Ошибка при компилировании:[Linker Error] Unresolved external '_mirexit'

#include <iostream.h>
#include <elliptic.h>
#include <big.h>
#include <crt.h>
#include <time.h>

#include <vcl.h>
#pragma hdrstop

//---------------------------------------------------------------------------

#pragma argsused
/* large 1024 bit prime p for which (p-1)/2 is also prime */
char *primetext=
"1553155263514823959911559963512318072201696448283 78937433223838972232518351958838087073321845624756 55014694524600379010804594038319477343949605191701 98923701023413789901139595618958910197168732905128 15434724157588460613638202017020672756091067223336 19439491076530983087606624648015661749216414009542 7773547319";

/* Use elliptic curve of the form y^2=x^3-3x+B */

/* 168 bit elliptic curve prime 2#168-2#8-1 */

char *ecp="37414441915671114706014331717536845303191873 1001599";

/* elliptic curve parameter B */

char *ecb="47118247490432258461368463772056856484209695 203684";

/* elliptic curve - point of prime order (x,y) */

char *ecx="2";
char *ecy="25472391824616698863304500988247869413410992 0361326";

char *text="MIRACL - Fastest multi-precision library in the World!\n";

Miracl precision(120,0);


int main(int argc, char* argv[])
{
int ia,ib;
long seed;
Big a,b,p,q,n,phi,pa,pb,key,e,d,m,c,x,y,k;
Big primes[2],pm[2];
ECn g,ea,eb;
miracl *mip=&precision;

time(&seed);
irand(seed); /* change parameter for different values */

cout << "First Diffie-Hellman Key exchange .... " << endl;

p=primetext;

/* offline calculations could be done quicker using Brickell's idea
- See brick.cpp. Note use of "truncated exponent" of 160 bits -
could be output from hash function SHA (see mrshs.c) */

cout << "\nAlice's offline calculation" << endl;
a=rand(160,2);

/* 3 generates the prime sub-group of size (p-1)/2 */

pa=pow(3,a,p); // pa =3^a mod p

cout << "Bob's offline calculation" << endl;
b=rand(160,2);
pb=pow(3,b,p);

cout << "Alice calculates Key=" << endl;
key=pow(pb,a,p);
cout << key << endl;

cout << "Bob calculates Key=" << endl;
key=pow(pa,b,p);
cout << key << endl;

cout << "Alice and Bob's keys should be the same!" << endl;

/*
Now Elliptic Curve version of the above.
Curve is y^2=x^3+Ax+B mod p, where A=-3, B and p as above
"Primitive root" is the point (x,y) above, which is of large prime order q.
In this case actually
q= 37414441915671114706014327879091018088790755806326 7

*/
cout << "\nLets try that again using elliptic curves...." << endl;
a=-3;
b=ecb;
p=ecp;
ecurve(a,b,p,MR_PROJECTIVE);
x=ecx;
y=ecy;
g=ECn(x,y);
ea=eb=g;

cout << "Alice's offline calculation" << endl;
a=rand(160,2);
ea*=a;
ia=ea.get(pa); /* <ia,pa> is compressed form of public key */

cout << "Bob's offline calculation" << endl;
b=rand(160,2);
eb*=b;
ib=eb.get(pb); /* <ib,pb> is compressed form of public key */

cout << "Alice calculates Key=" << endl;
eb=ECn(pb,ib); /* decompress eb */
eb*=a;
eb.get(key);
cout << key << endl;

cout << "Bob calculates Key=" << endl;
ea=ECn(pa,ia); /* decompress ea */
ea*=b;
ea.get(key);
cout << key << endl;

cout << "Alice and Bob's keys should be the same! (but much smaller)" << endl;


/* El Gamal's Method */

cout << "\nTesting El Gamal's public key method" << endl;
p=primetext;
x=rand(160,2);
y=pow(3,x,p);
do
{
k=rand(160,2);
} while (gcd(p-1,k)!=1);

mip->IOBASE=256;
a=pow(3,k,p);
b=modmult(pow(y,k,p),(Big)text,p);
mip->IOBASE=10;
cout << "Ciphertext= \n" << a << "\n" << b << endl;

m=modmult(b,pow(a,p-1-x,p),p);
mip->IOBASE=256;
cout << "Plaintext= \n" << m << endl;
mip->IOBASE=10;

/* RSA. Generate primes p & q. Use e=65537, and find d=1/e mod (p-1)(q-1) */

cout << "\nNow generating 512-bit random primes p and q" << endl;
for(;
{
p=rand(512,2); // random 512 bit number
if (p%2==0) p+=1;
while (!prime(p)) p+=2;

q=rand(512,2);
if (q%2==0) q+=1;
while (!prime(q)) q+=2;

n=p*q;

e=65537L;
phi=(p-1)*(q-1);
if (gcd(e,phi)!=1) continue;
d=inverse(e,phi);
break;
}
cout << p << endl;
cout << q << endl;
cout << "n = p.q = \n";
cout << n << endl;

/* set up for chinese remainder thereom */

primes[0]=p;
primes[1]=q;

Crt chinese(2,primes);

mip->IOBASE=256;

cout << "Encrypting test string" << endl;
c=pow((Big)text,e,n); // c=m^e mod n
mip->IOBASE=10;
cout << "Ciphertext= \n";
cout << c << endl;

cout << "Decrypting test string" << endl;


pm[0]=pow(c%p,d%(p-1),p); /* get result mod p */
pm[1]=pow(c%q,d%(q-1),q); /* get result mod q */
m=chinese.eval(pm); /* combine them using CRT */

mip->IOBASE=256;
cout << "Plaintext= \n";
cout << m << endl;

return 0;
}