How to use a 64-bit integer in C?

First, the ANSI c99 standard does not contain 64-bit integers. Secondly, many practical compilers support 64-bit integer types.
For specific analysis, see: http://blog.csdn.net/lychee007/archive/2010/04/04/5449237.aspx
The key is the table:

Variable definition	Output Mode	GCC (mingw32)	G ++ (mingw32)	GCC (Linux i386)	G ++ (Linux i386)	Microsoftvisual C ++ 6.0
Long long	"% LLD"	Error	Error	Correct	Correct	Unable to compile
Long long	"% I64d"	Correct	Correct	Error	Error	Unable to compile
_ Int64	"LLD"	Error	Error	Unable to compile	Unable to compile	Error
_ Int64	"% I64d"	Correct	Correct	Unable to compile	Unable to compile	Correct
Long long	Cout	Non-C ++	Correct	Non-C ++	Correct	Unable to compile
_ Int64	Cout	Non-C ++	Correct	Non-C ++	Unable to compile	Unable to compile
Long long	Printint64 ()	Correct	Correct	Correct	Correct	Unable to compile

We noticed that the support for 64-bit integers in VC ++ 6.0 is the most "that.

To publish a new bookCode: 2.4.3 64-bit integer

Question:Enter a positive integer to count the number of positive factors. N <= (12 power of 10)

Solution to the Standard C language:
Compiler: TCC (basically the international standard of c99, it's amazing !)

 0001 /* 0002 2.4.3, 64-bit integer 0003 Example 0004 Enter a positive integer n to count the number of positive factors. 0005 N <= (12 power of 10) 0006      0007 Solution: 0008 The exhaustion method starts from 1, and every possible factor X is lifted. If the condition (N % x = 0) is met, the counter count increments by 1. 0009 TIPS: 0010 You only need to enumerate the range of [1 .. SQRT (n; 0011 If X is a factor of N, (n/X) is also a factor of N. 0012 0013       0014 */ 0015 # Include "stdio. H" 0016 # Include "math. H" 0017 Void main (){ 0018 Long long n, 0019 X,/* hypothetical factor */ 0020 Count = 0;/* counter */ 0021  0022 Scanf ("% i64d", & N ); 0023 For (x = 1; x <= (long) SQRT (n); X = x + 1 ){ 0024 If (N % x = 0 ){ 0025 Count = count + 2;/* The reason for + 2 is that X and (n/x) are both factors of N */ 0026 /* Printf ("% i64d \ n", x );*/ 0027 If (N/x = x) Count = count-1; 0028 } 0029 } 0030 Printf ("number of factors: % i64d", count ); 0031  0032 Return; 0033 }

The above code is written in itself, but it will be told during compilation in VC ++ 6.0:

...... Test. C (4): Error c2632: 'long' followed by 'long' is illegal

This is the proof that VC ++ 6.0 cannot recognize long!

So,How does VC ++ 6.0 complete the 64-bit integer required in the above Code?

See the following code: (the only secret is to change long to _ int64. Note:There are two underscores on the left of int64!)

0001# Include "stdio. H"0002# Include "math. H"0003Void main (){0004_ Int64 N,0005X,/* hypothetical factor */0006Count = 0;/* counter */00070008Scanf ("% i64d", & N );0009Printf ("% i64d", N );0010For (x = 1; x <= (_ int64) SQRT (n); X = x + 1 ){0011If (N % x = 0 ){0012Count = count + 2;/* The reason for + 2 is that X and (n/x) are both factors of N */0013/* Printf ("% i64d \ n", x );*/0014If (N/x = x) Count = count-1;0015}0016}0017Printf ("number of factors: % i64d", count );0018Return;0019}
 

Please note that the above Code can correctly process the data input of an extremely large integer within the power of 12 of 10.

Believe it? Try it!