我想使用64位整数可以接受的全部范围。 如何在x86 Visual Studio 2008中做到这一点？

Question

I included the stdint.h in my solution and used uint64_t , but the result was not what i wanted. Here is the code that i used.

#include <stdio.h>
#include "./stdint.h"

void    main (void)
{

    //-- to get the maximum value that the 32-bit integer can take
    unsigned int test = 0 ;
    test-- ;

    //-- to see if the 64-bit integer can take the value that the 32-bit integer can't take
    uint64_t test2 = test ;
    test2++ ;

    printf("%u\n", test) ;

    printf("%u\n", test2) ;

    while(1) { }

}

And Here is the result.

4294967295
0

I want to use the full range that the 64-bit integer can take. How can i do it in x86 Visual Studio 2008? For your information, i'm using a 32-bit windows 7.

Answer 1

The %u format specifier for printf is for unsigned integers. Use %llu .

Since this is C++ though, you might as well use the type-safe std::cout to avoid programmer error:

std::cout << test2;

Answer 2

Use:

#include <inttypes.h>

/* ... */

printf("%" PRIu64 "\n", test2);

to print an uint64_t value.

u conversion specifier is used to print an unsigned int value.

Note that the PRIu64 macro is a C macro. In C++ the macro is not present and you may have to use %llu conversion specification.

Answer 3

由于您也将其标记为C ++，因此我将添加一种避免类型不匹配的明显方法，就像您在printf遇到的那样：改用ostream：

std::cout << test2;

Answer 4

In a book that i have, it says that program always convert any variables to integer when it calculates. Is it possible to make the default integer 64-bit?

This is 1) not correct, and 2) no, you can't tell (most) compilers what size int you want. You could of course use something like:

 typedef Int int64_t;
 typedef Uint uint64_t; 

But you can't rename int to something other than what the compiler things it should be - which may be 16, 32, 64, 36, or 72 bits - or some other number >= 16.