Converting Hex To Ascii Memory Access - c

So I have a memory block that is fill with hex values, and I want to read in those hex values as Ascii. Unfortunately, I can't seem to figure out how to do this.
char* charstring = malloc(sizeof(uint16_t));
uint8_t test = 0x65;
sprintf(charstring, "%x ", test);
fprintf(Output, "%s", charstring);
free(charstring);
This just gives me a ton of "65", not the "e" I was expecting.
I know it's a simple question, but I've literally clicked every link on the front page of google and can't figure out my problem!

You get 65 because you are using %x as the format specifier, change it to %c for the character value.
Note that you don't have enough space for charstring for the null terminator. Fix it.

Related

error when printing special characters in C. The problem is %c vs %s

I'm starting with C, I have found a little difference between %c and %s, when must print special characters. I dont know if I'm doing something wrong, or it's a C limitation:
unsigned char str1[]="á";
printf("str1 c (%c)\n", str1[0]);
printf("str1 s (%s)\n", &str1[0]);
unsigned char str2[]="áéíóúñ";
printf("str2 s (%s)\n", str2);
And the output is:
str1 c ( )
str1 s (á)
str2 s (áéíóúñ)
In conclusion: when I try write special characters with %c, I cannot see it.
The %c format string for printf causes the corresponding argument to be converted to and interpreted as an unsigned char. An unsigned char is 1 byte long. One byte from your non-ASCII string does not necessarily correspond to anything you would recognize as a character.
It is likely that your editor, which you used to place some representation of those strings into your source code, encodes the two strings with some Unicode encoding scheme. This SO answer has some information to get you started on dealing with Unicode in C.
The reason things work fine with the %s formatting string is that printf will just start dumping out bytes until it hits the null byte terminator. Your output terminal is probably set to the same encoding scheme as your editor, so it's able to correctly interpret those bytes the way you intended.
Ok, now I understand it.
If I write:
unsigned char str3[]="a";
printf("%d\n", strlen((char *)str3));
The output is:
1
But I write:
unsigned char str1[]="á";
printf("%d\n", strlen((char *)str1));
The output is:
2
I understood that the character sizes can be 1, or 2 if are special characters , no?

Convert hash to 4 hex numbers

I would like to convert my MD5 hash to 4 hex numbers. What's wrong in my code?
//hash = 8ce4b16b22b58894aa86c421e8759df3
char *hash = argv[1];
unsigned int parts[4];
sscanf(&hash[0], "%x", &parts[0]);
sscanf(&hash[8], "%x", &parts[1]);
sscanf(&hash[16], "%x", &parts[2]);
sscanf(&hash[24], "%x", &parts[3]);
printf("Part[0]: %x\n", parts[0]);
printf("Part[1]: %x\n", parts[1]);
printf("Part[2]: %x\n", parts[2]);
printf("Part[3]: %x\n", parts[3]);
In your code, you wanted but forgot to limit the input for the HEX value to 8 (in characters). You need to use the field width specifier with the format specifier. You will need something like
sscanf(&hash[0], "%8x", &parts[0]);
sscanf(&hash[8], "%8x", &parts[1]);
sscanf(&hash[16], "%8x", &parts[2]);
sscanf(&hash[24], "%8x", &parts[3]);
In your code, you are trying to tell sscanf() where to start parsing the string (stored in the variable named "hash"), but you aren't telling sscanf() where it should stop parsing it. Try inserting whitespace characters into "hash" where you want the 4 numbers to be divided, first, and then your code may work. Also, I would write "&(hash[0])" instead of "&hash[0]" because I don't want to clutter my memory banks with trivia such as the operator precedence rules for C - easier just to use parentheses to force what I want.

why i have exc_bad_access doing cast?

i am programming in C and i have a problem while casting an int into a char.
I am using my mac with Xcode to program c.
The code is:
int main(){
int t = 2;
printf("test %s\n", (char)t); //EXC_BAD_ACCESS
return 0;
}
I tried all I found in many post, I really don't know what is going on... any suggestion?
Please include the goal of your code in your question not in a comment below.
If you want the output
test 2
your have to change %s to %d
printf("test %d\n", t);
I guess you got the wrong idea about the %s. It does not tell the printf that you want to have the int as string, it does tell printf that the parameter is a string! It is obviously not, so you got the exception.
if you use %c you tell the printf function that you want to output your number as character from your current ASCII table. For example 65 is 'A'.
If you have a concatenation situation like
strcpy(str_buscar, "controlID='");
strcat(str_buscar, (char) t);
strcat(str_buscar, "'");
you need itoa instead of the cast:
strcat(str_buscar, (char) t);
you need the follow:
char buffer[32]; // enough space for a "number as string"
itoa(t,buffer,10);
strcat(str_buscar, buffer);
a (IMHO) shortcut is to "print" to a buffer with sprintf
sprintf(str_buscar,"controlID='%d'",t);
instead of printing to a console sprintf prints into the given buffer. Make shure that your buffer str_buscar is big enough.
The %s format specifier represents a string, not an individual character. Printf thinks that the number 2 you're passing it is a string's address. It tries to access the memory address 2 and fails, because that address doesn't exist.
If you want to print a character, you'll want the %c specifier. This tells printf to print the character whose ASCII code is 2. The ASCII character number 2 is, according to the ASCII table, a control character that cannot be printed, which is why you're getting strange output.
If you actually want to print the character '2' (which has a different code, 50), you will want to use something like:
printf("test: %c", (char)('0' + c));
This example leverages the fact that all ASCII characters have consecutive codes, starting with 48 ('0'). This way, if you wanted to print the digit 0, you'd end up printing the '0' character (ASCII code 48 = 48 + 0). If you want to print the digit 2, you'll end up printing the '2' character (50 = 48 + 2).
This way, however, is a bit clunky and fails when encountering numbers larger than 9 (i.e. it only works with digits). The easier way consists of no longer working with characters at all and, instead, using the '%d' specifier (used for printing whole number):
int t = 0;
printf("test: %d", t);

How to create array with hex chars from array of numbers

I have array of chars:
char macChars=[12];
The content of it is e.g. macChars[0]=53, macChars[1]=66 ...
I need to convert these numbers to hex chars, so i would have another array:
macCharsHex[0]=5 //value 53 in hex
macCharsHex[1]=B //value 66 in hex
Thank you.
Assuming ASCII, your example already does contain the values you want them to contain. So you don't have to convert anything. Maybe you want to print them?
This should work:
char hex[255] = {0}; // Varible to hold the hex value
int dec = 1234; // Decimal number to be converted
sprintf(hex,"%X", dec);
printf("%s", hex); // Print hex value
Use sprintf(), for instance. Note that it will take more space, if you have 12 bytes you will need 24 + 1 bytes for the string representation, since each byte requires two characters in hex and then the terminating '\0'-byte.
I suspect that I don't understand the question at all, especially not the example given.
If you have macChars[0] == 53, which is 0x35 in hex, then I would expect to get maxCharsHex[0] == '3' and macCharsHex[1] == '5' after the first char has been converted. This is done like so with sprintf():
sprintf(maxCharsHex, "%02x", (int) macChars[0] & 0xff);
The cast and mask is to be on the safe side for signed characters.
They are already converted, since character in C are represented by their corresponding character codes.
Therefore, as far as storing things in arrays is concerned there is nothing you need to do and if you want "5" and "B" to show up correctly when printing or doing something like that its a matter of using the correct printing function (putchar, printf with %c, etc).

memcpy() to copy integer value to char buffer

I am trying to copy the memory value of int into the char buffer. The code looks like below,
#define CPYINT(a, b) memcpy(a, &b, 4)
............
char str1[4];
int i = 1;
CPYINT(str1, i);
printf("%s",s);
...........
When I print str1 it’s blank. Please clarify.
You are copying the byte representation of an integer into a char array. You then ask printf to interpret this array as a null terminating string : str1[0] being zero, you are essentially printing an empty string (I'm skipping the endianness talk here).
What did you expect ? Obviously, if you wanted to print a textual representation of the integer i, you should use printf("%d", i).
try
printf("%02X %02X %02X %02X\n", str1[0], str1[1], str1[2], str1[3]);
instead.
The binary representation of the integer 1, probably contains leading NULs, and so your current printf statement terminates earlier than you want.
What is your intention here? Right now you are putting arbitrary byte values into the char array, but then interpreting them as a string, as it happens the first byte is probably a zero (null) and hence your print nothing, but in all probability many of the characters will be unprintable, so printf is the wrong tool to use to check if the copy worked.
So, either: loop through the array and print the numeric value of each byte, %0xd might be useful for that or if your intention is actually to create a string representation of the int then you'll need a larger buffer, and space for a null terminator.
Maybe you need convert intger to char* in that way tou can use itoa function
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