The problem of float accuracy in C language

3.1415926 (10 binary) = =

11.00100100001111110110100110100010010110110000100101 (2 binary) = =

1.100100100001111110110100110100010010110110000100101 * 2

Debug print with GdB the contents of the float variable are 0X40490FDA (16 binary), i.e. 01000000010010010000111111011010 (2 binary)

0 (Highest bit 0, indicates positive number) 10000000 (IEEE stipulates that the sub-square to be calculated here is the true exponent minus 127, where the exponent is 1) 10010010000111111011010 (this 23 bit represents the fractional part,

Because the hidden bit technology is used in the computer: the highest bit of 1 does not write memory, so the fractional part is actually 1.10010010000111111011010)

1.10010010000111111011010 (2 binary) ==1.5707962512969970703125 (10 binary)

1.5707962512969970703125 * 2 = = 3.141592502593994140625

float F = 3.1415926;

printf ("%f\n", f);

The printf function defaults to the decimal 6 bits, and the last one is rounded, so the result is 3.141593.

The following is validation:

printf ("%.7f\n", f); The result should be 3.1415925;

printf ("%.8f\n", f); The result should be 3.14159250.

printf ("%.9f\n", f); The result should be 3.141592503.

Because the mantissa of float is 23 bits, the decimal place retains 23 bits, i.e.

3.1415926 (10 binary) = = 1.1001001000011111101100 (2 binary)

Because the hidden bit technology is used in the computer: the highest bit of 1 does not write memory, so 1.1001001000011111101100 (2 binary) should be represented as 0.1001001000011111101100

Because the index can be negative, in order to facilitate the calculation, so the IEEE stipulates that the sub-square to be calculated here to subtract 127 is the true exponent, where the exponent is 1 (that is, 2 of the 1 power)

The problem of float accuracy in C language