#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
Hex:
printf("64bit: %llp", 0xffffffffffffffff);
输出:
64bit: FFFFFFFFFFFFFFFF
其他回答
Hex:
printf("64bit: %llp", 0xffffffffffffffff);
输出:
64bit: FFFFFFFFFFFFFFFF
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 !
使用ll (el-el) long-long修饰符和u (unsigned)转换。(适用于windows, GNU)。
printf("%llu", 285212672);
对于使用MSVS的long long(或__int64)，您应该使用%I64d:
__int64 a;
time_t b;
...
fprintf(outFile,"%I64d,%I64d\n",a,b); //I is capital i
%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
