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我正在嘗試測量線程切換開銷時間。 我有兩個線程，一個共享變量，一個互斥鎖和兩個條件變量。兩個線程會來回切換，將1或0寫入共享變量。這是衡量線程上下文切換開銷的正確解決方案嗎？
我假設pthread_cond_wait（& cond，&互斥量）等待時間大約等於2 x線程上下文切換時間。因爲如果一個thread1必須等待一個條件變量，它必須放棄互斥鎖到thread2->thread2 context switch - > thread2執行它的任務併發信號通知條件變量喚醒第一個線程 - >上下文切換回線程1 - > thread1重新獲取鎖。
我的假設是否正確？
我的代碼如下：
#include <sys/types.h>
#include <wait.h>
#include <unistd.h>
#include <errno.h>
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
#include <sys/resource.h>
#include <dirent.h>
#include <ctype.h>
#include<signal.h>
#include <stdio.h>
#include <stdint.h>
#include <time.h>
#include <pthread.h>
int var = 0;
int setToZero = 1;
int count = 5000;
pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t isZero = PTHREAD_COND_INITIALIZER;
pthread_cond_t isOne = PTHREAD_COND_INITIALIZER;
struct timespec firstStart;
unsigned long long timespecDiff(struct timespec *timeA_p, struct timespec *timeB_p)
{
return ((timeA_p->tv_sec * 1000000000) + timeA_p->tv_nsec) -
((timeB_p->tv_sec * 1000000000) + timeB_p->tv_nsec);
}
void* thread1(void* param)
{
int rc;
struct timespec previousStart;
struct timespec start; //start timestamp
struct timespec stop; //stop timestamp
unsigned long long result;
int idx = 0;
int measurements[count];
clock_gettime(CLOCK_MONOTONIC, &stop);
result = timespecDiff(&stop,&firstStart);
printf("first context-switch time:%llu\n", result);
clock_gettime(CLOCK_MONOTONIC, &previousStart);
while(count > 0){
//acquire lock
rc = pthread_mutex_lock(&mutex);
clock_gettime(CLOCK_MONOTONIC,&start);
while(setToZero){
pthread_cond_wait(&isOne,&mutex); // use condition variables so the threads don't busy wait inside local cache
}
clock_gettime(CLOCK_MONOTONIC,&stop);
var = 0;
count--;
setToZero = 1;
//printf("in thread1\n");
pthread_cond_signal(&isZero);
//end of critical section
rc = pthread_mutex_unlock(&mutex); //release lock
result = timespecDiff(&stop,&start);
measurements[idx] = result;
idx++;
}
result = 0;
int i = 0;
while(i < idx)
{
result += measurements[i++];
}
result = result /(2*idx);
printf("thread1 result: %llu\n",result);
}
void* thread2(void* param)
{
int rc;
struct timespec previousStart;
struct timespec start; //start timestamp
struct timespec stop; //stop timestamp
unsigned long long result;
int idx = 0;
int measurements[count];
while(count > 0){
//acquire lock
rc = pthread_mutex_lock(&mutex);
clock_gettime(CLOCK_MONOTONIC,&start);
while(!setToZero){
pthread_cond_wait(&isZero,&mutex);
}
clock_gettime(CLOCK_MONOTONIC,&stop);
var = 1;
count--;
setToZero = 0;
//printf("in thread2\n");
pthread_cond_signal(&isOne);
//end of critical section
rc = pthread_mutex_unlock(&mutex); //release lock
result = timespecDiff(&stop,&start);
measurements[idx] = result;
idx++;
}
result = 0;
int i = 0;
while(i < idx)
{
result += measurements[i++];
}
result = result /(2*idx);
printf("thread2 result: %llu\n",result);
}
int main(){
pthread_t threads[2];
pthread_attr_t attr;
pthread_attr_init(&attr);
clock_gettime(CLOCK_MONOTONIC,&firstStart);
pthread_create(&threads[0],&attr,thread1,NULL);
pthread_create(&threads[1],&attr,thread2,NULL);
printf("waiting...\n");
pthread_join(threads[0],NULL);
pthread_join(threads[1],NULL);
pthread_cond_destroy(&isOne);
pthread_cond_destroy(&isZero);
}
我得到以下時間：
first context-switch time:144240
thread1 result: 3660
thread2 result: 3770
除非您只有一個CPU內核，否則在這種情況下，線程可能不需要上下文切換*。 – EOF
我正在使用單核心機器 – JJTO
然後有一個事實，即有其他進程/線程要求cpu時間。上下文切換是狀態變量等的加載/卸載，我不認爲你可以測量它，你甚至可能遭受觀察者的影響:)操作系統負責上下文切換，在用戶空間中你不需要控制它，這一切都發生在你的進程正在睡覺。 –