家庭作业 - 使用信号量和互斥锁的队列

Question

...所以我固定了一下，但它现在说...加速度测试：程序交叉时间限制... 持续时间：5秒！（限制8秒）！ ... 这里是我的代码，如果任何人有任何想法...

#ifndef __PROGTEST__ 
#include <stdio.h> 
#include <stdlib.h> 
#include <pthread.h> 
#include <semaphore.h> 

#define MAX_TERRORISTS 32 

typedef struct TImage 
{ 
    int    m_W; 
    int    m_H; 
    unsigned char ** m_Data; 
} TIMAGE; 

void TerroristHunter (int  databaseSize, 
          TIMAGE ** database, 
          int  threads, 
          TIMAGE *(* scanner) (void), 
          void (* officer) (TIMAGE *, int)); 
#endif /* __PROGTEST__ */ 


/* Vase implementace/your implementation here */ 
pthread_mutex_t m1; 

typedef struct Par { 
TIMAGE *(* scanner) (void); 
TIMAGE ** terrorist; 
int number_of_terrorists; 
void (*officer) (TIMAGE *, int);} PAR; 

void * function (void * arg) { 

PAR * a = (PAR *) arg; 
TIMAGE * image; 
int i1, i2, i3, i4, i5, result = 0; 

while (1) { 
pthread_mutex_lock (& m1); 
if ((image = a->scanner()) == NULL) {pthread_mutex_unlock (& m1);break;} 
pthread_mutex_unlock (& m1); 
for (i5 = 0; i5 < a->number_of_terrorists; i5 ++) { 
    for (i1 = 0; i1 <= (image->m_H - a->terrorist[i5]->m_H); i1 ++) { 
     for (i2 = 0; i2 <= (image->m_W - a->terrorist[i5]->m_W); i2 ++) { 
      if ((a->terrorist[i5]->m_Data[0][0] == image->m_Data[i1][i2]) || (a->terrorist[i5]->m_Data[0][0] == 255)) { 
      for (i3 = 0; i3 < a->terrorist[i5]->m_H; i3 ++) { 
       for (i4 = 0; i4 < a->terrorist[i5]->m_W; i4 ++) 
        if ((a->terrorist[i5]->m_Data[i3][i4]!= image->m_Data[i1 + i3][i2 + i4]) && (a->terrorist[i5]->m_Data[i3][i4] != 255)) 
         break; 
       if (i4 != a->terrorist[i5]->m_W) 
        break; } 
      if (i3 == a->terrorist[i5]->m_H) { 
       result = result + (1 << i5); 
       break; } } } 
     if (i3 == a->terrorist[i5]->m_H) 
      break; } } 
a->officer (image, result); 
result = 0;} 
return NULL; 
} 

void TerroristHunter (int  databaseSize, 
          TIMAGE ** database, 
          int  threads, 
          TIMAGE *(* scanner) (void), 
          void (* officer) (TIMAGE *, int)) { 
PAR * pom = (PAR *) malloc (sizeof (* pom)); 
int i; 

pthread_t * thr = (pthread_t *) malloc (threads * sizeof(* thr)); 
pthread_attr_t Attr; pthread_attr_init (& Attr); pthread_attr_setdetachstate (& Attr, PTHREAD_CREATE_JOINABLE); 
pthread_mutex_init (& m1, NULL); 

pom->terrorist = database; 
pom->number_of_terrorists = databaseSize; 
pom->officer = officer; 
pom->scanner = scanner; 

for (i = 0; i < threads; i ++) 
    pthread_create (& thr[i], & Attr, function, (void *) pom); 
for (i = 0; i < threads; i ++) 
    pthread_join (thr[i], NULL); 

pthread_attr_destroy (& Attr); pthread_mutex_destroy (& m1); 
free (pom); 
return; } 
#ifndef __PROGTEST__ 
unsigned char t0r0[] = { 255, 255, 255, 255, 255, 255, 255, 255 }; 
unsigned char t0r1[] = { 255, 50, 60, 70, 255, 255, 255, 255 }; 
unsigned char t0r2[] = { 255, 50, 60, 70, 255, 255, 255, 255 }; 
unsigned char t0r3[] = { 255, 255, 50, 60, 70, 80, 255, 255 }; 
unsigned char t0r4[] = { 255, 255, 50, 60, 70, 80, 255, 255 }; 
unsigned char t0r5[] = { 255, 255, 255, 255, 255, 255, 255, 255 }; 
unsigned char t0r6[] = { 255, 255, 255, 255, 255, 255, 255, 255 }; 
unsigned char * t0all[] = { t0r0, t0r1, t0r2, t0r3, t0r4, t0r5, t0r6 }; 
TIMAGE t0 = { 8, 7, t0all }; /* one sample */ 

unsigned char t1r0[] = { 10, 20, 30 }; 
unsigned char t1r1[] = { 20, 50, 255 }; 
unsigned char t1r2[] = { 80, 50, 255 }; 
unsigned char * t1all[] = { t1r0, t1r1, t1r2 }; 
TIMAGE t1 = { 3, 3, t1all }; /* another sample */ 

