我已经阅读了model answer on unresolved externals并发现它非常有用,并且已经将其归结为最后两个超出我的顽固错误。
我已附上所有代码以防万一,如果您想查看标题或其他任何内容请说。
// Stokes theory calculations
#include <math.h>
#include <stdio.h>
#include <process.h>
#include <string.h>
#include <conio.h>
#include <stdlib.h>
#define ANSI
#include "../Allocation.h"
#define Main
#define Char char
#define Int int
#define Double double
#include "../Allocation.h"
#include "../Headers.h"
Double
kH, skd, ckd, tkd, SU;
Double
ss[6], t[6], C[6], D[6], E[6], e[6];
// Main program
int main(void)
{
int i, Read_data(void), iter, Iter_limit = 40;
double F(double), kd1, kd2, kFM, omega, delta, accuracy = 1.e-6, F1, F2, Fd;
void CDE(double), AB(void), Output(void);
Input1 = fopen("../Data.dat", "r");
strcpy(Convergence_file, "../Convergence.dat");
strcpy(Points_file, "../Points.dat");
monitor = stdout;
strcpy(Theory, "Stokes");
strcpy(Diagname, "../Catalogue.res");
Read_data();
z = dvector(0, 2 * n + 10);
Y = dvector(0, n);
B = dvector(0, n);
coeff = dvector(0, n);
Tanh = dvector(0, n);
monitor = stdout;
H = MaxH;
iff(Case, Wavelength)
{
kd = 2. * pi / L;
kH = kd * H;
CDE(kd);
}
// If period is specified, solve dispersion relation using secant method
// Until February 2015 the bisection method was used for this.
// I found that in an extreme case (large current) the bracketting
// of the solution was not correct, and the program failed,
// without printing out a proper error message.
iff(Case, Period)
{
fprintf(monitor, "\n# Period has been specified.\n# Now solving for L/d iteratively, printing to check convergence\n\n");
omega = 2 * pi / T;
// Fenton & McKee for initial estimate
kFM = omega*omega*pow(1 / tanh(pow(omega, 1.5)), (2. / 3.));
kd1 = kFM;
kd2 = kFM*1.01;
CDE(kd2);
F2 = F(kd2);
for (iter = 1; iter <= Iter_limit; ++iter)
{
CDE(kd1);
F1 = F(kd1);
Fd = (F2 - F1) / (kd2 - kd1);
delta = F1 / Fd;
kd2 = kd1;
kd1 = kd1 - delta;
fprintf(monitor, "%8.4f\n", 2 * pi / kd1);
if (fabs(delta / kd1) < accuracy) break;
F2 = F1;
if (iter >= Iter_limit)
{
printf("\n\nSecant for solution of wavenumber has not converged");
printf("\nContact John Fenton johndfenton@gmail.com");
getch();
exit(1);
}
}
kd = kd1;
kH = kd * H;
}
z[1] = kd;
z[2] = kH;
SU = 0.5*kH / pow(kd, 3);
printf("\n# Stokes-Ursell no.: %7.3f", SU);
if (SU > 0.5)
printf(" > 1/2. Results are unreliable");
else
printf(" < 1/2, Stokes theory should be valid");
e[1] = 0.5 * kH;
for (i = 2; i <= n; i++) e[i] = e[i - 1] * e[1];
// Calculate coefficients
AB();
z[7] = C[0] + e[2] * C[2] + e[4] * C[4]; // ubar
z[8] = -e[2] * D[2] - e[4] * D[4];
z[9] = 0.5 * C[0] * C[0] + e[2] * E[2] + e[4] * E[4];
if (Current_criterion == 1)
{
z[5] = Current*sqrt(kd);
z[4] = z[7] + z[5];
z[6] = z[4] + z[8] / kd - z[7];
}
if (Current_criterion == 2)
{
z[6] = Current*sqrt(kd);
z[4] = z[6] - z[8] / kd + z[7];
z[5] = z[4] - z[7];
}
iff(Case, Wavelength) z[3] = 2 * pi / z[4];
iff(Case, Period) z[3] = T * sqrt(kd);
for (i = 1; i <= n; i++)
Tanh[i] = tanh(i*z[1]);
// Output results and picture of wave
Solution = fopen("Solution.res", "w");
Elevation = fopen("Surface.res", "w");
Flowfield = fopen("Flowfield.res", "w");
Output();
fflush(NULL);
printf("\nTouch key to continue "); getch();
printf("\n\nFinished\n");
}
我收到以下错误消息:
LNK2019未解析的外部符号“void __cdecl Output(void)”(?输出@@ YAXXZ)在函数_main Stokes中引用
LNK2019未解析的外部符号“double * __cdecl dvector(long,long)”(?dvector @@ YAPANJJ @ Z)在函数_main Stokes中引用
我检查了列表中的所有内容,试图找出这些错误的来源,并将其削减到这两个左边。
到目前为止所做的事情:
非常感谢任何帮助!
答案 0 :(得分:1)
未解析的外部符号表示您的代码无法找到您尝试使用的方法或类的定义。这通常意味着发生了几件事中的一件(或多件):
编辑:我还要证实其他人在原始评论帖子中所说的内容:确保代码中定义了Output()和dvector()的代码
还有其他一些选择,但那些是最常发生的最重要的选择。