简单问题..入门问题啊。。帮我看看

[huanhuan30]

#include <stdio.h>
#include "stdafx.h"
#include <string.h>
char *BinToStr(char *, int);
int BinToInt (char *);
int main(void)
{
	int nNum1, nNum2;
	char lpszBinStr[sizeof(int) * 8 + 1];
	char lpszBin[2][33];
	gets(lpszBin[0]);
	gets(lpszBin[1]);
    nNum1 = BinToInt(lpszBin[0]);
	nNum2 = BinToInt(lpszBin[1]);
	printf("~Number1 :%s \n", BinToStr(lpszBinStr, ~nNum1));
	printf("~Number2 :%s \n", BinToStr(lpszBinStr, ~nNum2));
	printf("Number1&Number : %s\n", BinToStr(lpszBinStr, nNum1&nNum2)); //这三条条语句打印出来为空。。。为什么？？
	printf("Number1|Number : %s\n", BinToStr(lpszBinStr, nNum1|nNum2));
	printf("Number1^Number : %s\n", BinToStr(lpszBinStr, nNum1^nNum2));
	return 0;
}
int BinToInt (char *lpszBin)          // 将输入的二进制码转化为数字
{
	int nSum = lpszBin[strlen(lpszBin) -1] - '0';
	int nSubSum = 1;
	int cNum, cNum1, cCount;
	for (cNum = strlen(lpszBin)-2, cCount = 0; cNum>=0; cNum--, cCount++)
	{
		for (cNum1 = 0; cNum1 <= cCount; cNum1++)
		{
			nSubSum *= 2;
		}
		nSum += (lpszBin[cNum]-'0') * nSubSum;
		nSubSum = 1;
	}
	//printf("%d \n", nSum);
	return nSum;
}
char *BinToStr(char *lpszBinString, int nNum)   //讲数字转化为2进制码
{
	int cNum;
	int nMark = 01;
	 int nSize = sizeof(int) * 8;
	for (cNum = nSize-1; cNum >= 0; cNum--)
	{
		lpszBinString[cNum] = nNum & nMark + '0';
		nNum >>= 1;
	}
	lpszBinString[nSize] = '\0';
	return lpszBinString;
}

题目在注释当中。。为什么nNum1&nNum2 和 nNum1|nNum2 nNum1^nNum2 这三个作为参数 显示为 空..。。无语啊

[mengde007]

[Quote=引用楼主 huanhuan30 的帖子:]
gets(lpszBin[0]);
gets(lpszBin[1]);
[/Quote]不能这样写吧！

[mengde007]

gets()是输入串的

[cyldf]

char *BinToStr(char *lpszBinString, int nNum)   //讲数字转化为2进制码
{
    int cNum;
    int nMark = 01;
     int nSize = sizeof(int) * 8;
    for (cNum = nSize-1; cNum >= 0; cNum--)
    {
        lpszBinString[cNum] = nNum & nMark + '0';//改成lpszBinString[cNum] = (nNum & nMark) + '0';
        nNum >>= 1;
    }
    lpszBinString[nSize] = '\0';
    return lpszBinString;
}

[liliangbao]

差了半天，原来是优先级的问题！
& 没有 +优先级高！

char *BinToStr(char *lpszBinString, int nNum)   //讲数字转化为2进制码
{
    int cNum;
    int nMark = 01;
    int nSize = sizeof(int) * 8;
    for (cNum = nSize-1; cNum >= 0; cNum--)
    {
        lpszBinString[cNum] = (nNum & nMark) + '0';//这样修改
        nNum >>= 1;
    }
    lpszBinString[nSize] = '\0';
    return lpszBinString;
}