Question

This is my first post and I have to admit, I am terrible at programming. I am that guy in the class that works his tail off, but can never seem to grasp programming as well as the rest of my classmates. So please be nice, I will try to explain my problem below.

I have the following code (comments removed), but when I run it I get a warning similar to the one listed below. Also, when I run the program, the first user inputted value is allowed, but then all of the sudden, it jumps to the end of the program, not allowing me to input the values for the other variables (e.g. the variable “beta”). I have an image of the output (https://i.stack.imgur.com/yc3jq.jpg) and you can see that I enter alpha, but then the program runs to the end. Any thoughts?

Thank You so very much for your help!
-Spencer

—————————–CODE—————-

    #include<stdio.h>
#include<stdlib.h>
#include<math.h>

float alpha, beta, h; 
float slope_k (float, float, float, float); 
float slope_q (float, float, float, float); 
float slope_p (float, float, float, float); 

int main (void)
{

float t0=0, tf, h, S0, I0, R0, k1, k2, k3, k4, q1, q2, q3, q4, p1, p2, p3, p4;
int N;
char sim_file[1000];  
FILE *out_file;
float *time_out, *s_out, *i_out, *r_out;

printf("Enter the value of the rate constant for infection (alpha) \n");
scanf("&f", &alpha);

printf("Enter the value of the rate constant for recovery or death (beta) \n");
scanf("&f", &beta);

printf("Enter the value of number of persons susceptible to the given contagion [S] at the initial  time zero [i.e. S(t)=S(0) = ? ] \n");
scanf("&f", &S0);

printf("Enter the value of the number of persons infected [I] at the intial time zero [i.e. I(t) = I(0) = ?] \n");
scanf("&f", &I0);

printf("Enter the value of the number of persons that have already been infected but have recovered [or died] [R] at the initial time zero [i.e. R(t) = R(0) = ?] \n");
scanf("&f", &R0); 

printf("Enter the final time for solution \n");
scanf("&f", &tf);

printf("Enter the solution step size (H) \n");
scanf("&f", &h);

N = (int)(tf/h);

printf("Enter file solution to store solution to simulation \n");
scanf("&s", sim_file);

out_file = fopen(sim_file, "w");

time_out = (float *)calloc(sizeof(float), N);
s_out = (float *)calloc(sizeof(float), N);
i_out = (float *)calloc(sizeof(float), N);
r_out = (float *)calloc(sizeof(float), N);


time_out[0]= 0; 
s_out[0] = S0;
i_out[0] = I0;
r_out[0] = R0;

for(int i = 0; i < N; ++i);
{
int i = 0;
time_out[i+1] = (i+1)*h;

k1 = h*slope_k(time_out[i], s_out[i], i_out[i], r_out[i]);
q1 = h*slope_q(time_out[i], s_out[i], i_out[i], r_out[i]);
p1 = h*slope_p(time_out[i], s_out[i], i_out[i], r_out[i]);

k2 = h*slope_k(time_out[i]+(h/2), s_out[i]+(k1/2), i_out[i]+(q1/2), r_out[i]+(p1/2));
q2 = h*slope_q(time_out[i]+(h/2), s_out[i]+(k1/2), i_out[i]+(q1/2), r_out[i]+(p1/2));
p2 = h*slope_p(time_out[i]+(h/2), s_out[i]+(k1/2), i_out[i]+(q1/2), r_out[i]+(p1/2));

k3 = h*slope_k(time_out[i]+(h/2), s_out[i]+(k2/2), i_out[i]+(q2/2), r_out[i]+(p2/2));
q3 = h*slope_q(time_out[i]+(h/2), s_out[i]+(k2/2), i_out[i]+(q2/2), r_out[i]+(p2/2));
p3 = h*slope_p(time_out[i]+(h/2), s_out[i]+(k2/2), i_out[i]+(q2/2), r_out[i]+(p2/2));

k4 = h*slope_k((time_out[i] + h), (s_out[i]+k3), (i_out[i]+q3), (r_out[i]+p3));
q4 = h*slope_q((time_out[i] + h), (s_out[i]+k3), (i_out[i]+q3), (r_out[i]+p3));
p4 = h*slope_p((time_out[i] + h), (s_out[i]+k3), (i_out[i]+q3), (r_out[i]+p3));

s_out[i+1] = s_out[i] + (1.0/6)*(k1 + (2*k2) + (2*k3) + k4);
i_out[i+1] = i_out[i] + (1.0/6)*(q1 + (2*q2) + (2*q3) + q4);
r_out[i+1] = r_out[i] + (1.0/6)*(p1 + (2*p2) + (2*p3) + p4);

}

return 0;
}

float slope_k(float t, float s, float i, float r)
{
float slope_k_out;
slope_k_out = -alpha*s*i;
return slope_k_out;
}

float slope_q(float t, float s, float i, float r)
{
float slope_q_out;
slope_q_out = (alpha*s*i)-(beta*i);
return slope_q_out;
}

float slope_p(float t, float s, float i, float r)
{
float slope_p_out;
slope_p_out = beta*i;
return slope_p_out;
}

---

Example warning:

warning C4996: 'scanf': This function or variable may be unsafe. Consider using scanf_s instead. To disable deprecation, use _CRT_SECURE_NO_WARNINGS. See online help for details.

Answer 1

What the compiler is telling you here is that the function scanf is not safe. scanf has a bug that, if exploited, can cause a system to become compromised (called a buffer overflow attack). In brief, the bug is that one does not tell scanf how many bytes to read for input. Thus scanf will read until it “believes” it is done reading the input. In a char array, this end is usually the null character '\0'. However, if one leaves off '\0' from a string, scanf will continue reading until it finds that byte — usually, scanf will reach a memory location that is outside of its own virtual memory space. This action will cause the OS to send your program a segmentation fault (seg fault) which will summarily end your program’s existence.

The newer function, scanf_s,(_s for secure), lets you determine the max size of the input, which you can use to more effectively prevent buffer overflow attacks. If this is for a HW assignment, which it looks like it is, you can leave scanf there. However, to get rid of the compilier warning AND try and become a better programmer, fix it! Use sscanf_s and have a global variable (or something…) that determines maximum input size (e.g. int SCANF_INPUT_SIZE = 1000 ).

Good luck!

EDIT — Change those "&f" to "%f" that’s the error!