具有信号量和线程的生产者和消费者

问题描述

我目前遇到此代码的问题。唯一出错的代码部分是“io_production_report(thread_args[0].Produced,pointer);”。

终端提错误如下"

driver.cpp: In function ‘int main(int,char**)’:
driver.cpp:83:51: error: cannot convert ‘int (*)[2]’ to ‘int**’
   83 |     io_production_report(thread_args[0].Produced,pointer);
      |                                                   ^~~~~~~
      |                                                   |
      |                                                   int (*)[2]
In file included from driver.cpp:2:
io.h:49:48: note:   initializing argument 2 of ‘void io_production_report(int*,int**)’
   49 | void io_production_report(int produced[],int *consumed[]);
      |                                           ~~~~~^~~~~~~~~~
make: *** [Makefile:8: driver.o] Error 1

我在实现指针数组时遇到问题。 如果有人可以帮助我，那就太好了！！ :

驱动程序.cpp

  // Print Production Report
    int pointer[2][2];
    pointer[0][0]=thread_args[0].Consumed[0]; //Ethel
    pointer[0][1]=thread_args[0].Consumed[1]; //Lucy
    pointer[1][0]=thread_args[1].Consumed[0];
    pointer[1][1]=thread_args[1].Consumed[1];

    io_production_report(thread_args[0].Produced,pointer);  //NEED HELP ON THIS LINE

驱动程序.h

struct args {
    sem_t *mutexOnBelt;
    sem_t *frogMax;
    sem_t *beltMax;
    sem_t *candiesOnBelt;
    sem_t *produceMax;
    sem_t *consumeMax;

    string *belt;
    int *head;
    int *tail;
    int wait_time;
    int produced;
    string *name;
    int Consumed[2];
    int onBelt[2];
    int Produced[2];
};

int option = 0;
int flagValues[4] = {0,0};

io.c

/*
 * void io_production_report(int produced[],int *consumed[])
 * Show how many candies of each type produced.  Show how many candies consumed by
 * each consumer.
 * produced[] - count for each ProductType
 * *consumed[] - array of pointers to consumed arrays
 *    e.g. consumed[Lucy] points to an array that is indexed by product name
 */
void io_production_report(int produced[],int *consumed[]) {
  int p,c;  /* array indices */
  int total;

  printf("\nPRODUCTION REPORT\n----------------------------------------\n");

  /* show number produced for each producer / candy type */
  for (p=0; p < ProductTypeN; p++) {
    printf("%s producer generated %d candies\n",ProducerNames[p],produced[p]);
  }
  
  /* show number consumed by each consumer */
  for (c=0; c < ConsumerTypeN; c++) {
    printf("%s consumed ",ConsumerNames[c]);
    total = 0;
    for (p=0; p < ProductTypeN; p++) {
      if (p > 0)
    printf(" + ");
      total += consumed[c][p];
      printf("%d %s",consumed[c][p],ProducerAbbrevs[p]);
    }
    printf(" = %d total\n",total);
  }

  printf("Elapsed time %.3f s\n",elapsed_s());
}

解决方法

如注释中所述，您需要指向数组的第一个元素 id est：您将数组 [N][M] 指定为 int (*)[M]，工作示例：

#include <stdio.h>

void io_production_report(int (*)[2]);

void io_production_report(int (*consumed)[2]) {
    for (int i = 0; i <4; i++)
        printf("%d\n",consumed[i/2][i%2]);
}

int main(void) {
    int pointer[2][2];
    pointer[0][0]=1; //Ethel
    pointer[0][1]=2; //Lucy
    pointer[1][0]=3;
    pointer[1][1]=4;

    io_production_report(pointer);
}