Question

Why does the following code, produce the following output?

UPDATING MY CODE TO THIS:
i get basically the same thing.

#ifdef _WIN32
    #include <windows.h>
    #include <direct.h>
    #define GetCurrentDir _getcwd
#else
    #include <unistd.h>
    #define GetCurrentDir getcwd
#endif

//==============================MAIN=======================================
#ifdef _WIN32   
    int main(int argc, char **argv) 
    {
        char *path = (char*)malloc(sizeof(char)*FILENAME_MAX);
        GetCurrentDir(path, sizeof(path));
        printf("path: %s\n", path);
        //other stuff
    }

//==============================END========================================

OUTPUT
path : –

Answer 1

You’re returning a pointer to a local variable declared in get_current_path, i.e., a variable that will likely be cleaned up once the function returns. You should accept a buffer as an argument and fill it for the caller, i.e.,

void char* get_current_path(char* outDir)

PER YOUR EDIT:

int main(int argc, char **argv) 
{
    char *path = (char*)malloc(sizeof(char)*FILENAME_MAX);
    GetCurrentDir(path, sizeof(path));
    printf("path: %s\n", path);
    //other stuff
}

sizeof(path) is going to be 4 or 8 (32-bit or 64-bit) as it is now just a pointer, not an array. You need to pass in the actual size, i.e., sizeof(char) * FILENAME_MAX, so…

int main(int argc, char **argv) 
{
    size_t bufSize = sizeof(char) * FILENAME_MAX;
    char *path = (char*)malloc(bufSize);
    GetCurrentDir(path, bufSize);
    printf("path: %s\n", path);
    //other stuff
}