我正在阅读有关共享内存的信息,而我正在阅读的操作系统书籍提供了以下生产者/消费者程序:
制片人:
#include <windows.h>
#include <stdio.h>
int main(int argc, char *argv[])
{
HANDLE hFile, hMapFile;
LPVOID lpMapAddress;
hFile = CreateFile("temp.txt",
GENERIC_READ | GENERIC_WRITE,
0,
NULL,
OPEN_ALWAYS,
FILE_ATTRIBUTE_NORMAL,
NULL);
hMapFile = CreateFileMapping(hFile,
NULL,
PAGE_READWRITE,
0,
0,
TEXT("SharedObject"));
lpMapAddress = MapViewOfFile(hMapFile,
FILE_MAP_ALL_ACCESS,
0,
0,
0);
sprintf(lpMapAddress, "Shared memory message");
UnmapViewOfFile(lpMapAddress);
CloseHandle(hFile);
CloseHandle(hMapFile);
}
消费者:
#include <windows.h>
#include <stdio.h>
int main(int argc, char *argv[])
{
HANDLE hMapFile;
LPVOID lpMapAddress;
hMapFile = OpenFileMapping(FILE_MAP_ALL_ACCESS,
FALSE,
TEXT("SharedObject"));
lpMapAddress = MapViewOfFile(hMapFile,
FILE_MAP_ALL_ACCESS,
0,
0,
0);
printf("Read message %s", lpMapAddress);
UnmapViewOfFile(lpMapAddress);
CloseHandle(hMapFile);
}
问题是它不编译。Visual C++ 2008 Express 在生产者部分给出了这个错误:
错误 C2664:“sprintf”:无法将参数 1 从“LPVOID”转换为“char *”
有什么问题?