TIMAGE * terrorists[2] = { &t0, &t1}; 

unsigned char i0r0[] = { 1, 2, 3, 4, 5, 6, 7, 8 }; 
unsigned char i0r1[] = { 2, 50, 60, 70, 6, 7, 8, 1 }; 
unsigned char i0r2[] = { 3, 50, 60, 70, 7, 10, 20, 30 }; 
unsigned char i0r3[] = { 4, 50, 60, 70, 8, 20, 50, 3 }; 
unsigned char i0r4[] = { 5, 6, 50, 60, 70, 80, 50, 4 }; 
unsigned char i0r5[] = { 6, 7, 50, 60, 70, 80, 4, 5 }; 
unsigned char i0r6[] = { 7, 8, 1, 2, 3, 4, 5, 6 }; 
unsigned char i0r7[] = { 8, 1, 2, 3, 4, 5, 6, 7 }; 
unsigned char * i0all[] = { i0r0, i0r1, i0r2, i0r3, i0r4, i0r5, i0r6, i0r7 }; 
TIMAGE i0 = { 8, 8, i0all }; /* t0 and t1 here */ 

unsigned char i1r0[] = { 1, 2, 3, 4, 5, 6, 7, 8 }; 
unsigned char i1r1[] = { 2, 3, 4, 5, 6, 7, 8, 1 }; 
unsigned char i1r2[] = { 3, 4, 5, 6, 7, 8, 1, 2 }; 
unsigned char i1r3[] = { 10, 20, 30, 7, 8, 1, 2, 3 }; 
unsigned char i1r4[] = { 20, 50, 7, 8, 1, 2, 3, 4 }; 
unsigned char i1r5[] = { 80, 50, 8, 1, 2, 3, 4, 5 }; 
unsigned char i1r6[] = { 7, 8, 1, 2, 3, 4, 5, 6 }; 
unsigned char * i1all[] = { i1r0, i1r1, i1r2, i1r3, i1r4, i1r5, i1r6 }; 
TIMAGE i1 = { 8, 7, i1all }; /* t1 here */ 

unsigned char i2r0[] = { 1, 50, 60, 70, 5, 6 }; 
unsigned char i2r1[] = { 2, 50, 60, 70, 6, 7 }; 
unsigned char i2r2[] = { 3, 4, 50, 60, 70, 80 }; 
unsigned char i2r3[] = { 10, 20, 50, 50, 70, 81 }; 
unsigned char i2r4[] = { 20, 50, 7, 8, 1, 2 }; 
unsigned char * i2all[] = { i2r0, i2r1, i2r2, i2r3, i2r4 }; 
TIMAGE i2 = { 6, 5, i2all }; /* no terrorists here */ 

TIMAGE * scans[3] = { &i0, &i1, &i2 }; 

TIMAGE * dummyScanner (void) 
{ 
    static int idx = 0; 

    if (idx < 3) return (scans[idx ++]); 
    return (NULL); 
} 

void  dummyOfficer (TIMAGE * img, int found) 
{ 
    int i; 

    printf ("Image: %d x %d ", img -> m_W, img -> m_H); 
    if (found) 
    { 
     printf ("TERRORISTS:"); 
     for (i = 0; i < (int)sizeof (found) * 8; i ++) 
     if (found & (1 << i)) 
     printf (" %d", i); 
     printf ("\n"); 
    } 
    else 
    printf ("no terrorists found\n"); 
} 

int main (int argc, char * argv[]) 
{ 
    TerroristHunter (2, terrorists, 20, dummyScanner, dummyOfficer); 
    return (0); 
} 
#endif /* __PROGTEST__ */ 

Answer 1

饥饿是不是一个问题，因为当扫描仪（）返回NULL，该线程将结束。

因此，在我看来，问题是当定义PROGTEST。换句话说，你的程序中没有列出的其他地方有什么问题;可能带有scanner（）和/或officer（）函数。