与PherricOxide的建议相同，但在C中

#include <sys/stat.h>
int exist(const char *name)
{
  struct stat   buffer;
  return (stat (name, &buffer) == 0);
}

all_of (begin(R), end(R), [](auto&p){ exists(p); })

其中R是你的路径序列，exists()来自未来std或当前boost。如果你自己卷，简单点，

bool exists (string const& p) { return ifstream{p}; }

分支解决方案并不是绝对可怕的，它不会吞噬文件描述符，

bool exists (const char* p) {
    #if defined(_WIN32) || defined(_WIN64)
    return p && 0 != PathFileExists (p);
    #else
    struct stat sb;
    return p && 0 == stat (p, &sb);
    #endif
}

与PherricOxide的建议相同，但在C中

#include <sys/stat.h>
int exist(const char *name)
{
  struct stat   buffer;
  return (stat (name, &buffer) == 0);
}

inline bool exist(const std::string& name)
{
    ifstream file(name);
    if(!file)            // If the file was not found, then file is 0, i.e. !file=1 or true.
        return false;    // The file was not found.
    else                 // If the file was found, then file is non-0.
        return true;     // The file was found.
}

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++.

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。