dynamically defined array size in a function

This is a discussion on dynamically defined array size in a function within the C Programming forums, part of the General Programming Boards category; I'm writing a function to read in a number of words from a binary file. the size of each word ...

  1. #1
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    dynamically defined array size in a function

    I'm writing a function to read in a number of words from a binary file. the size of each word and the number of words to be read are defined by the function input parameters. I cannot allocate the size of the array to contain all th words properly.

    Code:
    void ReadSW01HEAD(FILE *ptr, long n /*Number of Parameters*/, long Size /*Section Size*/)
    {
    	char Param[n];
    	int FieldLength = Size/n; 
    	
    	printf("Parameters Present\n\n");
    
    	fread(&Param,sizeof(char),FieldLength,ptr);
    	printf("%s   ",Param);
    }
    Do I need a pointer of some sort?

  2. #2
    Gawking at stupidity
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    When asking a question, try to say what the program is doing rather than what it it's not doing.

    Bad:
    I cannot allocate the size of the array to contain all th words properly.
    Good:
    It gives me the error message "<error message here>" when I try to allocate the size of the array to contain all the words properly.
    If you understand what you're doing, you're not learning anything.

  3. #3
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    The error message that I get is:
    Code:
    error C2133: 'Param' : unknown size

  4. #4
    Gawking at stupidity
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    It looks like you're using a compiler that doesn't support C99. You can get around the limitation by using malloc() to get memory for Param:
    Code:
    char *Param = malloc(n);
    Don't forget to free(Param); when you're doing using that memory.
    If you understand what you're doing, you're not learning anything.

  5. #5
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    Code:
    void ReadSW01HEAD(FILE *ptr, long n /*Number of Parameters*/, long Size /*Section Size*/)
    {
    	char *Param = malloc(n);
    	int FieldLength = Size/n; 
    	
    	printf("Parameters Present\n\n");
    
    	fread(&Param,sizeof(char),FieldLength,ptr);
    	printf("%s   ",Param);
    }
    Gives me an error message like this:

    Code:
    error C2440: 'initializing' : cannot convert from 'void *' to 'char *'
            Conversion from 'void*' to pointer to non-'void' requires an explicit cast
    Error executing cl.exe.
    Why is it asking for a cast?

  6. #6
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    Ignore last post, this corrected code seems to do the trick.

    Code:
    void ReadSW01HEAD(FILE *ptr, long n /*Number of Parameters*/, long Size /*Section Size*/)
    {
    	char *Param;
    	int FieldLength = Size/n; 
    
    	Param = (char *) calloc(n, FieldLength);
    	
    	printf("Parameters Present\n\n");
    
    	fread(&Param,sizeof(char),FieldLength,ptr);
    	printf("%s   ",Param);
    }

  7. #7
    Gawking at stupidity
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    It looks like you're using a C++ compiler to compile C code. The "broken" malloc() code was actually the correct one. Your compiler is just wrong.

    But you don't want to use fread(&Parm,...). fread(Param,...) is correct.
    If you understand what you're doing, you're not learning anything.

  8. #8
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    Another problem with the calloc:

    Once I try to write to the array elements, an error message pops up saying that the <memory could not be"read">. Is the syntax incorrect?

  9. #9
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    Can I use calloc inside a structure, ie:

    Code:
    struct ParamName    // An array of char for one param name
       {
    	   char *Name;
    	   Name = (char *)calloc(NameLength, sizeof(char));
    	   
       };
    and then

    Code:
    ParamName Params[NumParam];
    Right now it's giving these errors:

    Code:
    illegal pure syntax, must be '= 0'
    H:\Code\Complete File Parcel\Main Fn.cpp(117) : error C2501: 'Name' : missing storage-class or type specifiers
    H:\Code\Complete File Parcel\Main Fn.cpp(122) : error C2057: expected constant expression
    H:\Code\Complete File Parcel\Main Fn.cpp(122) : error C2466: cannot allocate an array of constant size 0
    H:\Code\Complete File Parcel\Main Fn.cpp(122) : error C2133: 'Params' : unknown size

  10. #10
    Gawking at stupidity
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    No, you can't. And you're still compiling your C program as C++. If you're going to be programming in C, compile it as C. C and C++ are not the same language.
    If you understand what you're doing, you're not learning anything.

  11. #11
    Frequently Quite Prolix dwks's Avatar
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    Just save to file as a .c file and it should compile as a C program.

  12. #12
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    Is there any way to dynamically allocate an array inside a structure.

    Code:
     struct ParamName    // An array of char for one param name
       {
    	   char Name[8];
       };
    If say [8] is not always 8 and it is a variable.

  13. #13
    Just Lurking Dave_Sinkula's Avatar
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    Code:
    #include <stdio.h>
    #include <stdlib.h>
    #include <string.h>
    
    struct ParamName
    {
       char *Name;
    };
    
    int main(void)
    {
       struct ParamName param;
       param.Name = malloc(8);
       if ( param.Name )
       {
          strcpy(param.Name, "Hello");
          puts(param.Name);
          free(param.Name);
       }
       return 0;
    }
    
    /* my output
    Hello
    */
    7. It is easier to write an incorrect program than understand a correct one.
    40. There are two ways to write error-free programs; only the third one works.*

  14. #14
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    I've gottna so confused with the proper variable declarations. here's the complete chunk of the program with the array structure in it. It gives a run-time error : memory could not be 'written'

    Code:
    struct ParamName    // An array of char for one param name
       {
    	   char *NameP;  
       };
                            // Allocate the array of size nameLengthP
       ParamName SizedParameterName;
       SizedParameterName.NameP = (char *)calloc(NameLengthP, sizeof (char));
    
       int j;
    
       ParamName *Params;
       Params = (ParamName *) calloc(NumParam, NameLengthP); // An array of names for all of the param's
    
       for(i=0;i<NumParam;i++)
       {
    	   for(j=0;j<NameLengthP;j++)
    	  	   fread(&Params[i].NameP[j],sizeof (char),1,inp);
    		   
       }

  15. #15
    ATH0 quzah's Avatar
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    Stop typecasting malloc / calloc. Read the FAQ is you want to know why it's wrong, I'm getting tired of explaining to everyone.

    Next make a nice little test program to get what you're doing working:
    Code:
    #include <stdio.h>
    #include <stdlib.h>
    #include <string.h>
    #define MAGICNUMBER 5
    
    struct foo
    {
        char *bar;
    };
    
    int main( void )
    {
        struct foo *array;
        int x;
    
        array = malloc( MAGICNUMBER * sizeof( struct foo ) );
        if( array == NULL )
        {
            printf( "malloc failure\n" );
            return 0;
        }
    
    
        for( x = 0; x < MAGICNUMBER; x++ )
        {
            array[ x ].bar = malloc( strlen( "Hello World!" ) + 1 );
            strcpy( array[ x ].bar, "Hello World!" );
        }
    
        for( x = 0; x < MAGICNUMBER; x++ )
        {
            printf("array[ %d ].bar = %s\n", x, array[ x ].bar );
            free( array[ x ].bar );
        }
        free( array );
    
        return 0;
    }
    The main problem I think you're having is that you keep using the & operator when you shouldn't be.


    Quzah.
    Hope is the first step on the road to disappointment.

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