題目大意

有1個資源區，2個緩沖區（大小為1），4個吃貨，總共有5個線程，其中一個線程從資源區取字母放到緩沖區，哪個緩沖區是空的就往哪放；4個吃貨線程，從緩沖區吃字母。

該題目要求實現對緩沖區的互斥訪問，緩沖區和資源區必須滿足“緩沖區為空，才能從資源區取字母”的同步要求，緩沖區和吃貨線程要滿足“緩沖區不空，吃貨才能從緩沖區取字母”的同步要求；除此之外，還要求使用信號量控制吃貨線程最大并發數為2.

運行結果

代碼

// EatChar.cpp : Defines the entry point for the application. //#include "stdafx.h"HANDLE evtBuffer1Empty, evtBuffer2Empty, evtBuffer1Full, evtBuffer2Full; HANDLE hSemaphore; char szBufferRes[100]; char szBuffer1[2], szBuffer2[2]; char szBufferEat[4][100]; HWND hEditEat[4]; HWND hEditRes, hEditBuffer1, hEditBuffer2; // 負責從資源區取字符，放到緩沖區 // 使用事件實現對緩沖區的互斥訪問 // 不使用MUTEX或CRITICAL_SECTION是因為事件能實現同步 // 同步的意思是要確保緩沖區為空才往里面存字符 DWORD WINAPI ThreadCharToBuffer(LPVOID) {// 只要資源區還有東西，就一直取while (szBufferRes[0] != 0){Sleep(500);// 先看看緩沖區1是否為空，不等待立即返回DWORD dwRet = WaitForSingleObject(evtBuffer1Empty, 0);// 如果1空，那么往1放一個字符if (dwRet == WAIT_OBJECT_0){szBuffer1[0] = szBufferRes[0];//strcpy(szBufferRes, szBufferRes+1);memcpy(szBufferRes, szBufferRes+1, strlen(szBufferRes));SetWindowText(hEditRes, szBufferRes);SetWindowText(hEditBuffer1, szBuffer1);SetEvent(evtBuffer1Full);continue;}else{// 1還有東西，嘗試往2里放dwRet = WaitForSingleObject(evtBuffer2Empty, 0);if (dwRet == WAIT_OBJECT_0){szBuffer2[0] = szBufferRes[0];//strcpy(szBufferRes, szBufferRes+1);memcpy(szBufferRes, szBufferRes+1, strlen(szBufferRes));SetWindowText(hEditRes, szBufferRes);SetWindowText(hEditBuffer2, szBuffer2); SetEvent(evtBuffer2Full);}} }return 0; }// 4個線程同時只能有兩個被調度，用信號量實現 DWORD WINAPI ThreadEat(LPVOID p) {// 現在這樣寫應該有兩個線程永遠不執行size_t id = (size_t)p;// 信號量控制并發數// 事件控制同步while (szBuffer1[0] != 0 || szBuffer2[0] != 0 || szBufferRes[0] != 0){Sleep(1000);WaitForSingleObject(hSemaphore,INFINITE);// 嘗試從1緩沖區取DWORD dwRet = WaitForSingleObject(evtBuffer1Full, 0);if (dwRet == WAIT_OBJECT_0){GetWindowText(hEditEat[id], szBufferEat[id], 100);strcat(szBufferEat[id], szBuffer1);memset(szBuffer1, 0, 2);SetWindowText(hEditBuffer1, szBuffer1);SetWindowText(hEditEat[id], szBufferEat[id]);SetEvent(evtBuffer1Empty);}// 嘗試從緩沖區2取else{dwRet = WaitForSingleObject(evtBuffer2Full, 0);if (dwRet == WAIT_OBJECT_0){GetWindowText(hEditEat[id], szBufferEat[id], 100);strcat(szBufferEat[id], szBuffer2);memset(szBuffer2, 0, 2);SetWindowText(hEditBuffer2, szBuffer2);SetWindowText(hEditEat[id], szBufferEat[id]);SetEvent(evtBuffer2Empty); }}ReleaseSemaphore(hSemaphore, 1, NULL); }return 0; }DWORD WINAPI ThreadStart(LPVOID p) {// 必要的初始化工作memset(szBufferRes, 0, 100);memset(szBuffer1, 0, 2);memset(szBuffer2, 0, 2);GetWindowText(hEditRes, szBufferRes, 100);// 初始化事件：緩沖區為空// 事件句柄應當在資源區為空時關閉evtBuffer1Empty = CreateEvent(0, FALSE, TRUE, NULL);evtBuffer2Empty = CreateEvent(0, FALSE, TRUE, NULL);evtBuffer1Full = CreateEvent(0, FALSE, FALSE, NULL);evtBuffer2Full = CreateEvent(0, FALSE, FALSE, NULL);// 創建1個資源線程HANDLE hThreadCharToBuffer = CreateThread(0,0,ThreadCharToBuffer,0,0,0);// 創建信號量控制并發數和4個吃貨線程hSemaphore = CreateSemaphore(0,0,4,0); // 最大資源數4HANDLE hThreadEat[4];hThreadEat[0] = CreateThread(0,0,ThreadEat,(LPVOID)0,0,0);hThreadEat[1] = CreateThread(0,0,ThreadEat,(LPVOID)1,0,0);hThreadEat[2] = CreateThread(0,0,ThreadEat,(LPVOID)2,0,0);hThreadEat[3] = CreateThread(0,0,ThreadEat,(LPVOID)3,0,0);// 釋放兩個資源，這樣同時只能有2個線程被調度ReleaseSemaphore(hSemaphore, 2, 0);WaitForSingleObject(hThreadCharToBuffer, INFINITE); WaitForMultipleObjects(4, hThreadEat, true, INFINITE); CloseHandle(evtBuffer1Empty);CloseHandle(evtBuffer2Empty);MessageBox(0,"所有線程結束，所有事件關閉", "", MB_OK);return 0; }BOOL CALLBACK MainProc(HWND hDlg, UINT uMsg, WPARAM wParam, LPARAM lParam) {switch(uMsg){case WM_INITDIALOG:{hEditRes = GetDlgItem(hDlg, IDC_EDIT_RES);hEditBuffer1 = GetDlgItem(hDlg, IDC_EDIT_BUFFER1);hEditBuffer2 = GetDlgItem(hDlg, IDC_EDIT_BUFFER2);hEditEat[0] = GetDlgItem(hDlg, IDC_EDIT_A);hEditEat[1] = GetDlgItem(hDlg, IDC_EDIT_B);hEditEat[2] = GetDlgItem(hDlg, IDC_EDIT_C);hEditEat[3] = GetDlgItem(hDlg, IDC_EDIT_D);return TRUE;}case WM_COMMAND:{switch(LOWORD(wParam)){case IDC_BUTTON_START:{HANDLE hThread = CreateThread(0, 0, ThreadStart, 0, 0, 0);CloseHandle(hThread);return TRUE;}}return TRUE;}case WM_CLOSE:{EndDialog(hDlg, 0);return TRUE;}}return FALSE; }int APIENTRY WinMain(HINSTANCE hInstance,HINSTANCE hPrevInstance,LPSTR lpCmdLine,int nCmdShow) {DialogBox(hInstance, MAKEINTRESOURCE(IDD_DIALOG_MAIN), 0, MainProc);return 0; }

總結

以上是生活随笔為你收集整理的WIN32 多线程吃字母练习的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。