将指针/数组从函数传递给main（）

Question

I'm learning functions/pointers, and having a bit of an issue. What I need to write is a C program with main() and two other functions.

Requirements:

1. read_funct() must allocate enough memory using malloc() to store the data.

2. Function prototype for read_funct must be:

    int read_funct(int *data_num, double **data_vals, char *filename) 

How it's meant to work:

1. main() calls the first function: read_funct()

2. read_num() reads binary data from a file. Two values have to be extracted: the no. of values (first 4 bytes), then the values themselves (8 bytes each, so contained in the next 8*no. of values). These correspond to data_num and data_vals . They have to be printed, the program then returns to main() .

3. main() performs operations to edit the data from the first file.

4. main() calls the second function: write_funct() , which writes the edited data into a new file.

Where I am:

The first function reads data_num correctly, and reads/prints data_vals . This is all working properly. However, I'm trying to print these in main() to verify that I'm performing operations on the correct data, but I can't get it working.

Note: I'm not trying to get it working with write_funct() at the moment, just taking it step-by-step.

Here's my current code for main() :

int read_funct(int *data_num, double **data_vals, char *filename);
int main()
{
  int data_num;
  double **data_vals;

  //Reads file using read_funct()
  read_funct(&data_num, data_vals, filename);

  //Check: print data_num
  printf("\nCheck No. of Values: %d\n", data_num);

  //Check: print data_vals
  for(int i = 0; i<data_num; i++)
  {
        printf("%0.3lf\t", data_vals[i]);
  }
  return(0);
}

Here's read_funct( )

int read_funct (int *data_num, double **data_vals, char *filename)
{
    FILE * fp = fopen(filename, "rb");                          //Opening file      
   //There's code here to check valid file open

   //There's code here to determine size and check valid length

  //Proceed with reading data_num if file is large enough
  char buffer_n[4];
  fread(buffer_n, 4, 1, fp);
  int res = buffer_n[0]|(buffer_n[1] << 8)|(buffer_n[2] << 16)|(buffer_n[3] << 24);     //Convert endian

  data_num = &res;          //Passes results back to data_num
  printf("Number of Data Values: %d \n", *data_num);    //Printing results

  //Proceeds with calculating data_vals
  data_vals = malloc(8*res);                //Allocating memory
  fread(data_vals, 8, res, fp); 

    //Prints data_vals
    for(int i=0; i<res; i++)
    {
        printf("%0.3f\t", data_vals[i]);
    }
    printf("\nEnd of File Read.\n\n");
    fclose(fp); 
    free(data_vals);                //Free allocated memory
    return(0);
}

Desired output:

Basically, I want it to print out the values from inside read_file() and then print a check in main() , so the output will be something like:

No. of values: 3            //From printf in read_file()
2 4 6

Check No. of values: 3      //From printf in main()
2 4 6

Where I think I'm going wrong:

1. Fairly sure that the main issue is that I've messed up my pointers and how I've initialised things in main() . I've been trying to fix this by myself, but I think I need some more experienced help to figure this out.

2. I know that every malloc() call must have a subsequent free() , but I'm worried that by doing so the way that I have, maybe I've made it so that I can't retrieve it in main() . Does it instead need to have an intermediate buffer to which memory is allocated instead?

Help to get this code working would be very greatly appreciated. Thank you!

Answer 1

Apart from freeing the data too soon, you have another problem here:

double **data_vals;
read_funct(&data_num, data_vals, filename);

If you want data_vals to be filled (written to, modified) by a function, you must pass its address, exactly as you do with data_num .

Here is another, slightly different, explanation. You see, you declare data_vals but you don't assign a value to it - it contains garbage. So it is a non-sense to pass data_vals to any function, or use it in any expression. It has a sense instead, to assign something to it, either via direct assignment or passing its address to a function, for the function to fill the variable.

Then, your usage of data_vals depicts a vector, or an array. So you really need to declare an array with [] , or may be a pointer (pointers and arrays are quite related/interchangeable in C). The logic of your main() function requires a pointer, not a pointer to pointer . Hence, this is appropriate:

double *data_vals;

The function which writes to your pointer variable, instead, needs the address of the variable to write to; in other words: a pointer to a pointer . This is why your function has this (correct) signature:

read_funct(..., double **data_vals, ...)

To understand easily, let see the other (simpler) thing you wrote correctly:

int data_num;
read_funct(&data_num, ...);  // declaration: read_funct(int *data_num, ...)

You declare data_num as integer in main(); you declare read_funct() with a formal parameter of pointer to integer , then you call read_funct() passing the address of your variable data_num . Perfect. Now, do the same with the other variable, data_vals . You declare it as pointer to double , pass its address to read_funct() using the notation &data_vals , and your function read_funct() declares that parameter as a pointer to pointer to double (and writes to it using *data_vals = ... . You can see the parallelism between the two variables, right?

May be I've been too pedantic, but your question was really clear and well formed, so I tried to do the same.

Answer 2

Yes, you are free-ing the buffer too soon. After you have freed it, there is not guarantee as to what it contains. You can free it at the end, in main.