要求用C++编写个运算器 可实现 逻辑运算（两个数或 与 非） 位运算 以及进制转换 怎么实现 ？

void main()
{
	int a,b;
	cin>>a;
	cin>>b;
	cout<< (a | b) <<endl; //或
	cout<< (~a)<<endl;//非
	cout<<hex<<a<<endl; //十六进制
	cout<<oct<<a<<endl; //八进制
	char binary[100];  
	itoa(a, binary, 2); //转换，这里的2表示2进制（可以是其他） 
	cout<<binary<<endl;
}

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <locale.h>
int main() {
    int i,v;
    char bs[33];
    char b[33];
    char hs[9];
    char h[9];
    char s[4];
    char *e;
// 十进制整数转二进制串；
    i=1024;
    ltoa(i,b,2);
    sprintf(bs,"%032s",b);
    printf("i=%d,bs=%s\n",i,bs);
// 十进制整数转十六进制串；
    i=1024;
    ltoa(i,h,16);
    sprintf(hs,"%08s",h);
    printf("i=%d,hs=%s\n",i,hs);
// 十六进制字符串转成十进制数
    strcpy(hs,"00000400");
    sscanf(hs,"%x",&i);
    printf("hs=%s,i=%d\n",hs,i);
// 二进制字符串转化为十六进制字符串；
    strcpy(bs,"00000000000000000000010000000000");
    i=strtol(bs,&e,2);
    ltoa(i,h,16);
    sprintf(hs,"%08s",h);
    printf("bs=%s,hs=%s\n",bs,hs);
// 二进制字符串转化为十进制数；
    strcpy(bs,"00000000000000000000010000000000");
    i=strtol(bs,&e,2);
    printf("bs=%s,i=%d\n",bs,i);
// 十六进制字符串转成二进制串
    strcpy(hs,"00000400");
    sscanf(hs,"%x",&i);
    ltoa(i,b,2);
    sprintf(bs,"%032s",b);
    printf("hs=%s,bs=%s\n",hs,bs);
// ASC\GBK字符串转十六进制串
    strcpy(s,"a汉");
    i=0;
    while (1) {
        if (0==s[i]) break;
        sprintf(hs+i*2,"%02X",(unsigned char)s[i]);
        i++;
    }
    setlocale(LC_ALL,"chs");
    printf("s=%s,hs=%s\n",s,hs);
// 十六进制字符串转成汉字（GBK）及字符(ASC)
    strcpy(hs,"61BABA");
    i=0;
    while (1) {
        if (1!=sscanf(hs+i*2,"%2x",&v)) break;
        s[i]=(char)v;
        i++;
    }
    s[i]=0;
    printf("hs=%s,s=%s\n",hs,s);
    return 0;
}
//i=1024,bs=00000000000000000000010000000000
//i=1024,hs=00000400
//hs=00000400,i=1024
//bs=00000000000000000000010000000000,hs=00000400
//bs=00000000000000000000010000000000,i=1024
//hs=00000400,bs=00000000000000000000010000000000
//s=a汉,hs=61BABA
//hs=61BABA,s=a汉

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <locale.h>
const int  DIGIT=1;
const int MAXLEN=16384;
char s[MAXLEN],*endss;
int calc(char *expr,char **addr) {
    int ST[10]={0}; //数字栈
    char op[10]={""+""}; //运算符栈
    char c,*pp;
    int ist=1,iop=1,last;
    char cc;
    pp=expr;
    last = !DIGIT;
    if ((c=*pp)==""!"") {
        op[0] = c;
        pp++;
    }
    while (1) {
        c=*pp;
        if (c==0) break;
        if (c==""("") {
            ST[ist++]=calc(++pp,addr);
            pp = *addr;
            last = DIGIT;
        } else if (c=="")"") {
            pp++;
            break;
        } else if (c==""+""||c==""*"") {
            if (last != DIGIT) {
                printf("运算符粘连\n");exit(5);
            }
            pp++;
            if (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] =1-ST[ist];break;
                    }
                } while (iop);
                op[iop++] = c;
            } else if (c==""*"") {
                cc = op[iop-1];
                if (cc==""+"") {
                    op[iop++] = c;
                } else {
                    --ist;
                    op[iop-1] = c;
                    switch (cc) {
                    case ""*"":  ST[ist-1] &= ST[ist];break;
                    case ""!"":  ST[ist-1] =1-ST[ist];break;
                    }
                }
            }
            last = !DIGIT;
        } else if (c==""!"") {
            op[iop++] = c;
            ST[ist++] =0;
            pp++;
        } else {
            if (last == DIGIT) {
                printf("两数字粘连\n");exit(7);
            }
            if (!(pp[0]==""0""||pp[0]==""1"")) {
                printf("非法字符\n");exit(8);
            }
            ST[ist++]=pp[0]-""0"";
            pp++;
            last = DIGIT;
        }
    }
    *addr=pp;
    if (iop>=ist) {
        printf("表达式有误\n");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] =1-ST[ist];break;
        }
    }
    return ST[0];
}
int main() {
    setlocale(LC_ALL,"chs");
    while (1) {
        printf("请输入布尔表达式，例如!(1+1)*1：");fflush(stdout);
        rewind(stdin);
        gets(s);
        if (s[0]==0) break;//
        printf("%d\n",calc(s,&endss));
    }
    return 0;
}