Function definition not allowed / [closed] - c

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My professor gave us this code in class to show how a program works and said "go home and try it and you'll see it works".... well after 30 minutes I cannot get it to run. can someone please help me and point me in the right direction. Thank you!
-I get function definition on the end "double g(double x)"
-On the first else where x_left = x_mid control reaches end of non-void function
#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);
};
}

Answer 1

I get an error on this line:
printf("\nRoot found at x = %.7f , midpoint of [%.7f, %.7f] ", x_mid, x_leftx_right
saying that x_leftx_right is undeclared.
If I change this to x_left, x_right then it compiles OK except for "undefined reference to g" and "undefined reference to h".
The reason for the undefined reference to g is that you never provided a function definition for the function g that was prototyped by double g(double);. You did provide a nested function g within bisect. Nested functions are a non-standard extension, and bisect::g is a different function to g. Similarly for h.
To fix this, move the definitions of g and h to be after the end of the bisect function; instead of inside that function.
The reason for your "control reaches end of non-void function" warning is probably because there is no return statement after the while loop.
Your line return ((x_left + x_right)/2.0); line is within the loop begun by while (fabs(x_right - x_left) > epsilon && !root_found) {. If this loop finishes by the loop condition no longer being true, then the execution hits the end of the function without returning anything.
NB. If you indent your code properly so that you line up { then you are less likely to have this sort of problem. Your editor should have a key that you can use to find matching curly-braces. Also, operating your compiler in strict standard mode would have given an error about the use of nested function.
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