Why pow(10,5) = 9,999 in C++

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忘掉有多难
忘掉有多难 2020-11-22 08:55

Recently i write a block of code:

const int sections = 10;

for(int t= 0; t < 5; t++){
   int i = pow(sections, 5- t -1);  
   cout << i << en         


        
8条回答
  •  逝去的感伤
    2020-11-22 09:17

    When used with fractional exponents, pow(x,y) is commonly evaluated as exp(log(x)*y); such a formula would mathematically correct if evaluated with infinite precision, but may in practice result in rounding errors. As others have noted, a value of 9999.999999999 when cast to an integer will yield 9999. Some languages and libraries use such a formulation all the time when using an exponentiation operator with a floating-point exponent; others try to identify when the exponent is an integer and use iterated multiplication when appropriate. Looking up documentation for the pow function, it appears that it's supposed to work when x is negative and y has no fractional part (when x is negative and `y is even, the result should be pow(-x,y); when y is odd, the result should be -pow(-x,y). It would seem logical that when y has no fractional part a library which is going to go through the trouble of dealing with a negative x value should use iterated multiplication, but I don't know of any spec dictating that it must.

    In any case, if you are trying to raise an integer to a power, it is almost certainly best to use integer maths for the computation or, if the integer to be raised is a constant or will always be small, simply use a lookup table (raising numbers from 0 to 15 by any power that would fit in a 64-bit integer would require only a 4,096-item table).

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