Number of super-long digits processed

 

Float and double data types are single-precision and double-precision data types. Their values are respectively from the negative power of 3.4E + 10 to the 38 power of 3.4E + 10, and the negative 308 power of 1.7E + 10 to the 308 power of 1.7E + 10.

How can a float 6-7 effective number be loaded with a power of 3.4E + 10, a power of 38, and a power of 15-16, a double character? How can a power of 308, a power of, or a power of, be loaded with a power of 1.7E + 10? number. Isn't that a conflict?

 

Answer: No conflict. Float 6-7 digits refer to the number of digits (precision) of valid numbers rather than the value size. For example, 3.14159267 has nine valid digits, but the value is 3 ~ 4, while 350 has three digits, the size is 300 ~ In the range of 400. Therefore, the float number can reach 3.4E + 10, but the number of valid digits can only reach 6-7. If 3.14159267 is assigned to a float variable, the accuracy will be reduced. For example

Float a = 3234567.1;

Float B = 3234567;

If (a = B)

Printf ("YES ");

Else

Printf ("NO ");

YES will be output, because the 11 at the end of a exceeds the float and can only reach 6-7 digits. (If a = 1234567.1; B = 3234567) the input result is NO. Why? This requires us to analyze: how to deal with the excess precision? It is not rounding, but the loss of binary digits. Therefore, sometimes the precision can reach 6 bits, and sometimes the precision can reach 7 bits, depending on the binary representation of the number.

 

So how can we express super-long digits and super-high-precision digits? For example, 123456789123456789123456789 (a large integer with a length of 30 digits), for example 3.14159012345678901234567890123 (a decimal number with a precision of 30 digits), long float cannot be stored for such a long number, this requires the help of "string" or "character array.

Unsigned _ int64 n;

The unsigned _ int64 Type Variable n has a maximum value of more than 1234567892345678912 (20 bits) and can reach about 1.8E + 19. Generally, it should be enough. _ Int64 type data cannot be output using cout in C ++. It should be that cout does not reload this type. If printf is used for output, it is clear that % d, % f, % l cannot meet the accuracy of 20 bits. printf ("% I64d \ n", n) can be used in VC6. However, the supported bits cannot exceed 20 bits, if the output exceeds 9.23E + 18, an error occurs. The best way is to convert the "long digits" into a string, as shown below:

Char buffer [65];

Printf ("% s", _ ui64toa (n, buffer, 10 ));

Function _ ui64toa is responsible for converting n into a string and storing it in the character Array buffer [65]. 10 indicates converting to a 10th hexadecimal format.

Convert a number to a string:

 

# Include <stdlib. h>

# Include <stdio. h>

Int main (void)

{

Char buffer [65];

Int r;

For (r = 10; r> = 2; -- r)

{

_ Itoa (-1, buffer, r );

Printf ("base % d: % s (% d chars) \ n", r, buffer, strlen (buffer ));

}

Printf ("\ n ");

For (r = 10; r> = 2; -- r)

{

_ I64toa (-1L, buffer, r );

Printf ("base % d: % s (% d chars) \ n", r, buffer, strlen (buffer ));

}

Printf ("\ n ");

For (r = 10; r> = 2; -- r)

{

_ Ui64toa (0 xffffffffffffffl, buffer, r );

Printf ("base % d: % s (% d chars) \ n", r, buffer, strlen (buffer ));

}

}

 

 

Output

Base 10:-1 (2 chars)

Base 9: 12068657453 (11 chars)

Base 8: 37777777777 (11 chars)

Base 7: 211301422353 (12 chars)

Base 6: 1550104015503 (13 chars)

Base 5: 32244002423140 (14 chars)

Base 4: 3333333333333333 (16 chars)

Base 3: 102002022201221111210 (21 chars)

Base 2: 11111111111111111111111111111111 (32 chars)

 

Base 10:-1 (2 chars)

Base 9: 145808576354216723756 (21 chars)

Base 8: 1777777777777777777777 (22 chars)

Base 7: 45012021522523134134601 (23 chars)

Base 6: 3520522010102100444244423 (25 chars)

Base 5: 2214220303114400424121122430 (28 chars)

Base 4: 33333333333333333333333333333333 (32 chars)

Base 3: 11112220022122120101211020120210210211220 (41 chars)

Base 2: 1111111111111111111111111111111111111111111111111111111111111111 (64 chars)

 

Base 10: 18446744073709551615 (20 chars)

Base 9: 145808576354216723756 (21 chars)

Base 8: 1777777777777777777777 (22 chars)

Base 7: 45012021522523134134601 (23 chars)

Base 6: 3520522010102100444244423 (25 chars)

Base 5: 2214220303114400424121122430 (28 chars)

Base 4: 33333333333333333333333333333333 (32 chars)

Base 3: 11112220022122120101211020120210210211220 (41 chars)

Base 2: 1111111111111111111111111111111111111111111111111111111111111111 (64 chars)

 

PS: You can use this function to convert a decimal integer into a binary string;

Int main (void)

{

Char buffer [65];

_ Itoa (12, buffer, 2 );

Printf ("base % d: % s (% d chars) \ n", r, buffer, strlen (buffer ));

}

 

Another way is to define a character array to store the number of extra-long digits, and the decimal point can also be solved. Then, define the algorithm between these strings and overload operators, it is said that this operation is still very efficient.

 

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