#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

使用VS2005将其编译为x64:

%llu运行良好。

如何使用printf格式化unsigned long long int ?

由于C99在转换说明符o,u,x, x之前使用了“ll”(ell-ell)。

很多答案中除了10进制选项外,还有16进制和8进制选项:

选择包括

unsigned long long num = 285212672;
printf("Base 10: %llu\n", num);
num += 0xFFF; // For more interesting hex/octal output.
printf("Base 16: %llX\n", num); // Use uppercase A-F
printf("Base 16: %llx\n", num); // Use lowercase a-f
printf("Base  8: %llo\n", num);
puts("or 0x,0X prefix");
printf("Base 16: %#llX %#llX\n", num, 0ull); // When non-zero, print leading 0X
printf("Base 16: %#llx %#llx\n", num, 0ull); // When non-zero, print leading 0x
printf("Base 16: 0x%llX\n", num); // My hex fave: lower case prefix, with A-F

输出

Base 10: 285212672
Base 16: 11000FFF
Base 16: 11000fff
Base  8: 2100007777
or 0x,0X prefix
Base 16: 0X11000FFF 0
Base 16: 0x11000fff 0
Base 16: 0x11000FFF

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。

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。

%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