无符号长长resTestBad = 0xffffffff + 1;使用 %llu 打印 0，但 unsigned long long resTestAgain = 0xffffffffff + 1；用 %llu 打印 68719476736

Question

我发现了一个Bug，当%llu的(0xffffffff + 1)的打印值为0时，见下面的代码：

unsigned long long resTestBad = 0xffffffff + 1;  // number of f is 8
printf("resTestBad is %llu, sizeof(unsigned long long) is %ld\n", resTestBad, sizeof(unsigned long long));

将输出：

resTestBad is 0, sizeof(unsigned long long) is 8

我认为输出是错误的，应该输出或者说正确的输出：

resTestBad is 4294967296, sizeof(unsigned long long) is 8

但是下面的代码没有问题：

unsigned long long resTestAgain = 0xfffffffff + 1;   // number of f is 8
printf("resTestAgain is %llu, sizeof(unsigned long long) is %ld\n", resTestAgain, sizeof(unsigned long long));

它会输出（我认为是对的）：

resTestAgain is 68719476736, sizeof(unsigned long long) is 8

我的.c 文件如下：

#include <stdio.h>
#include <limits.h>
int main(void)
{

    int s = 0xfffe;
    printf("s is %d\n", s);

    unsigned int ww = 0xfffe;
    printf("ww is %d\n", ww);

    unsigned long long resTestOk = 0xfffffffe + 1; 
    printf("resTestOk is %llu, sizeof(unsigned long long) is %ld\n", resTestOk, sizeof(unsigned long long));

    unsigned long long resTestBad = 0xffffffff + 1;  
    printf("resTestBad is %llu, sizeof(unsigned long long) is %ld\n", resTestBad, sizeof(unsigned long long));

    unsigned long long resTestAgain = 0xfffffffff + 1; 
    printf("resTestAgain is %llu, sizeof(unsigned long long) is %ld\n", resTestAgain, sizeof(unsigned long long));

    unsigned long long resA = 4294967295 + 1; 
    printf("resA is %llu, sizeof(unsigned long long) is %ld\n", resA, sizeof(unsigned long long));

    unsigned long long resTestUUU = 0xffffffffffffff + 1; 
    printf("resTestUUU is %llu, sizeof(unsigned long long) is %ld\n", resTestUUU, sizeof(unsigned long long));

    unsigned long long resH = (0xffffffff+1) / 1024 / 1024 / 1024;
    printf("%llu\n", resH);

    unsigned long long resD = (4294967296) / 1024 / 1024 / 1024;
    printf("%llu\n", resD);


    return 0;

}

运行会输出：

s is 65534
ww is 65534
resTestOk is 4294967295, sizeof(unsigned long long) is 8
resTestBad is 0, sizeof(unsigned long long) is 8
resTestAgain is 68719476736, sizeof(unsigned long long) is 8
resA is 4294967296, sizeof(unsigned long long) is 8
resTestUUU is 72057594037927936, sizeof(unsigned long long) is 8
0
4

可能你不相信，我也不相信，所以我录了一个视频来证明我说的。

参见https://imgur.com/a/RAQVPxS

是cpu的Bug还是VMWare Workstation的Bug？我该如何解决？

Answer 1

正如@rici对另一个问题的回答中所解释的那样，无符号类型是十六进制整数文字的候选者。 0xffffffff 适合

unsigned long

，因此其类型为

unsigned long

，根据您的编译器，它是 32 位。然后将

int

1

转换为

unsigned long

进行加法，导致溢出。最后将

unsigned long

0

转换为

unsigned long long

，成为

resTestBad

的初始值。使用

0xffffffffULL

强制文字从一开始就为

unsigned long long

。（当您添加另一个

f

时，该值不再适合

unsigned long

，因此文字已经具有类型

long long

并且可以工作。）

Answer 2

unsigned long long resTestBad = 0xffffffff + 1;

无符号十六进制常量 

0xffffffff

可以用

unsigned int

类型的对象表示。所以它的类型是

unsigned int

。所以在这个表达中0xffffffff + 1

对 

unsigned int

类型的对象使用算术（整数 由于通常的算术转换，常量

1

也会转换为类型

unsigned int

）。结果会出现溢出，因为结果可以用

unsigned int

类型的对象表示。表达式的值为

0

.至于这个声明

unsigned long long resTestAgain = 0xfffffffff + 1;

则十六进制常量

0xfffffffff

不能用unsigned int类型的对象表示（前提是

sizeof( unsigned int )

不大于

4

）。如果

sizeof( long int )

等于

8

（并且在您的程序中它确实等于

8

），则该常量可以用

long int

类型的对象表示。所以在这个表达中

0xfffffffff + 1

对于

long int

类型的对象使用算术..并且表达式的结果也可以用这种类型的对象表示。