无符号长long在C中表现为奇数

Question

在运行为MIPS编译的程序时,我偶然发现了一个非常奇怪的问题.下面的代码片段是获取时间纪元并将其以微秒精度存储在unsigned long long变量中.

该变量能够存储我检查过的8个字节sizeof(unsigned long long).

这段代码打印出来很奇怪:

unsigned long long microtime=0;
struct timeval time_camera = { .tv_sec=0, .tv_usec=0 };
gettimeofday(&time_camera,NULL);
microtime = time_camera.tv_sec * 1000000 + time_camera.tv_usec;
printf("Times is now %llu , time since epoch in seconds is: %lu\n", microtime, time_camera.tv_sec);

它给了我以下输出:

>> Times is now 484305845 , time since epoch in seconds is: 1357751315

但是,当我将计算分解为不同的行时,它才有效!:

unsigned long long microtime=0;
struct timeval time_camera = { .tv_sec=0, .tv_usec=0 };
gettimeofday(&time_camera,NULL);
microtime = time_camera.tv_sec;
microtime = microtime * 1000000;
microtime = microtime  + time_camera.tv_usec;
printf("Times is now %llu , time since epoch in seconds is: %lu\n", microtime, time_camera.tv_sec);

输出是:

Times is now 1357751437422143 , time since epoch in seconds is: 1357751437

现在这只是一个巧合吗？例如,我已经损坏了内存或者实际上超过了它吗？也许这是MIPS编译器？任何帮助表示赞赏!

Answer 1

microtime = time_camera.tv_sec * 1000000 + time_camera.tv_usec;

tv_sec是一个较小的整数类型(time_t可能int或long),所以

time_camera.tv_sec * 1000000

溢出.使用后缀为常量提供适当的类型

time_camera.tv_sec * 1000000ULL

在

microtime = time_camera.tv_sec;
microtime = microtime * 1000000;

乘法是在unsigned long long,因为一个操作数(microtime)已经具有该类型,所以另一个转换为该类型.