Return multiple pointers from function

Return multiple pointers from function

I'm trying to return 4 pointers which are stored in another pointer from a function in C, but I get segmentation fault. Do anyone know how to do this?

That's the way I tried to do this:

//declaration of 5 pointers (int * ptr0, float * ptr1, ..., int * ptr4)    int * function()  {      int * ptr;      ptr = malloc(sizeof(int)*4);        float * ptr1;      ptr1 = malloc(sizeof(float)*4);        float * ptr2;      ptr2 = malloc(sizeof(float)*4);        float * ptr3;      ptr3 = malloc(sizeof(float)*4);        int * ptr4;      ptr4 = malloc(sizeof(int)*4);        retunr ptr;  }    ptr0 = function();  ptr1 = ptr0[0];  ptr2 = ptr0[1];  //and so on...  

Ok, I changed my program but now I can not write in the pointers anymore. I know this is a realy 'stupid' question but I realy dont know. Can someone help ?

Answer by lorro for Return multiple pointers from function

You didn't fill *ptr, but even if you did, don't do this. Use std::tuple<> or pair of std::pair<>s - or return a struct.

Edit: since it's no more C++, just return a struct with the four fields.

Answer by user2079303 for Return multiple pointers from function

You can only return a singe value from a function. A simple solution is to return a struct that contains the desired pointers.

struct pointers {      float* ptr1;      float* ptr2;      float* ptr3;      // or perhaps an array instead:      // float* f_ptrs[3];      int*   ptr4;  }    struct pointers function() {      struct pointers p;      // initialize the members here      return p;  }  

Answer by mvidelgauz for Return multiple pointers from function

void ** function()  {      void ** ptr;      ptr = (void **)malloc(sizeof(void *)*4);        ptr[0] = malloc(sizeof(float)*4);      ptr[1] = malloc(sizeof(float)*4);      ptr[2] = malloc(sizeof(float)*4);      ptr[3] = malloc(sizeof(int)*4);        return ptr;  }  

Answer by Vatine for Return multiple pointers from function

There are three choices, you could return a void** (basically, a pointer to an array of void* that you then cast), you can declare a struct that contain the pointers, then allocate and return a pointer to such a struct (then let the caller de-allocate the struct) or you can use a func(float **ptr0, float **ptr1, float **ptr2, int **ptr3) and assign your malloced memory to *ptr0, ...

Neither of these are perfect, depending on circumstances I would pick either solution 2 or solution 3, on the basis that it's going to end up messy remembering all the relevant casts and allowing the compiler to help you with types is a good idea.

Answer by iedoc for Return multiple pointers from function

i saw it wasn't mentioned, but you could also pass double pointers into the function and have the function fill them out, another way to return multiple values:

void function(int** p1, float** p2, float** p3, float** p4, int** p5)  {      *p1 = malloc(sizeof(int)*4);        *p2 = malloc(sizeof(float)*4);        *p3 = malloc(sizeof(float)*4);        *p4 = malloc(sizeof(float)*4);        *p5 = malloc(sizeof(int)*4);  }    int* ptr1;  float* ptr2;  float* ptr3;  float* ptr4;  int* ptr5;    function(&ptr1, &ptr2, &ptr3, &ptr4, &ptr5);  

Answer by alk for Return multiple pointers from function

What about this:

#include   #include   /* for perror() */      int function(void ** ptrs, size_t n)  {    int result = -1;      pptrs[0] = malloc(...);    if (NULL == ptrs[0])    {      goto err;    }      ...      ptrs[n - 1] = malloc(...);    if (NULL == ptrs[n - 1])    {      goto err;    }      result = 0;    err:      return result;  }    int main(void)  {    size_t n = 5;    void * ptrs[n];    for (size_t i = 0; i < n; ++n)    {      ptrs[n] = NULL;    }      if (-1 == function(ptrs, n))    {       perror("function() failed");       exit(EXIT_FAILURE);    }      /* Do stuff. */      /* Free memory where ptrs' are pointing to. */      return EXIT_SUCCESS;  }  

If VLAs are not supported change the beginning of main() to be:

#define N (5)    int main(void)     {    size_t n = N;    void * ptrs[N] = {0};      if (-1 == function(ptrs, n))    {      ...  

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