你如何printf一个无符号的long long int(unsigned long long int的格式说明符)?

#include <stdio.h> int main() { unsigned long long int num = 285212672; //FYI: fits in 29 bits int normalInt = 5; printf("My number is %d bytes wide and its value is %ul. A normal number is %d.\n", sizeof(num), num, normalInt); return 0; } 

输出:

 My number is 8 bytes wide and its value is 285212672l. A normal number is 0. 

我假设这个意外的结果是从打印unsigned long long int 。 你如何printf()一个unsigned long long int

使用ll(el-el)long-long修饰符与u(无符号)转换。 (在Windows,GNU工作)。

 printf("%llu", 285212672); 

您可能想尝试使用inttypes.h库,它提供了诸如int32_tint64_tuint64_t等types。然后可以使用它的macros,例如:

 uint64_t x; uint32_t y; printf("x: %"PRId64", y: %"PRId32"\n", x, y); 

这是“保证”,不会给你longunsigned long long等相同的麻烦,因为你不必猜测每个数据types有多less位。

很长一段时间(或__int64)使用MSVS,你应该使用%I64d:

 __int64 a; time_t b; ... fprintf(outFile,"%I64d,%I64d\n",a,b); //I is capital i 

这是因为%llu在Windows下无法正常工作,%d无法处理64位整数。 我build议使用PRIu64,而且你会发现它也可以移植到Linux上。

试试这个:

 #include <stdio.h> #include <inttypes.h> int main() { unsigned long long int num = 285212672; //FYI: fits in 29 bits int normalInt = 5; /* NOTE: PRIu64 is a preprocessor macro and thus should go outside the quoted string. */ printf("My number is %d bytes wide and its value is %"PRIu64". A normal number is %d.\n", sizeof(num), num, normalInt); return 0; } 

产量

 My number is 8 bytes wide and its value is 285212672. A normal number is 5. 

%d – >为int

%ld – > long int

%lld – > long long int

%llu – >为unsigned long long int

用VS2005编译为x64:

%llu工作得很好。

在Linux中是%llu ,在Windows中是%I64u

虽然我发现它在Windows 2000中不起作用,但似乎有一个错误!

非标准的东西总是很奇怪:)

在GNU下的很长一段时间,它是Lllq

并在Windows下,我相信它只会

hex:

 printf("64bit: %llp", 0xffffffffffffffff); 

输出:

 64bit: FFFFFFFFFFFFFFFF 

除了几年前人们写的东西之外,

  • 你可能会在gcc / mingw上得到这个错误:

main.c:30:3: warning: unknown conversion type character 'l' in format [-Wformat=]

printf("%llu\n", k);

那么你的版本的mingw不会默认为c99。 添加这个编译器标志: -std=c99

那么,一种方法是用VS2008将其编译为x64

这可以像你期望的那样运行:

 int normalInt = 5; unsigned long long int num=285212672; printf( "My number is %d bytes wide and its value is %ul. A normal number is %d \n", sizeof(num), num, normalInt); 

对于32位代码,我们需要使用正确的__int64格式说明符%I64u。 所以它成为。

 int normalInt = 5; unsigned __int64 num=285212672; printf( "My number is %d bytes wide and its value is %I64u. A normal number is %d", sizeof(num), num, normalInt); 

此代码适用于32位和64位VS编译器。