在不使用其他库的情况下，我喜欢使用以下代码片段:

#ifdef _WIN32
   #include <io.h> 
   #define access    _access_s
#else
   #include <unistd.h>
#endif

bool FileExists( const std::string &Filename )
{
    return access( Filename.c_str(), 0 ) == 0;
}

这适用于Windows和posix兼容系统的跨平台。

你也可以使用bool b = std::ifstream('filename').good();。如果没有分支指令(比如if)，它必须执行得更快，因为它需要被调用数千次。

It depends on where the files reside. For instance, if they are all supposed to be in the same directory, you can read all the directory entries into a hash table and then check all the names against the hash table. This might be faster on some systems than checking each file individually. The fastest way to check each file individually depends on your system ... if you're writing ANSI C, the fastest way is fopen because it's the only way (a file might exist but not be openable, but you probably really want openable if you need to "do something on it"). C++, POSIX, Windows all offer additional options.

While I'm at it, let me point out some problems with your question. You say that you want the fastest way, and that you have thousands of files, but then you ask for the code for a function to test a single file (and that function is only valid in C++, not C). This contradicts your requirements by making an assumption about the solution ... a case of the XY problem. You also say "in standard c++11(or)c++(or)c" ... which are all different, and this also is inconsistent with your requirement for speed ... the fastest solution would involve tailoring the code to the target system. The inconsistency in the question is highlighted by the fact that you accepted an answer that gives solutions that are system-dependent and are not standard C or C++.

如果你需要区分一个文件和一个目录，考虑下面这两个都使用stat的最快标准工具，如PherricOxide所演示的:

#include <sys/stat.h>
int FileExists(char *path)
{
    struct stat fileStat; 
    if ( stat(path, &fileStat) )
    {
        return 0;
    }
    if ( !S_ISREG(fileStat.st_mode) )
    {
        return 0;
    }
    return 1;
}

int DirExists(char *path)
{
    struct stat fileStat;
    if ( stat(path, &fileStat) )
    {
        return 0;
    }
    if ( !S_ISDIR(fileStat.st_mode) )
    {
        return 0;
    }
    return 1;
}

虽然有几种方法可以做到这一点，但对您的问题最有效的解决方案可能是使用fstream的预定义方法之一，例如good()。使用此方法可以检查指定的文件是否存在。

fstream file("file_name.txt");

if (file.good()) 
{
    std::cout << "file is good." << endl;
}
else 
{
    std::cout << "file isnt good" << endl;
}

我希望这对你有用。

这里有一个简单的例子!

#include <iostream>
#include <fstream>
using namespace std;
    
void main(){
   SearchFile("test.txt");
}

bool SearchFile(const char *file)
{
   ifstream infile(file);
   if (!infile.good())
   {
    // If file is not there
    exit(1);
   }
}