请各位大侠帮帮忙!

[OneMoreFreeMan]

构造一个函数,传入的字符串参数是文件路径,要求将目录中的所有文件名显示出来.
在windows的VC++上加以实现!

[kugou123]

http://www.softhouse.com.cn/html/200411/2004111007561200001712.html

[社会栋梁]

//.cpp

#include<windows.h>
#include<iostream.h>

void PrintFindData(WIN32_FIND_DATA* findData)
{
//if it's a directory ,print the name
if(findData->dwFileAttributes & FILE_ATTRIBUTE_DIRECTORY)
{
cout<<"directory :"<<findData->cFileName<<endl;
}
else
{
cout<<findData->cFileName;
cout<<"\t("<<findData->nFileSizeLow<<")";
cout<<endl;
}
}

//lists the contents of the current directory
void ListDirectoryContents(char* fileMask)
{
HANDLE fileHandle;
WIN32_FIND_DATA findData;

//get first file
fileHandle = FindFirstFile(fileMask,&findData);
if(fileHandle!=INVALID_HANDLE_VALUE)
{
PrintFindData(&findData);

//loop on all remeaining entries in dir
while (FindNextFile(fileHandle,&findData))
{
PrintFindData(&findData);
}
}
FindClose(fileHandle);
}

int main(int argc,char* argv[])
{
ListDirectoryContents("*.*");
return 0;
}