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。

还有一个更简单的方法

#include <fstream>
#include <iostream>

void FileExists(std::string myfile){
std::ifstream file(myfile.c_str());

if (file) {
    std::cout << "file exists" << std::endl;
}
else {
    std::cout << "file doesn't exist" << std::endl;
}
}

int main() {
FileExists("myfile.txt");

return 0;
}

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;
}

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);
   }
}

只有一种更快的方法来检查文件是否存在，如果你有权限读取它，这种方法是使用C语言，希望更快，也可以在c++的任何版本中使用

解决方案:在C语言中有一个errno.h库，它有一个外部(全局)整数变量errno，它包含一个可以用来识别错误类型的数字

    #include <stdio.h>
    #include <stdbool.h>
    #include <errno.h>

    bool isFileExist(char fileName[]) {
        FILE *fp = fopen(fileName, "r");
        if (fp) {
            fclose(fp);
            return true;
        }
        return errno != ENOENT;
    }

    bool isFileCanBeRead(char fileName[]) {
        FILE *fp = fopen(fileName, "r");
        if (fp) {
            fclose(fp);
            return true;
        }
        return errno != ENOENT && errno != EPERM;
    }