一个简单的将二进制转换为十进制的程序--- 位操作练习题

# include <stdio.h>
# include <string.h>
# define LEN 32            //规定字符串的长度为32，对应32位

int be_val(const char* ps);
char* s_gets(char* str, int n);

int main()
{
    char pbin[LEN];

    puts("Please enter the binary.Support 32 bits .");
    while (s_gets(pbin, LEN) != NULL && pbin[0] != '\0')
    {
        int dec;
        dec = be_val(pbin);

        printf("%d\n", dec);
        puts("Please enter next binary");
        puts(" Press 【enter】 at the beginnning of the new line to quit");
    }
    puts("bye");
    

}

int be_val(const char* ps)
{
    int i = strlen(ps)-1;      //计算字符串长度，并减去1，作为循环条件用

    int sum = 0;
    int j = 1;
    int num;                     //位的情况：1或0
    int val；                 int k = strlen(ps)-1;      //同int i
    while (*ps!='\0')              
    {
        i = k;
        num = (*ps) - '0';
        if (num)     //为1，则进入；不然跳过此判定    
        {
            val = 1;
            while (i)
            {
                val = val * 2;
                i--;
            };
            sum += val;  
        }
        k--;
        ps++;
    }
    
    return sum;
}

char* s_gets(char* str, int n)
{
    char* ret_val;
    char* find;

    ret_val = fgets(str, n, stdin);
    if (ret_val)
    {
        find = strchr(str, '\n');
        if (find)
            *find = '\0';
        else
            while (getchar() != '\n')
                continue;
    }
    return ret_val;
}