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[英]error: invalid conversion from ‘int’ to ‘void*’ [-fpermissive] in the given set of commands
[英]error: invalid conversion from 'int' to 'void*' [-fpermissive]
為實驗室構建一個程序,我必須使用線程,我對此有點迷惑,但是我接近將其編譯。 我有2個錯誤:標題中提到了一個錯誤,另一個錯誤是相同的,但是它表示從'void *'到'int'的無效轉換。
該錯誤發生在生產者線程和使用者線程中的第98和124行上,我已在代碼中標記了它們。 這是很多代碼,但是抱歉,我不知道如何真正縮小它。
// syncA.cpp for lab2
#include <pthread.h>
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
using namespace std;
#define BSIZE 10
#define NUM_ITEMS 10
#define NUM_THREADS 2
int buf[BSIZE];
int nextin=0, nextout=0;
void * producer(void *); // function for producer thread
void * consumer(void *); // function for consumer thread
pthread_mutex_t lock;
pthread_t tid[NUM_THREADS]; // array of thread IDs
int main( int argc, char *argv[] )
{
int i;
cout << "Creating threads" << endl;
pthread_create(&tid[1], NULL, consumer, (void *) buf[BSIZE]);
pthread_create(&tid[0], NULL, producer, (void *) buf[BSIZE]);
for (i = 0; i < NUM_THREADS; i++){
pthread_join(tid[i], NULL);
}
cout << "All threads have been terminated" << endl << endl;
// Finding minimum
int minimum = buf[1];
for (i = 1; i <= BSIZE; i ++){
if (minimum > buf[i + 1])
minimum = buf[i + 1];
}
// Finding maximum
int maximum = buf[1];
for (i = 1; i <= BSIZE; i++){
if (maximum < buf[i + 1])
maximum = buf[i + 1];
}
// Finding average
int average;
int sum = 0;
for (i = 1; i <= BSIZE; i++){
sum = sum + buf[i];
}
average = sum / BSIZE;
// Outputting claculated data
cout << "Minimum value: " << minimum << endl;
cout << "Maximum value: " << maximum << endl;
cout << "Average value: " << average << endl;
return 0;
} /* main */
void * producer(void * buf[])
{
int product; // For multiplying inside the for loop for the "wait"
int num;
cout << "Producer started" << endl;
// Locking thread
pthread_mutex_lock(&lock);
// Producing 10 items and putting them in the buffer
for (int i = 0; i < BSIZE; i++){
num = rand() % 1000;
// Using a for loop 1000 times to act as the wait
for (int k = 0; k < 1000; k++){
product = 8 * 9;
}
// Putting the num in the buffer at pos 1, 2, 3, etc
buf[nextin++] = num; <---------------- ***ERROR***
}
// Unlocking thread
pthread_mutex_unlock(&lock);
// Exiting the producer
pthread_exit(0);
}
void * consumer(void * buf[])
{
int num;
int product;
cout << "Consumer started" << endl << endl;
// Locking thread
pthread_mutex_lock(&lock);
// Waiting before accessing buffer
for (int k = 0; k < 1000; k++){
product = 8 * 9;
}
//consuming items
for (int i = 0; i < BSIZE; i++){
num = buf[nextout++]; <---------------- ***ERROR***
cout << "Consuming item: " << num << endl;
// TODO: Consume item
}
// Unlocking thread
pthread_mutex_unlock(&lock);
// Exiting consumer
pthread_exit(0);
}
我已經在與此類似的其他線程上進行了一些閱讀,但找不到所需的答案。 感謝您的幫助,對於這種類型的編程我還是很陌生。
您將函數聲明為
void * producer(void *); // function for producer thread
void * consumer(void *); // function for consumer thread
但是他們的定義還有另一個原型:
void * producer(void * buf[]);
void * consumer(void * buf[]);
在這種情況下, buf
是指向void
的指針數組。 而且您正在嘗試將數字指向指向void
指針:
buf[nextin++] = num;
函數原型的聲明和定義必須相同。
顯然,您想將數字寫入緩沖區。 但是您不能具有void
的數組。 因此將buf
為int *
:
static_cast<int*>(buf)[nextin++] = num;
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