计划做有优先级的计算器，vc五个lnk2001错误，求助呀。

[行的吧]

hanshushixian.obj : error LNK2001: unresolved external symbol "void __cdecl DestroyStack(struct SqStackN * &)" (?DestroyStack@@YAXAAPAUSqStackN@@@Z)
hanshushixian.obj : error LNK2001: unresolved external symbol "bool __cdecl GetTop(struct SqStackN *,double &)" (?GetTop@@YA_NPAUSqStackN@@AAN@Z)
hanshushixian.obj : error LNK2001: unresolved external symbol "bool __cdecl Push(struct SqStackN * &,double)" (?Push@@YA_NAAPAUSqStackN@@N@Z)
hanshushixian.obj : error LNK2001: unresolved external symbol "bool __cdecl Pop(struct SqStackN * &,double &)" (?Pop@@YA_NAAPAUSqStackN@@AAN@Z)
hanshushixian.obj : error LNK2001: unresolved external symbol "void __cdecl initStack(struct SqStackN * &)" (?initStack@@YAXAAPAUSqStackN@@@Z)
Debug/y8210181709.exe : fatal error LNK1120: 5 unresolved externals
执行 link.exe 时出错.



基本想法是搞个栈，利用栈进行计算，在栈计算后缀表达式时出现错误，上面是vc报错,下面是我的栈计算后缀函数。

double calculate(char* postexp)
{
double a,b,c,d,e;
SqStackN *o;initStack(o);
while(*postexp!='\0')
{
switch(*postexp)
{
case'+':
Pop(o,a);
Pop(o,b);
c= b+a;
Push(o,c);
break;
case'-':
Pop(o,a);
Pop(o,b);
c=b-a;
Push(o,c);
break;
case'*':
Pop(o,a);
Pop(o,b);
c=b*a;
Push(o,c);
break;
case '/':
Pop(o,a);
Pop(o,b);
if(a!=0)
{
c=b/a;
Push(o,c);
break;
}
else
{
exit(0);
}
break;
default:
d=0;
while (*postexp>='0'&&*postexp<='9')
{
d=d*10+(*postexp-'0');
postexp++;
}
Push(o,d);
break;
}
postexp++;
}
GetTop(o,e);
DestroyStack(o);
return(e);
}

[赵4老师]

仅供参考：

#pragma warning(disable:4996)
/*---------------------------------------
函数型计算器(VC++6.0,Win32 Console)
功能：
目前提供了10多个常用数学函数:
⑴正弦sin
⑵余弦cos
⑶正切tan
⑷开平方sqrt
⑸反正弦arcsin
⑹反余弦arccos
⑺反正切arctan
⑻常用对数lg
⑼自然对数ln
⑽ｅ指数exp
⑾乘幂函数^
⑿向上取整ceil
⒀向下取整floor
⒁四舍五入取整round
用法：
如果要求2的32次幂，可以打入2^32<回车>
如果要求30度角的正切可键入tan(Pi/6)<回车>
注意不能打入：tan(30)<Enter>
如果要求1.23弧度的正弦，有几种方法都有效：
sin(1.23)<Enter>
sin 1.23 <Enter>
sin1.23  <Enter>
如果验证正余弦的平方和公式,可打入sin(1.23)^2+cos(1.23)^2 <Enter>或sin1.23^2+cos1.23^2 <Enter>
此外两函数表达式连在一起,自动理解为相乘如：sin1.23cos0.77+cos1.23sin0.77就等价于sin(1.23)*cos(0.77)+cos(1.23)*sin(0.77)
当然你还可以依据三角变换，再用sin(1.23+0.77)也即sin2验证一下。
本计算器充分考虑了运算符的优先级因此诸如：2+3*4^2 实际上相当于：2+(3*(4*4))
另外函数名前面如果是数字,那么自动认为二者相乘.
同理，如果某数的右侧是左括号，则自动认为该数与括弧项之间隐含一乘号。
如：3sin1.2^2+5cos2.1^2 相当于3*sin2(1.2)+5*cos2(2.1)
又如：4(3-2(sqrt5-1)+ln2)+lg5 相当于4*(3-2*(√5 -1)+loge(2))+log10(5)
此外，本计算器提供了圆周率Pi键入字母时不区分大小写,以方便使用。
16进制整数以0x或0X开头。
----------------------------------------*/
#include <iostream>
#include <iomanip>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <stdio.h>
#include <string.h>
#include <windows.h>
using namespace std;
const char Tab = 0x9;
const int  DIGIT = 1;
const int MAXLEN = 16384;
char s[MAXLEN], *endss;
int pcs = 15;
double round(double dVal, short iPlaces) {//iPlaces>=0
	char s[30];
	double dRetval;

	sprintf(s, "%.*lf", iPlaces, dVal);
	sscanf(s, "%lf", &dRetval);
	return (dRetval);
}
double fun(double x, char op[], int *iop) {
	while (op[*iop - 1]<32) //本行使得函数嵌套调用时不必加括号,如 arc sin(sin(1.234)) 只需键入arc sin sin 1.234<Enter>
		switch (op[*iop - 1]) {
		case  7: x = sin(x);    (*iop)--; break;
		case  8: x = cos(x);    (*iop)--; break;
		case  9: x = tan(x);    (*iop)--; break;
		case 10: x = sqrt(x);   (*iop)--; break;
		case 11: x = asin(x);   (*iop)--; break;
		case 12: x = acos(x);   (*iop)--; break;
		case 13: x = atan(x);   (*iop)--; break;
		case 14: x = log10(x);  (*iop)--; break;
		case 15: x = log(x);    (*iop)--; break;
		case 16: x = exp(x);    (*iop)--; break;
		case 17: x = ceil(x);   (*iop)--; break;
		case 18: x = floor(x);  (*iop)--; break;
		case 19: x = round(x, 0); (*iop)--; break;
		}
	return x;
}
double calc(char *expr, char **addr) {
	static int deep; //递归深度
	static char *fname[] = { "sin","cos","tan","sqrt","arcsin","arccos","arctan","lg","ln","exp","ceil","floor","round",NULL };
	double ST[10] = { 0.0 }; //数字栈
	char op[10] = { '+' }; //运算符栈
	char c, *rexp, *pp, *pf;
	int ist = 1, iop = 1, last, i, n;
	__int64 i64;

	if (!deep) {
		pp = pf = expr;
		do {
			c = *pp++;
			if (c != ' '&& c != Tab)
				*pf++ = c;
		} while	(c != '\0');
	}
	pp = expr;
	if ((c = *pp) == '-' || c == '+') {
		op[0] =	c;
		pp++;
	}
	last = !DIGIT;
	while ((c =	*pp) !=	'\0') {
		if (c == '(') {//左圆括弧
			deep++;
			ST[ist++] =	calc(++pp, addr);
			deep--;
			ST[ist - 1]	= fun(ST[ist - 1], op, &iop);
			pp = *addr;
			last = DIGIT;
			if (*pp	== '(' || isalpha(*pp) && strnicmp(pp, "Pi", 2)) {//目的是：当右圆括弧的右恻为左圆括弧或函数名字时，默认其为乘法
				op[iop++] =	'*';
				last = !DIGIT;
				c =	op[--iop];
				goto operate;
			}
		}
		else if	(c == ')') {//右圆括弧
			pp++;
			break;
		}
		else if	(isalpha(c)) {
			if (!strnicmp(pp, "Pi", 2)) {
				if (last ==	DIGIT) {
					wcout <<	L"π左侧遇）" << endl; exit(1);
				}
				ST[ist++] =	3.14159265358979323846264338328;
				ST[ist - 1]	= fun(ST[ist - 1], op, &iop);
				pp += 2;
				last = DIGIT;
				if (!strnicmp(pp, "Pi", 2)) {
					wcout <<	L"两个π相连" << endl; exit(2);
				}
				if (*pp	== '(') {
					wcout <<	L"π右侧遇（" << endl; exit(3);
				}
			}
			else {
				for	(i = 0;	(pf	= fname[i])	!= NULL; i++)
					if (!strnicmp(pp, pf, strlen(pf))) break;
				if (pf != NULL)	{
					op[iop++] =	07 + i;
					pp += strlen(pf);
				}
				else {
					wcout <<	L"陌生函数名" << endl; exit(4);
				}
			}
		}
		else if	(c == '+' || c == '-' || c == '*' || c == '/' || c == '%' || c == '^') {
			char cc;
			if (last !=	DIGIT) {
				wcout <<	L"运算符粘连" << endl; exit(5);
			}
			pp++;
			if (c == '+' || c == '-') {
				do {
					cc = op[--iop];
					--ist;
					switch (cc)	{
					case '+':  ST[ist - 1] += ST[ist]; break;
					case '-':  ST[ist - 1] -= ST[ist]; break;
					case '*':  ST[ist - 1] *= ST[ist]; break;
					case '/':  ST[ist - 1] /= ST[ist]; break;
					case '%':  ST[ist - 1] = fmod(ST[ist - 1], ST[ist]); break;
					case '^':  ST[ist - 1] = pow(ST[ist - 1], ST[ist]); break;
					}
				} while	(iop);
				op[iop++] =	c;
			}
			else if	(c == '*' || c == '/' || c == '%') {
			operate:		cc = op[iop	- 1];
				if (cc == '+' || cc == '-') {
					op[iop++] =	c;
				}
				else {
					--ist;
					op[iop - 1]	= c;
					switch (cc)	{
					case '*':  ST[ist - 1] *= ST[ist]; break;
					case '/':  ST[ist - 1] /= ST[ist]; break;
					case '%':  ST[ist - 1] = fmod(ST[ist - 1], ST[ist]); break;
					case '^':  ST[ist - 1] = pow(ST[ist - 1], ST[ist]); break;
					}
				}
			}
			else {
				cc = op[iop	- 1];
				if (cc == '^') {
					wcout <<	L"乘幂符连用" << endl; exit(6);
				}
				op[iop++] =	c;
			}
			last = !DIGIT;
		}
		else {
			if (last ==	DIGIT) {
				wcout <<	L"两数字粘连" << endl; exit(7);
			}
			if (pp[0] == '0' && (pp[1] == 'x' || pp[1] == 'X')) {
				sscanf(pp +	2, "%I64x%n", &i64, &n);
				rexp = pp +	2 +	n;
				ST[ist++] =	(double)i64;
			}
			else ST[ist++] = strtod(pp,	&rexp);
			ST[ist - 1]	= fun(ST[ist - 1], op, &iop);
			if (pp == rexp)	{
				wcout <<	L"非法字符" << endl; exit(8);
			}
			pp = rexp;
			last = DIGIT;
			if (*pp	== '(' || isalpha(*pp)) {
				op[iop++] =	'*';
				last = !DIGIT;
				c =	op[--iop];
				goto operate;
			}
		}
	}
	*addr =	pp;
	if (iop	>= ist)	{
		wcout <<	L"表达式有误" << endl; exit(9);
	}
	while (iop)	{
		--ist;
		switch (op[--iop]) {
		case '+':  ST[ist - 1] += ST[ist]; break;
		case '-':  ST[ist - 1] -= ST[ist]; break;
		case '*':  ST[ist - 1] *= ST[ist]; break;
		case '/':  ST[ist - 1] /= ST[ist]; break;
		case '%':  ST[ist - 1] = fmod(ST[ist - 1], ST[ist]); break;
		case '^':  ST[ist - 1] = pow(ST[ist - 1], ST[ist]); break;
		}
	}
	return ST[0];
}
int main(int argc, char **argv) {
	int	a;

