#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来自于输出unsigned long long int。你如何打印()一个unsigned long long int?
当前回答
%d——>为int
%u——>为unsigned int
%ld—>为长int或长
%lu——>用于unsigned long int或long unsigned int或unsigned long
%lld—>表示long long int或long long
%llu——>用于unsigned long long int或unsigned long long
其他回答
使用VS2005将其编译为x64:
%llu运行良好。
这是因为%llu不能在Windows下正常工作，而%d不能处理64位整数。我建议使用PRIu64，你会发现它也可以移植到Linux上。
试试这个吧:
#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;
}
输出
My number is 8 bytes wide and its value is 285212672. A normal number is 5.
您可能想尝试使用inttypes.h库，它可以为您提供诸如 Int32_t, int64_t, uint64_t等。 然后你可以使用它的宏，如:
#include <inttypes.h>
uint64_t x;
uint32_t y;
printf("x: %"PRIu64", y: %"PRIu32"\n", x, y);
这“保证”不会给您带来与long、unsigned long等相同的麻烦，因为您不必猜测每种数据类型中有多少位。
在Linux中是%llu，在Windows中是%I64u
虽然我发现它不能在Windows 2000中工作，但似乎有一个bug !
除了人们多年前写的:
你可能会在gcc/mingw上得到这个错误:
main.c:30:3:警告:未知转换类型字符'l'的格式[-Wformat=]
printf（“%llu\n”， k）;
那么你的mingw版本不会默认为c99。添加编译器标志:-std=c99。
