The output is
x=1000300 y=1000000, z=1000300
I can understand how I got x and z but c's y's output makes no sense.
#include <stdio.h>
int main()
{ int i=0;
float a = 100;
a = a*a*a*a*a;
float c = 3;
float x = 1000000*c + a;
float y = a;
float z = 0;
for (i=0; i<1000000; i++)
{ y += c;
z += c;
}
z += a;
x /= 10000;
y /= 10000;
z /= 10000;
printf("x=%.0f y=%.0f, z=%.0f\n", x, y, z);
}
The value in y starts out at 1E10 (from the assignment to a). You add 3 to this a million times.
The trouble is that a float has at most 7 significant decimal digits, so you effectively do not change y each time, hence the result divided by 10,000 is 10,000,000 1,000,000 as displayed.
If you coded it with double, you would see more nearly the result you expect.
Related
I am pretty sure I got this right just want to make sure though... In the following code, we have omitted the definitions of constants M and N
here are the two functions in question arith and optarith:
#define M /* Mystery number 1 */
#define N /* Mystery number 2 */
unsigned int arith(unsigned int x, unsigned int y){
unsigned int result = 0;
result = x * M + y/N;
return result;
}
We compiled this code for particular values of M and N. The compiler optimized the multiplication and
division. The following is a translation of the generated machine code back into C:
unsigned int optarith(unsigned int x, unsigned int y){
unsigned int t = 3 * x;
x <<= 6;
x -= 2 * t;
y >>= 4;
y = y/3;
return x + y;
}
so here's my work
so to find value M for the operation M * x
x <<= 6; // same as x = x * 64
x-= 2 * t; //same as x = x - 2 * (3 * x)
so basically x = 64x - 6x which is just 58x therefore M = 58
and to find the value N for operation y/N
y >>= 4; same as y = y/16;
y = y/3; same as y = (y/16)/3;
that makes y/48 so N = 48
Did I do this correctly?
Second part of Q: Then solve the integral between 0 and y of (x^2)(e^(-x^2))dx=0.1 for y using bracketing and bisection.
Here's what I have done so far:
#include <stdio.h>
#include <math.h>
double f(double x, double y);
int main(void) {
int i, steps;
double a, b, y, h, m, lower, upper, x, simp, val;
/*
* Integrate (x^2)(e^(-x^2)) from 0 to y
*/
steps = 20000;
a = 0;
b = y;
h= (b-a)/steps;
/*
* now apply Simpson's rule. Note that the steps should be even.
*/
simp = -f(a, y);
x = a;
for (i =0; i < steps; i += 2) {
simp += 2.0*f(x,y)+4.0*f(x+h, y);
x += 2*h;
}
simp += f(b, y);
simp *= h/3.0;
/*
* print out the answer
*/
printf("The integral from 0 to y with respect to x by Simpson's Rule is %f\n", simp);
/*
* Now we need to bracket and bisect to find y
*/
lower = 0;
/*
* Lower bound is from turning point
*/
upper = 100;
/*
*Upper value given.
*/
while (upper - lower > 10E-10){
m = (lower + upper)/2;
val = f(m, y);
if (val >=0)
upper = m;
if (val <=0)
lower = m;
}
m = (lower + upper)/2;
printf("The value for y is: %lf\n", m);
return 0;
}
double f(double x, double y) {
return pow(x,2)*exp(pow(-x,2))-0.1;
}
Output: The integral from 0 to y with respect to x by Simpson's Rule is -0.000000
The value for y is: 0.302120
It runs but doesn't do exactly what I need it to do. I need to be able to continue working with the integral once I've used 0 and y as the limits. I can't do this. Then continue on and solve for y. It gives me a value for y but is not the same one I get if i solve using online calculators. Also, the output gave zero for the integral even when I changed the equation to be integrated to x^2. Can anyone help explain in as simple terms as possible?
I've written some code to Integrate the function 5x^4 + 4x^3 + 3x^2 + 2x + 1.
#include<stdio.h>
#include<math.h>
float func(float x){
float a;
a = 5*pow(x,4) + 4*pow(x,3) + 3*pow(x,2) + 2*x +1;
return a;
}
int main(){
float numberOfXValues;
float a = 0; //lower limit
float b = 1; //upper limit
float numberOfStrips;
float stripSize;
float finalAnswer;
float sumFirstAndLast; //summation of first and last x value
while(1){
printf("Input number of X values:");
scanf("%f", &numberOfXValues);
numberOfStrips = numberOfXValues - 1;
stripSize = (b - a)/(numberOfStrips);
sumFirstAndLast = 0.5*func(a) + 0.5*func(b);
for (float z = stripSize; z < b; z += stripSize ){
sumFirstAndLast += func(z);
}
finalAnswer = sumFirstAndLast * stripSize;
printf("%f\n", finalAnswer);
}
return 0;
}
And it works for the majority of values, but the output for 13 and 20 is giving the wrong answer. I've looked through it a few times but can't see what could be causing this.
