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cheroky
Asked: 2020-11-02 06:33:35 +0800 CST 2020-11-02 06:33:35 +0800 CST

Interpret float as unsigned int

  • 772

Can someone explain why interpreting a float as an integer as I understand data would be lost, is it a valid C casting?

float sqrt7(float x)
{
   unsigned int i = *(unsigned int*) &x;
   i  += 127 << 23;
   i >>= 1; 
   return *(float*) &i;
}

In addition, after it is cast again to float, it is the first time that I see this type of code

c

1 Answers

  1. Best Answer

    Floating point numbers can, on the one hand, have decimal places, while an integer cannot. If you convert a number with decimals to a representation that does not support decimals, you will lose the decimal part of the number, so you will lose precision.

    float num_decimal = 10.5;
int num_entero = (int)num_decimal;
printf("%d",num_entero); // imprime 10

    Another problem with this conversion is that the range of values ​​supported by floatis much larger than that supported by intor unsigned int. Forcing a conversion when the range of the target type is exceeded results in numbers that have nothing to do with it... A positive number that becomes a negative...

    And as a final note, the above especially affects unsigned intsince everything floatwith a negative value that is converted to unsigned intwill result in a positive number that has little or nothing to do with the original number.

    Having said that:

    Is it bad to do this kind of conversions?

    Absolutely. The bad thing is doing them crazy and without being very clear that the conversion is safe. On many occasions you will find that it is unavoidable to make a conversion to communicate your program with a library... You will only have to guarantee that the conversion does not become something harmful.

    Does what I comment on affect the example?

    The specific example is not affected at all.

    Let me explain: the example performs a conversion from floatto unsigned intonly because binary shift instructions are not available for the type float.

    The code is assuming that unsigned intand floatwill occupy the same number of bytes. To avoid doing a data conversion (which would involve loss of lookahead), create a pointer to unsigned intand point that pointer to float. In this way, from the point of view of the pointer, what is stored there is of type unsigned int, so the binary shift operations are automatically activated.

    If instead of a pointer a conversion of the data were made, the effects mentioned at the beginning would occur.

    All the best

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