = y;
}
}
When I am trying to pass x, and y to the program, it doesn't change value, and is 0.00000.
Can you help me?
You've declared x and y as doubles, but then went on to use the format specifier for floats (%f). So the end result is reading them in as if they were regular floats, then jamming the result into a double leaving you with unexpected values when trying to actually use them.
You need to use the format specifier specifically for doubles (%lf) here.
scanf("%lf %lf", &x, &y);
See format string specifications for more information about different format specifiers.
I am very new to C, and while working on a project which requires pulling an indeterminate amount of values from the console, I am finding that it is not pulling the correct values. It seems like addresses, which I believe means it is a pointer issue, but I can't seem to find it.
int getVals(int degree){
double sum;
double x;
double coefs[degree];
for(int counter = 0; counter<=degree; counter = counter+1){
double nxt;
scanf(" %d", &nxt);
coefs[counter] = nxt;
printf("coefs[%d] = %d\n", counter, coefs[counter]);
}
printf(" x ? ");
scanf(" %d", &x);
printf("degree %d x %d\n", degree, x);
sum = poly(x, degree, coefs);
printf ("polynomial evaluate to: %lf\n", sum);
int newDegree;
scanf(" %d", &newDegree);
degree = newDegree;
if(degree>-1){
getVals(degree);
}
else
return degree;
}
Note: poly returns a double result of the evaluated polynomial
I am getting the following infinite loop after entering a degree of 1 and a coefficient of 1.5. It does not allow me to enter an x.
Infinite loop
In scanf(" %d", &newDegree); you should use the "%lf" format specifier (since your values is a double, not an int). Change the format specifier in all your calls to scanf() and "%f" in calls to printf().
Please refer to the documentation at this links printf(3), scanf(3).
I have a question in C where I need to insert coefficients of a quadratic equation into a function and return the number of solutions and result.
Write a program that accepts a series of 3 real numbers, which are the
coefficients of a quadratic equation, and the program will print out
some solutions to the equation and the solutions themselves.
Guidelines:
Functions must be worked with one of the functions that
returns the number of solutions as a returned value, and returns the
solutions themselves through output parameters.
3 numbers must be
received each time. The input will be from a file (will end in EOF)
In the meantime I built the function without reading from a file just to see that it works for me, I built the function that returns the number of solutions but I got entangled in how to return the result as output parameter
here is my code for now:
int main ()
{
double a, b, c, root1,root2,rootnum;
printf("Enter coefficients a, b and c: ");
scanf("%lf %lf %lf",&a, &b, &c);
rootnum=(rootnumber(a,b,c);
printf("the number of roots for this equation is %d ",rootnum);
}
int rootnumber (double a,double b, double c)
{
formula=b*b - 4*a*c;
if (formula<0)
return 0;
if (formula==0)
return 1;
else
return 2;
}
In C, providing an "output parameter" usually amounts to providing an argument that is a pointer. The function dereferences that pointer and writes the result. For example;
int some_func(double x, double *y)
{
*y = 2*x;
return 1;
}
The caller must generally provide an address (e.g. of a variable) that will receive the result. For example;
int main()
{
double result;
if (some_func(2.0, &result) == 1)
printf("%lf\n", result);
else
printf("Uh oh!\n");
return 0;
}
I've deliberately provided an example that illustrates what an "output parameter" is, but has not relationship to the code you actually need to write. For your problem, you will need to provide two (i.e. a total of five arguments, three that you are providing already, and another two pointers that are used to return values to the caller).
Since this is a homework exercise, I won't explain WHAT values your function needs to return via output parameters. After all, that is part of the exercise, and the purpose is for you to learn by working that out.
Apart from a wayward parenthesis in the call and some other syntax errors, what you have so far looks fine. To print out the number of roots, you need to put a format specifier and an argument in your printf statement:
printf("the number of roots for this equation is %d\n", rootNum);
The %d is the format specifier for an int.
Here is your working code:
#include <stdio.h>
int rootnumber (double a,double b, double c)
{
double formula = (b*b) - (4*(a)*(c));
if (formula > 0) {
return 2;
}
else if (formula < 0) {
return 0;
}
else {
return 1;
}
}
int main (void)
{
double a, b, c;
printf("Enter coefficients a, b and c: ");
scanf("%lf %lf %lf",&a, &b, &c);
printf("The number of roots for this equation is %d ", rootnumber(a,b,c));
return 0;
}
It just need some sanity checking, its working now:
#include<stdio.h>
int rootnumber(double a, double b, double c);
int main ()
{
double a, b, c, root1,root2;
int rootnum;
printf("Enter coefficients a, b and c: ");
scanf("%lf %lf %lf",&a, &b, &c);
rootnum=rootnumber(a,b,c);
printf("the number of roots for this equation is %d", rootnum);
return 0;
}
int rootnumber(double a, double b, double c)
{
int formula= (b*b) - (4*a*c);
if (formula<0)
return 0;
if (formula==0)
return 1;
else
return 2;
}
Hello guys I started learning C few weeks ago and I am trying to make my first useful program, I am trying to create a squared equation calculator but no matter what I type in the input it always outputs "no solution2" can anyone take a look and help me ?
examples for input :
0 0 0=
1 4 1=
#include <stdio.h>
#include <math.h>
int main()
{
printf("enter a\n");
double a; scanf("%1f", &a);
printf("\nenter b\n");
double b; scanf("%1f", &b);
printf("\nenter c\n");
double c; scanf("%1f", &c);
if (a==0)
{
if (b==0)
{
if (c==0)
printf("x can be every number\n");
else
printf("no solution1\n");
}
else
{
printf("x equals %.2f\n", ((-1)*c) / b);
}
}
else
{
double delta = (b*b)-(4*(a*c));
if (delta>0)
{
double sqrtDlt = sqrt(delta);
printf("x1 = %4.2f\n", (((-1)*b) + sqrtDlt) / 2 * a);
printf("x2 = %4.2f\n", (((-1)*b) - sqrtDlt) / 2 * a);
}
else
printf("no solution2\n");
}
return 0;
}
For double use specifier %l(ell)f not %1(one)f in scanf.
Note that this is one place that printf format strings differ substantially from scanf (and fscanf, etc.) format strings. For output, you're passing a value, which will be promoted from float to double when passed as a variadic parameter. For input you're passing a pointer, which is not promoted, so you have to tell scanf whether you want to read a float or a double, so for scanf, %f means you want to read a float and %lf means you want to read a double.