#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
一种方法是用VS2008将其编译为x64
这运行如你所料:
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。所以它变成了。
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);
这段代码适用于32位和64位VS编译器。
其他回答
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。
Hex:
printf("64bit: %llp", 0xffffffffffffffff);
输出:
64bit: FFFFFFFFFFFFFFFF
一种方法是用VS2008将其编译为x64
这运行如你所料:
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。所以它变成了。
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);
这段代码适用于32位和64位VS编译器。
使用ll (el-el) long-long修饰符和u (unsigned)转换。(适用于windows, GNU)。
printf("%llu", 285212672);
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
