我需要构建一个函数来返回 (float)x 的位级等价物,而不使用任何 float 数据类型、操作或常量。我想我有它,但是当我运行测试文件时,它返回有一个无限循环。任何调试帮助将不胜感激。
我可以使用任何整数/无符号运算,包括 ||、&&、if、while。 另外,我只能使用 30 个操作
unsigned float_i2f(int x) {
printf("\n%i", x);
if (!x) {return x;}
int mask1 = (x >> 31);
int mask2 = (1 << 31);
int sign = x & mask2;
int complement = ~x + 1;
//int abs = (~mask1 & x) + (mask1 & complement);
int abs = x;
int i = 0, temp = 0;
while (!(temp & mask2)){
temp = (abs <<i);
i = i + 1;
}
int E = 32 - i;
int exp = 127 + E;
abs = abs & (-1 ^ (1 << E));
int frac;
if ((23 - E)>0)
frac = (abs << (23 - E));
else
frac = (abs >> (E - 23));
int rep = sign + (exp << 23) + frac;
return rep;
}
为了回应非常有帮助的评论和答案,这里是更新后的代码,现在仅针对 0x80000000 失败:
unsigned float_i2f(int x) {
int sign;
int absX;
int E = -1;
int shift;
int exp;
int frac;
// zero is the same in int and float:
if (!x) {return x;}
// sign is bit 31: that bit should just be transferred to the float:
sign = x & 0x80000000;
// if number is < 0, take two's complement:
if (sign != 0) {
absX = ~x + 1;
}
else
absX = x;
shift = absX;
while ((!!shift) && (shift != -1)) {
//std::cout << std::bitset<32>(shift) << "\n";
E++;
shift = (shift >> 1);
}
if (E == 30) { E++;}
exp = E + 127+24;
exp = (exp << 23);
frac = (absX << (23 - E)) & 0x007FFFFF;
return sign + exp + frac;
}
任何人都知道修改后的代码中的错误在哪里?再次感谢大家!
最佳答案
您可以做很多事情来改进您的代码并清理它。对于初学者,添加评论!其次,(为了减少操作次数),您可以组合某些东西。第三 - 区分“可以精确表示的整数”和“不能精确表示的整数”。
下面是一些示例代码,可以将其中的一些内容付诸实践;我实际上无法编译和测试它,所以可能存在一些错误 - 我正在尝试展示一种方法,而不是为你做作业......
unsigned float_i2f(int x) {
// convert integer to its bit-equivalent floating point representation
// but return it as an unsigned integer
// format:
// 1 sign bit
// 8 exponent bits
// 23 mantissa bits (plus the 'most significant bit' which is always 1
printf("\n%i", x);
// zero is the same in int and float:
if (x == 0) {return x;}
// sign is bit 31: that bit should just be transferred to the float:
sign = x & 0x8000;
// if number is < 0, take two's complement:
int absX;
if(sign != 0) {
absX = ~x + 1;
}
else
absX = x;
}
// Take at most 24 bits:
unsigned int bits23 = 0xFF800000;
unsigned int bits24 = 0xFF000000;
unsigned E = 127-24; // could be off by 1
// shift right if there are bits above bit 24:
while(absX & bits24) {
E++; // check that you add and don't subtract...
absX >>= 1;
}
// shift left if there are no bits above bit 23:
// check that it terminates at the right point.
while (!(absX & bits23))
E--; // check direction
absX <<= 1;
}
// now put the numbers we have together in the return value:
// check that they are truncated correctly
return sign | (E << 23) | (absX & ~bits23);
关于c - 如何将无符号整数转换为 float ?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/19529356/