Question

我的代码如下：

#include <iostream>
#include <sys/time.h>
using namespace std;

int main(int argc, char** argv) {
                if(argv[0])
                        argc++;

                struct timeval m_timeEnd, m_timeCreate, m_timeStart;
        long mtime, alltime, seconds, useconds;

                gettimeofday(&m_timeStart,NULL);
                sleep(3);
                gettimeofday(&m_timeCreate,NULL);
                sleep(1);

        gettimeofday(&m_timeEnd, NULL);
        seconds  = m_timeEnd.tv_sec  - m_timeStart.tv_sec;
        useconds = m_timeEnd.tv_usec - m_timeStart.tv_usec;

        mtime = (long) (((seconds) * 1000 + useconds/1000.0) + 0.5);
        seconds = useconds = 0;
        seconds  = m_timeEnd.tv_sec  - m_timeCreate.tv_sec;
        useconds = m_timeEnd.tv_usec - m_timeCreate.tv_usec;
        alltime = (long) (((seconds) * 1000 + useconds/1000.0) + 0.5);

        printf("IN=%ld ALL=%ld milsec.\n", mtime, alltime);

}

我正在编译

g ++ -W -Wall -Wno-unknown-pragmas -Wpointer-arith -Wcast-align -Wcast-qual -Wsign-compare -Wconversion -O -fno-strict-aliasing

我有一些警告，我需要消除。怎么样？

a1.cpp:21: warning: conversion to 'double' from 'long int' may alter its value
a1.cpp:21: warning: conversion to 'double' from 'long int' may alter its value
a1.cpp:25: warning: conversion to 'double' from 'long int' may alter its value
a1.cpp:25: warning: conversion to 'double' from 'long int' may alter its value

Answer 1

如果你不真的需要将值四舍五入到最接近的毫秒 - 也就是说，如果你可以使用高达1毫秒而不是1/2毫秒的不准确度 - 你可以简单地写

mtime = seconds * 1000 + useconds / 1000;

否则，它必须是

mtime = seconds * 1000 + (useconds / 500 + 1) / 2;

修改：是否。见评论。

Answer 2

也改变它：

mtime = seconds * 1000 + useconds/1000;

差别仅在于它没有四舍五入到最接近的微秒（它向下舍入）
无论如何都没有那么准确的计时器。

如果你真的必须有额外的准确性（四舍五入到最接近而不是四舍五入到地板上）。

// Add 500 to useconds so that when we divide by 1000 we effectively
// round to nearest rather than truncate thus rounding to floor
mtime = seconds * 1000 + (useconds + 500) / 1000;

Answer 3

hacky方式是施展......

mtime = (long) (((double)seconds * 1000.0 + (double)useconds/1000.0) + 0.5);

这会删除所有警告......

Answer 4

这应该有效：

mtime = (long)(((long long)seconds*1000000 + useconds + 500)/1000);

以相同的方式转换alltime的表达式。

您看到警告的原因是您的表达式从long转换为double并返回以进行数学运算。您可以通过重新调整表达式来避免它，以完全保持在整数类型中。请注意转换为long long以避免溢出（谢谢，尼克）。

编辑您可以进一步简化此操作并取消转换：

mtime = seconds*1000 + (useconds + 500)/1000;

警告：从'long int'转换为'double'可能会改变其值

4 个答案: