Problem with memory access error in C with char*

Question

I can run a c programm without problems on my Linux Mint system on a other Linux system I get a memory access error.

void digest_message(const unsigned char *message, size_t message_len, unsigned char **digest, unsigned int *digest_len)
{
    EVP_MD_CTX *mdctx;

    if((mdctx = EVP_MD_CTX_create()) == NULL)
        handleErrors();
    if(1 != EVP_DigestInit_ex(mdctx, EVP_sha1(), NULL))
        handleErrors();
    if(1 != EVP_DigestUpdate(mdctx, message, message_len))
        handleErrors();
    if((*digest = (unsigned char *)OPENSSL_malloc(EVP_MD_size(EVP_sha1()))) == NULL)
        handleErrors();
    if(1 != EVP_DigestFinal_ex(mdctx, *digest, digest_len))
        handleErrors();

    EVP_MD_CTX_destroy(mdctx);
}

void main ()

{
...
    const unsigned char *message= (const unsigned char*) decryptedtext;
    size_t mlen=x;
    unsigned char *digest;
    unsigned int dlen;
    digest_message(message,mlen,&digest,&dlen);

    // if i printf in a for loop i get this memmory access error 
    for(int i=0;i<dlen;i++)
        printf("%x ",digest[i]);
    }

I tried it with:

printf("%x ",digest[0]);
printf("%x ",digest[1]);
printf("%x ",digest[2]);
...

and got no error

Why does this happen? How can I change it?

Thanks in advance

Answer 1

I can spot the following (potential) problems.

---

I can run a c programm without problems on my Linux Mint

I think that it is just a happy coincidence, that you do not see the problems.

---

const unsigned char *message= (const unsigned char*) decryptedtext;

If decryptedtext is just a pointer, then accessing it further using message is quite a bad idea. Normally, you should have memory allocated, in order to access it.

---

size_t mlen=x;

Ideally, mlen should be related to the length of message. If mlen is bigger then the real length of message, you are in trouble (buffer overflow, memory corruption...).

---

unsigned int dlen;

How do you make sure that dlen is not bigger than the number of entries available in digest?

---

printf("%x ",digest[i]);

Why do you want to printf a string (digest) as a hexadecimal number?

---------------------

if(decryptedtext[0]==0x25 &&   //%
    decryptedtext[1]==0x50 &&  //P
    decryptedtext[2]==0x44 &&  //D
    decryptedtext[3]==0x46)    //F
{ printf("%s",decryptedtext);
  break;
}

This is ugly. decryptedtext is not a null-terminated string, therefore printf() will not print only 4 characters.

---

int x;
f = fopen("test.pdf", "w+b");
if(f == NULL) {
printf("Datei konnte nicht geoeffnet werden.\n");
}else {

while(x<decryptedtext_len)  {
    putc(decryptedtext[x],f);
    x++;
}

x is not initialized before being used.

---

unsigned char hasharrayone[20];
unsigned char hasharraytwo[20];
unsigned char hasharraythree[20];

You could have used:

unsigned char hasharray[3][20];

---

int xc;
file = fopen("sxxxxx-dest-cipher.bin", "w+b");
if(file == NULL) {
printf("Datei konnte nicht geoeffnet werden.\n");
}else {
while(xc<cpl)  {
    putc(cfbciphertext[xc],file);
    xc++;
}

xc is not initialized.

Answer 2

You have an undefined behavior using printf("%x ", digest[i]).

As mentionned in this answer using the wrong format specifier results in an undefined behavior. This is probably why the behavior of your program depends on the OS you are running on (it may also depend of your hardware and compiler).

As I don't have the error message, it's hard to know what happens in your case, but as "%x" format specifier usually works with unsigned int it should reads 4bytes in memory. As disgest[i] is a unsigned char, it is written on 1byte only and this is probably why you get a memory access error.

Converting your unsigned char variable digest[i] to an unsigned int before using printf should resolve your issue.

Also, this topic should help you printing hexadecimal characters