c - 如何将 char 转换为 unsigned int？

Question

程序如何运作

./my program 1 10000100 01000000001000001000000
// Reads the argv's
// Converts them into unsigned ints

如何将 char* 转换为 unsigned int？我假设我使用位操作，但我对 char 到 unsigned int 的转换有点迷失。

代码。

struct _float {
    unsigned int sign:1, exp:8, frac:23;
};

union _bits32 {
    float    fval;  // Bits as a float
    Word     xval;  // Bits as a word
    Float32 bits;  // manipulate individual bits
};

union _bits32 getBits(char *sign, char *exp, char *frac);

int main(int argc, char **argv) {
    u = getBits(argv[1], argv[2], argv[3]);
    return 0;
}

// Here I am converting the char's into unsigned ints
union _bits32 getBits(char *sign, char *exp, char *frac) {
    Union32 new;

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

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

    // convert char *frac into a 23-bit value in new.bits
    new.bits.frac = *frac;

    return new;
}

Answer 1

因此，您需要读取每一位(每位 1 个字符)并将其保存到正确位位置的整数中。下面是一个函数，可以对最多 32 位的任意长度执行此操作。

// Takes in the string with the bits (e.g. "01010100011")
// Returns the integer represented by those bits (leading bits are 0's)
int parseFromBinary(char* bitString) {
  int val = 0;
  int i = 0;
  // While there are more characters
  while (bitString[i] != 0) {
    // Shift everything left (multiply by 2) to make room for the new bit
    val <<= 1;
    // Add the new bit (no effect if it is a 0)
    val |= bitString[i] - '0';
    i++;
  }
  return val;
}

您可能希望为每个字段调用此函数一次，但实际上没有必要循环遍历 1 个符号位，这可能会让事情变得不那么清晰。