顺序栈的实现

[lrd408]

#ifndef STACK_H
#define STACK_H
#define NUM 100

class tack {
public:
int* stase;
int* top;
int sizestack;
tack(int* stase=0,int* top=0,int sizetack=0);

};
#endif

#ifndef AA_H
#define AA_H
#include "lrd.h"
#include <iostream>
using std::cout;
using std::endl;

double GetTop(tack& S) {
int e=0;
if(S.top==S.stase) exit(1);
e=*(S.top-1);
cout<<endl;
cout<<"栈的头元素是 "<<e<<endl;
return e;
}

void Push(tack& S,int e) {
if(S.top-S.stase>=S.sizestack) {
cout<<"栈元素已满 "<<endl;
exit(1);
}
else *S.top++=e;
}

double Pop(tack& S,int e=0) {
if(S.top ==S.stase) {
cout<<"栈中已无元素！"<<endl;
exit(1);
}
else e=*--S.top;
return e;
}
#endif


#include <iostream>
#include "lrd.h"

tack::tack(int* stase,int* top,int sizetack) {
stase=new(NUM*sizeof(int));
if(!stase) exit(1);
stase=top;
sizetack=NUM;
}

#include <iostream>
#include "lrd.h"
#include "aa.h"

showtack(tack& S);
int main() {
tack we;
while((!we.stase)&&(we.stase-we.top)<=NUM) {
static int i=0;
Push(we, ++i);
}

showtack(we);
return 0;
}

showtack(tack& S) {
int t=0;
while(S.stase!=S.top) {
if(++t%5==0) cout<<'\n'<<Pop(S);
else cout<<' '<<Pop(S);
}
}


编译时无错误
连接是：--------------------Configuration: stack - Win32 Debug--------------------
Compiling...
lrd.cpp
C:\yanlian\stack\lrd.cpp(5) : error C2059: syntax error : ';'
C:\yanlian\stack\lrd.cpp(9) : error C2143: syntax error : missing ';' before '}'




C:\yanlian\stack\lrd.cpp(9) : error C2143: syntax error : missing ';' before '}'C:\yanlian\stack\lrd.cpp(9) : fatal error C1003: error count exceeds 100; stopping compilation
执行 cl.exe 时出错.

stack.exe - 1 error(s), 0 warning(s)

[huangxiaofei]

你的"lrd.h"文件呢?一起贴上来撒，要不然根本无法调试的说。

[lrd408]

//lrd.h
#ifndef STACK_H
#define STACK_H
#define NUM 100

class tack {
public:
int* stase;
int* top;
int sizestack;
tack(int* stase=0,int* top=0,int sizetack=0);

};
#endif

//aa.h
#ifndef AA_H
#define AA_H
#include "lrd.h"
#include <iostream>
using std::cout;
using std::endl;

double GetTop(tack& S) {
int e=0;
if(S.top==S.stase) exit(1);
e=*(S.top-1);
cout < <endl;
cout < <"栈的头元素是 " < <e < <endl;
return e;
}


//lrd.cpp
#include <iostream>
#include "lrd.h"

tack::tack(int* stase,int* top,int sizetack) {
stase=new(NUM*sizeof(int));
if(!stase) exit(1);
stase=top;
sizetack=NUM;
}


//main
#include <iostream>
#include "lrd.h"
#include "aa.h"

showtack(tack& S);
int main() {
tack we;
while((!we.stase)&&(we.stase-we.top)<=NUM) {
static int i=0;
Push(we, ++i);
}

showtack(we);
return 0;
}

showtack(tack& S) {
int t=0;
while(S.stase!=S.top) {
if(++t%5==0) cout<<'\n'<<Pop(S);
else cout<<' '<<Pop(S);
}
}

[ttkk_2007]

tack::tack(int* stase,int* top,int sizetack) { 
stase=new(NUM*sizeof(int));   //new int[NUM];
if(!stase) exit(1);  //分配失败会抛异常，这句没用，除非上句式new (nothrow) int[NUM];
stase=top;         //怎么又让stase指向top？，你想内存泄漏？
sizetack=NUM; 
} 

[lrd408]

回3楼的：
我是想构造个空栈