#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
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。
其他回答
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.
非标准的东西总是奇怪的:)
对于长而长的部分 在GNU下是L, ll或q
在窗户下，我相信它只是
使用VS2005将其编译为x64:
%llu运行良好。
Hex:
printf("64bit: %llp", 0xffffffffffffffff);
输出:
64bit: FFFFFFFFFFFFFFFF
使用ll (el-el) long-long修饰符和u (unsigned)转换。(适用于windows, GNU)。
printf("%llu", 285212672);