	wcout.imbue(locale("chs"));
	if (argc<2)	{
		//if (GetConsoleOutputCP() != 936) system("chcp 936>NUL");//中文代码页
		wcout <<	L"计算函数表达式的值。" << endl << L"支持(),+,-,*,/,%,^,Pi,sin,cos,tan,sqrt,arcsin,arccos,arctan,lg,ln,exp,ceil,floor,round" << endl;
		while (1) {
			wcout <<	L"请输入表达式：";
			fgets(s,16384,stdin);
			if ('\n' == s[strlen(s)-1]) s[strlen(s) - 1] = 0;
			if (s[0] ==	0) break;//
			cout <<	s << "=";
			cout <<	setprecision(15) <<	calc(s,	&endss)	<< endl;
		}
	}
	else if	(argc == 2 && 0	== strcmp(argv[1], "/?")) {
		//if (GetConsoleOutputCP() != 936) system("chcp 936>NUL");//中文代码页
		wcout <<	L"计算由≥1个命令行参数给出的函数表达式的值。\n"
			"最后一个参数是.0～.15表示将计算结果保留小数0～15位\n"
			"最后一个参数是x表示将计算结果以16进制正整数格式输出\n"
			"支持(),+,-,*,/,%,^^,Pi,sin,cos,tan,sqrt,arcsin,arccos,arctan,lg,ln,exp,ceil,floor,round\n"
			"16进制整数以0x或0X开头\n";
	}
	else {
		strncpy(s, argv[1],	MAXLEN - 1); s[MAXLEN -	1] = 0;
		if (argc>2)	{
			for	(a = 2;	a<argc - 1;	a++) strncat(s,	argv[a], MAXLEN	- 1);//将空格间隔的各参数连接到s
			if (1 == sscanf(argv[a], ".%d", &pcs) && 0 <= pcs && pcs <= 15) {//最后一个参数是.0～.15表示将计算结果保留小数0～15位
				printf("%.*lf\n", pcs, calc(s, &endss));
			}
			else if	(argv[a][0]	== 'x' || argv[a][0] == 'X') {//最后一个参数是x表示将计算结果以16进制正整数格式输出
				printf("0x%016I64x\n", (__int64)calc(s, &endss));
			}
			else {
				strncat(s, argv[a],	MAXLEN - 1);
				printf("%.15lg\n", calc(s, &endss));
			}
		}
		else {
			printf("%.15lg\n", calc(s, &endss));
		}
	}
	return 0;
}

[行的吧]

老师这个是？？？、我是菜鸟，零基础，可以给我一个解题思路吗

[行的吧]

我不知道如何解决这个问题