如何将无符号整数转换为浮点数? [英] How to convert an unsigned int to a float?

问题描述

我需要构建一个函数，该函数返回 (float)x 的位级等价物，而不使用任何浮动数据类型、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.

I need to build a function that returns the bit-level equivalent of (float)x without using any floating data types, operations or constants. I think I have it, but when I run the test file, it returns that there's an infinite loop. Any debugging help would be appreciated.

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 次操作

I'm allowed to use any integer/unsigned operations including ||, &&, if, while. Also, I can only use 30 operations

unsigned float_i2f(int x) {
    printf("
%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;
}

响应非常有用的评论和答案，这里是更新的代码，现在只有 0x80000000 失败:

In response to the very helpful comments and answers, here is the updated code, now only failing for 0x80000000:

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) << "
";
        E++;
        shift = (shift >> 1);
    }
    if (E == 30) { E++;}
    exp = E + 127+24;
    exp = (exp << 23);
    frac = (absX << (23 - E)) & 0x007FFFFF;
    return sign + exp + frac;
}

有人知道修改后的代码中的错误在哪里吗?再次感谢大家！

Anyone have any idea where the bug is in the revised code? Thank you all again!

推荐答案

您可以做很多事情来改进和清理代码.对于初学者，添加评论！其次，(并减少操作次数)，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".

There is quite a lot you can do to improve your code and clean it up. For starters, add comments! Secondly, (and to reduce number of operations), you can combine certain things. Thirdly - differentiate between "integers that can be represented exactly" from "those that cannot".

这里有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 - 我试图展示一种方法，而不是为你做任务......

Here is some sample code to put some of these things into practice; I could not actually compile and test this, so it's possible there are some bugs - I am trying to show an approach, not do your assignment for you...

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("
%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);

}

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