我的教授在课堂上给了我们这段代码来展示一个程序如何工作,并说“回家试试看,你会看到它的作品”...... 30分钟后,我无法得到它跑步。有人可以帮助我,并指出我在正确的方向。谢谢!功能定义不允许/
-I得到在端“双克(双X)” 函数定义-On否则该第一其中x_left = x_mid控制到达非void函数的端
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
#define FALSE 0
#define TRUE 1
#define NO_ROOT -99999.0
//function prototypes
double bisect(double, double, double, double f(double farg));
// evaluation of function
double g(double);
double h(double);
int main(void) {
double x_left, x_right, epsilon, root; //declare variables
// get endpoint and error tolerance
printf("\nEnter interval endpoints > ");
scanf("%lf%lf", &x_left, &x_right);
printf("\nEnter tolerance > ");
scanf("%lf", &epsilon);
//use bisect function to look for roots of functions
printf("\n\n For function g(x)");
root = bisect(x_left, x_right, epsilon, g);
if (root != NO_ROOT)
printf("\n g(%.7f) = %e\n", root, g(root));
printf("\n\n For function h(x)");
root = bisect(x_left, x_right, epsilon, h);
if (root != NO_ROOT)
printf("\n h(%.7f) = %e\n", root, h(root));
system("pause");
return (0);
}
// bisection method program coding
double bisect(double x_left, double x_right, double epsilon, double f(double farg)){
double x_mid, f_left, f_right, f_mid;
int root_found;
// computes function at initial end points
f_left = f(x_left);
f_right = f(x_right);
// if no change in sign
if (f_left * f_right > 0) {
printf("\nmay not be no root in [%.7f, %.7f]\n\n", x_left, x_right);
return NO_ROOT;
}
// searches as long as interval size is large enough
root_found = FALSE;
while (fabs(x_right - x_left) > epsilon && !root_found) {
// compute the mid point
x_mid = (x_left + x_right)/2.0;
f_mid = f(x_mid);
if (f_mid == 0.0) {
root_found = TRUE;}
else if (f_left * f_mid < 0.0) {
x_right = x_mid;
} else {
x_left = x_mid;
}
// trace loop execution
if (root_found)
printf("\nRoot found at x = %.7f , midpoint of [%.7f, %.7f] ", x_mid, x_leftx_right);
else
printf("\nNew interval is [%.7f, %.7f] \n\n", x_left, x_right);
//if there is a root
return ((x_left + x_right)/2.0);
}
// functions for which roots are sought
double g(double x){
return (5 * pow(x, 3.0) - 2 * pow(x, 2.0) +3);
}
double h(double x){
return (pow(x, 4.0) - 3 * pow(x,2.0) - 8);
};
}
缺少}后'x_left = x_mid;}'' – chux
双克(双)'是坏的,此心不是所述函数的函数原型或评估。 –
@Ben'double g(double);'看起来像一个函数原型。 – chux