Input number of X values:10
5.039070
Input number of X values:11
5.031651
Input number of X values:12
5.026160
Input number of X values:13
6.271982
Input number of X values:14
5.018732
Input number of X values:15
5.016153
Input number of X values:16
5.014071
Input number of X values:17
5.012367
Input number of X values:18
5.010955
Input number of X values:19
5.009773
Input number of X values:20
5.798243
Input number of X values:21
5.007917
The problem is that in the exit condition of the inner loop, you compare two floating point numbers. After numberOfStrips iterations, it's not guaranteed that z == b because of floating point accuracy. It's entirely possible that z < b or z > b. In the z < b case, the loop it executed another time.
You should make numberOfStrips an integer and rewrite the loop like this:
float z = stripSize;
for (int i = 1; i < numberOfStrips; i++) {
sumFirstAndLast += func(z);
z += stripSize;
}
At a first glance this looks to precision and rounding issues of float.
Therefore change float to double.
For example on my machine I get 6.271982 for an input of 10 when using float while I get 5.021983 as result when using double.
Moreover you should use an epsilon in floating point comparisons.
Placing the line:
printf("%f, %d\n", z, z < b);
Inside the for loop will print the z values each iteration along with the loop condition result. The last output of this will be:
1.000000, 1
Which is pointing that we have already reached the right limit, but still performing the loop body. Literally it is thinking that 1.000 < 1.000. It can happen when working with float numbers, so in order to prevent it, consider looping on some integer number (numberOfXValues for example).
I was coding up a short program to compute right endpoints for a given graph (not shown here), so to avoid tedious computation, I decided to write up a short program to do the work for me. Yet my C program just prints out nan. I am very rusty on C, but I am not sure why I am getting NaN.
#include <stdio.h>
int main(void) {
int x;
float y, z;
for (x = 0; x <= 8; x++) {
y += 10.0 - (12.0 + (float)x) / 4.0;
printf("%f\n", y);
}
z = 0.5 * y;
printf("%f\n", z);
return 0;
}
y = 10.0 - (12.0 + (float)x) / 4.0;
Followed by
y = y+1;
This makes sense else you have y uninitialized which leads to undefined behavior because the value of y is undeterminate.
During declaration you can initialize y and use += operator.
Like
float y = 0;
So I have to make this formula "y = y / (3/17) - z + x / (a % 2) + PI" in C
I am having a problem with (a%2) as it is returning odd values. ie 1%2 = 0.000001
int assignment7()
{
#define PI 3.14
int a=0,amod2;
double Ny=0,y=0,z=0,x=0;
printf("Enter values for x,y,z and a: ");
scanf("%d%lf%lf%lf",&a,&y,&z,&x);
//printf("%d,%lf,%lf,%lf\n",a,y,z,x);
//amod2=1%2;
//printf("%lf",amod2);
Ny=y/(double)(3/17) - z+x / amod2 + PI;
printf("%lf\n",Ny);
When you say:
printf("%lf",amod2);
the compiler expects amod2 to be a "long float" (aka a double), but you defined it as:
int amod2;
Also your prompt says "x,y,z and a" but you read in the order "a,y,z,x":
printf("Enter values for x,y,z and a: ");
scanf("%d%lf%lf%lf",&a,&y,&z,&x);
that's awkward at best.
EDIT: cleaned up a bit and made some assumptions about order of operations:
#include <stdio.h>
#define PI 3.14
#define DIVSOR (3.0/17.0)
int assignment7 ( void );
int assignment7 ( void ) {
double x = 0.0;
double y = 0.0;
double z = 0.0;
int a = 0;
int amod2;
double Ny;
printf("Enter values for x,y,z and a: ");
scanf("%lf%lf%lf%d",&x,&y,&z,&a);
amod2 = a % 2;
Ny = (y / DIVSOR) - z + (x / amod2) + PI;
printf("%lf\n", Ny);
return 0;
}
int main ( void ) { return assignment7(); }
You don't say what inputs you are giving it, (a test case with inputs and the expected results would be super helpful), but I can point out that x / (a % 2) is going to be infinity when a is 2 or 4 or 6 or ...