Question

实际上，我必须将命令参数（三个字符串）转换为位字段（内部三个无符号整数）。该程序将从位转换为浮点数。我首先考虑使用数组存储三个参数，但我真的不知道如何从数组转换为无符号整数。我应该只使用atoi将arg更改为int，然后直接更改为unsigned int吗？它在我的电脑上没有声音。不知道。

Union32 getBits(char *sign, char *exp, char *frac)
{
    Union32 new;

   // this line is just to keep gcc happy
   // delete it when you have implemented the function
   //new.bits.sign = new.bits.exp = new.bits.frac = 0;

   new.bits.sign = *(unsigned int *)atoi(sign);
   new.bits.exp = *(unsigned int *)atoi(exp);
   new.bits.frac = *(unsigned int *)atoi(frac);
   //int i ;
   //int balah[8] = {};
   //for(i = 0; i < 8; i++){
   //balah[i] = sign[i];
   //}

   //int j ;
   //int bili[23] = {};
   //for(j = 0; j < 23; j++){
      //bili[j] = sign[j];
   //}

   //convert array into unsigned integer?
   printf("%u %u %u\n", new.bits.sign, new.bits.exp, new.bits.frac);


   // convert char *sign into a single bit in new.bits

   // convert char *exp into an 8-bit value in new.bits

   // convert char *frac into a 23-bit value in new.bits
enter code here
   return new;
}

以下是该程序所需的typedef和并集的详细信息，以及该程序中的四个功能。

typedef uint32_t Word;

struct _float {
   // define bit_fields for sign, exp and frac
   // obviously they need to be larger than 1-bit each
   // and may need to be defined in a different order
   unsigned int sign:1, exp:8, frac:23;
};
typedef struct _float Float32;

union _bits32 {
   float   fval;  // interpret the bits as a float
   Word    xval;  // interpret as a single 32-bit word
   Float32 bits;  // manipulate individual bits
};
typedef union _bits32 Union32;

void    checkArgs(int, char **);
Union32 getBits(char *, char *, char *);
char   *showBits(Word, char *);
int     justBits(char *, int);

getBits要求我们将位转换为浮点数，并且showBits要求我们将float转换为bit。

Answer 1

在代码中假设正确的typedef：

   new.bits.sign = (unsigned int)atoi(sign);
   new.bits.exp = (unsigned int)atoi(exp);
   new.bits.frac = (unsigned int)atoi(frac);

如何在C中将数字字符串存储为无符号整数

1 个答案: