Display the first characters of an unsigned char*

Question

I have the following code (I stripped down the useless parts):

unsigned char* decrypted= (unsigned char *) malloc(500);
bufSize = operations.RSADecrypt(newEncrypted, bufSize, key, decrypted);
printf("Test: %s", decrypted);

And I would like to display only the bufSize first characters of decrypted because actually it displays a lot of nonsense characters!

Answer 1

You can use the "%.*s" format specifier:

printf("Test: %.*s", bufSize, decrypted);

which instructs printf() to write the first bufSize characters from decrypted.

Answer 2

You can limit the length with the format specifier:

printf ("Test: %-20.20s", decrypted);

For a version using a variable bufSize:

printf ("Test: %-*.*s", bufSize, bufSize, decrypted);

Note that this forces the length to exactly that many characters, padded with spaces on the right if need be. If you want a shorter string to be shorter in the output (irrelebant in your case if the string is, as indicated, always longer than what you want output), you can use:

printf ("Test: %.*s", bufSize, decrypted);

Answer 3

If you are 'allowed' to modify the decrypted string. You can simply add a terminator to it:

decrypted[bufSize] = 0;

So printf() will only print the buffer contents.

If you are not allowed to add a custom char to the decrypted buffer you need to copy the contents to a temporary buffer and use that buffer in your printf():

unsigned char* tmp = (unsigned char *) malloc(bufSize + 1);
strncpy(tmp, decrypted, bufSize);
tmp[bufSize] = 0;

Answer 4

I don't like that you said the pointer contained nonsense. Its not nonsense, its residual memory. There is a good chance you expect and want this area to be set to zero. Try the following, where calloc sets the malloc bits to zero.

unsigned char* decrypted= (unsigned char *) calloc(500,sizeof(char));
bufSize = operations.RSADecrypt(newEncrypted, bufSize, key, decrypted);
printf("Test: %s", decrypted);