Help Needed!! Compiling C++ program - errors - c++

Question

This is a routine that I believe is for C. I copied it (legally) out a book and am trying to get it to compile and run in visual studio 2008. I would like to keep it as a C++ program. Lots of programming experience in IBM mainframe assembler but none in C++. Your help is greatly appreciated. I think just a couple of simple changes but I have read tutorials and beat on this for hours - getting nowhere. Getting a lot (4) of error C2440 '=' : cannot convert form 'void*' to to 'int*' errors in statements past the reedsolomon function: Thanks so much! Program follows:
#include <iostream>
using namespace std;
int wd[50] = {131,153,175,231,5,184,89,239,149,29,181,153,175,191,153,175,191,159,231,3,127,44,12,164,59,209,104,254,150,45};
int nd = 30, nc=20, i, j, k, *log, *alog, *c, gf=256, pp=301;
/* The following is routine which calculates the error correction codewords
for a given data codeword string of length "nd", stored as an integer array wd[].
The function ReedSolomon()first generates log and antilog tables for the Galois
Field of size "gf" (in the case of ECC 200, 28) with prime modulus "pp"
(in the case of ECC 200, 301), then uses them in the function prod(), first to
calculate coefficients of the generator polynomial of order "nc" and then to
calculate "nc" additional check codewords which are appended to the data in wd[].*/
/* "prod(x,y,log,alog,gf)" returns the product "x" times "y" */
int prod(int x, int y, int *log, int *alog, int gf)
{if (!x || !y)
return 0;
else
return alog[(log[x] + log[y]) % (gf-1)];
}
/* "ReedSolomon(wd,nd,nc,gf.pp)" takes "nd" data codeword values in wd[] */
/* and adds on "nc" check codewords, all within GF(gf) where "gf" is a */
/* power of 2 and "pp" is the value of its prime modulus polynomial */
void ReedSolomon(int *wd, int nd, int nc, int gf, int pp)
{int i, j, k, *log,*alog,*c;
/* allocate, then generate the log & antilog arrays: */
log = malloc(sizeof(int) * gf);
alog = malloc(sizeof(int) * gf);
log[0] = 1-gf; alog[0] = 1;
for (i = 1; i < gf; i++)
{alog[i] = alog[i-1] * 2;
if (alog[i] >= gf) alog[i] ^= pp;
log[alog[i]] = i;
}
/* allocate, then generate the generator polynomial coefficients: */
c = malloc(sizeof(int) * (nc+1));
for (i=1; i<=nc; i++) c[i] = 0; c[0] = 1;
for (i=1; i<=nc; i++)
{c[i] = c[i-1];
for (j=i-1; j>=1; j--)
{c[j] = c[j-1] ^ prod(c[j],alog[i],log,alog,gf);
}
c[0] = prod(c[0],alog[i],log,alog,gf);
}
/* clear, then generate "nc" checkwords in the array wd[] : */
for (i=nd; i<=(nd+nc); i++) wd[i] = 0;
for (i=0; i<nd; i++)
{k = wd[nd] ^ wd[i] ;
for (j=0; j<nc; j++)
{wd[nd+j] = wd[nd+j+1] ^ prod(k,c[nc-j-1],log, alog,gf);
}
}
free(c);
free(alog);
free(log);
return ();
}
int main ()
{reedsolomon (50,30,20,256,301);
for (i = 1; i < 51; i++)
{cout<< i; "="; wd[i];}
cout<<"HEY, you, I'm alive! Oh, and Hello World!\n";
cin.get();
return 1;
}

Answer 1

In C++, a void pointer can't be implicitly cast to a different pointer.
So instead of
int *pInt;
pInt = malloc(sizeof(int) * 5);
You need to say
int *pInt;
pInt = (int *) malloc(sizeof(int) * 5);
or preferably
int *pInt = new int[5];
(with a matching delete[] instead of free), or preferably preferably use a vector if it's intended to be dynamic.

Answer 2

At the beginning of the program type: #include <cstdlib> . If you do not include this library, then malloc will not work. In C++, void* to int* is not an automatic conversion, in lines: 31 32 and 40 you need to cast to int* e.g: log = (int *)malloc(sizeof(int) * gf);
At main funcion, line 63 you're calling the function as reedsolomon, it should be ReedSolomon, the way you declared it.
Also, in "void ReedSolomon(int *wd, int nd, int nc, int gf, int pp)" when you call the function in the main, you say ReedSolomon (50,30,20,256,301); so you are asigning an int value to a pointer to int, that's a type clash. I'm not sure what it is you want to do with wd.
Next time, please post the errors from the compiler so people dont have to compile the code themselves to check and see whats wrong.
Also a good technique which will save you a lot of time is to do a google search on the error the compiler gives you (it is very likely somebody already had that same mistake), and also read a C++ book to get acquainted with the language.
Cheers!

Answer 3

C++ requires that you cast the return value of malloc to whatever type of pointer you're assigning it to. So e.g. log = malloc(sizeof(int) * gf); needs to become log = (int *) malloc(sizeof(int) * gf);.

Answer 4

You should type cast when assigning a pointer to the return of malloc.
Example:
log = reinterpret_cast<int*>(malloc(sizeof(int) * gf));