函数从表返回函数指针作为参数

Question

I have been reading for a while, but today I can't figure someting out and find a solution.

How to return a function pointer from a function table as parameter? All similair solutions don't work for this one and end up not compiling.

I have tried a lot of methods but the compiler always returns with errors like:

function returning function is not allowed solution (when using typedef void (*func)(); )

As NO parameters have to be passed into the final routine it should be possible.

My simplified example:

void PrintOne(void) { printf("One")};
void PrintTwo(void) { printf("Two")};

struct ScanListStruct
{
    int Value;
    void (*Routine)(void);
}

const ScanListStruct DoList[] =
{ 
    {1, PrintOne},
    {2, PrintTwo}
}

bool GetRoutine(void *Ptr, int Nr)
{
    for (int x =0; x<=1; x++)
    {
        if (DoList[x].Value = Nr)   
        {
            Ptr = DoList[(x)].Routine;
            //((*DoList[(x)].Routine)());    // Original Working and executing version!
            return true;
        }
    }
    return false;
}

void main(void)
{
    int y = 1;
    void (*RoutineInMain)();     // Define
    if (GetRoutine( RoutineInMain, y) == true)    // get the address
    {   
        RoutineInMain();    // Execute the function
    }
}

Answer 1

There a few things wrong with the code;

• Syntax errors (missing ; etc.)
• main must return int
• GetRoutine should accept the function pointer by reference, not just a void* pointer to anything
• if condition should contain an equality test, not an assignment

As follows, works as expected;

void PrintOne(void) { printf("One"); };
void PrintTwo(void) { printf("Two"); };

struct ScanListStruct
{
    int Value;
    void (*Routine)(void);
};

const ScanListStruct DoList[] =
{ 
    {1, &PrintOne},
    {2, &PrintTwo}
};

bool GetRoutine(void (*&Ptr)(), int Nr)
{
    for (int x =0; x<=1; x++)
    {
        if (DoList[x].Value == Nr)   
        {
            Ptr = *DoList[(x)].Routine;
            //((*DoList[(x)].Routine)());    // Original Working and executing version!
            return true;
        }
    }
    return false;
}

int main(void)
{
    int y = 1;
    void (*RoutineInMain)();     // Define
    if (GetRoutine( RoutineInMain, y) == true)    // get the address
    {   
        RoutineInMain();    // Execute the function
    }
}

Prints One .

Answer 2

You have lots of errors in your code. Like here you put the comas at the wrong place:

void PrintOne(void) { printf("One")};
void PrintTwo(void) { printf("Two")};

It should be

void PrintOne(void) { printf("One");}
void PrintTwo(void) { printf("Two");}

And here you are using the wrong operator, = instead of ==.

if (DoList[x].Value = Nr) 

When the argument Ptr is a pointer, and that is passed by value, so the value assigned in the function will not be available when the function returns.

This is how your code should be:

void PrintOne(void) { printf("One"); }
void PrintTwo(void) { printf("Two"); }

typedef void(*prototype)();

struct ScanListStruct
{
   int Value;
   prototype Routine;
};

const ScanListStruct DoList[] =
{
   { 1, PrintOne },
   { 2, PrintTwo }
};

bool GetRoutine(prototype &Ptr, int Nr)
{
   for (int x = 0; x <= 1; x++)
   {
      if (DoList[x].Value == Nr)
      {
         Ptr = DoList[(x)].Routine;
         return true;
      }
   }
   return false;
}

int main()
{
   int y = 1;
   prototype RoutineInMain;     // Define
   if (GetRoutine(RoutineInMain, y) == true)    // get the address
   {
      RoutineInMain();    // Execute the function
   }

    return 0;
}