XOR two hexadecimal strings(unmasking) in C - c

I wanted to unmask a hex string using the following algorithm in C
j = i MOD 4 transformed-octet-i = original-octet-i XOR masking-key-octet-j
#include<stdlib.h>
#include<string.h>
int main()
{
char masked[]="951bfdcdc113ebca921fe9dc";
char masking_key[]="e17e8eb9";
char *unmasked;
int length=strlen(masked);
unmasked=malloc(sizeof(char)*(length+1));
int i=0;
for(i=0;i<length;i++)
{
unmasked[i]=masked[i]^masking_key[i%4];
}
printf("%s\n",unmasked);
return 0;
}
The output I am getting is \UT instead of 74657374206d657373616765. It would be really helpful if someone could help me fix the error here.

You need to convert from string format to raw integers before doing the XOR, then convert back before printing it as a string. Otherwise you'll XOR the symbol values, not the raw values.
You can convert the whole string in one go with strtol(data, 0, 16). But one of the common details that makes programmers different from the code monkeys is the ability to code trivial string-integer conversions without the help of library functions. So here is a a simplified code doing just that - be aware of the complete lack of error handling, as this is just quick & dirty code:
#include <stdio.h>
#include <stdlib.h>
char hexlify (char nibble)
{
return "0123456789abcdef"[nibble];
}
char unhexlify (char ch)
{
if(ch>='0' && ch<='9')
return ch - '0';
if(ch>='a' && ch<='f')
return ch - 'a' + 0xA;
return 0;
}
int main (void)
{
char masked[]="951bfdcdc113ebca921fe9dc";
char masking_key[]="e17e8eb9";
char *unmasked;
size_t length = sizeof masked - 1;
unmasked = malloc(length + 1);
for(size_t i=0;i<length;i++)
{
char op1 = unhexlify(masked[i]);
char op2 = unhexlify(masking_key[i%4]);
unmasked[i]= hexlify(op1 ^ op2);
}
unmasked[length]='\0';
printf("%s\n",unmasked);
free(unmasked);
return 0;
}
Output:
74651cb3206d0ab4736108a2

What you're doing here is not xoring the hexadecimal bytes, but xoring the characters representing them.
While the ideal process would be (0x95 ^ 0xe1) + (0xbf ^ 7e) + ..., what you're doing is xoring the ascii values of each character, meaning ('9' ^ 'b') + ('5' ^ '1') + ....
What you need to do is first transform the hexadecimal string to the bytes themselves (bytes.fromhex("951bfdcdc113ebca921fe9dc") in python), and only then XOR them. The way I would do that is with sscanf("%2x", ...).
For example:
#include <stdio.h>
int main()
{
unsigned char a = 0;
scanf("%2hhx", &a);
}

Related

Converting int to char in C

Right now I am trying to convert an int to a char in C programming. After doing research, I found that I should be able to do it like this:
int value = 10;
char result = (char) value;
What I would like is for this to return 'A' (and for 0-9 to return '0'-'9') but this returns a new line character I think.
My whole function looks like this:
char int2char (int radix, int value) {
if (value < 0 || value >= radix) {
return '?';
}
char result = (char) value;
return result;
}
to convert int to char you do not have to do anything
char x;
int y;
/* do something */
x = y;
only one int to char value as the printable (usually ASCII) digit like in your example:
const char digits[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
int inttochar(int val, int base)
{
return digits[val % base];
}
if you want to convert to the string (char *) then you need to use any of the stansdard functions like sprintf, itoa, ltoa, utoa, ultoa .... or write one yourself:
char *reverse(char *str);
const char digits[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
char *convert(int number, char *buff, int base)
{
char *result = (buff == NULL || base > strlen(digits) || base < 2) ? NULL : buff;
char sign = 0;
if (number < 0)
{
sign = '-';
}
if (result != NULL)
{
do
{
*buff++ = digits[abs(number % (base ))];
number /= base;
} while (number);
if(sign) *buff++ = sign;
if (!*result) *buff++ = '0';
*buff = 0;
reverse(result);
}
return result;
}
A portable way of doing this would be to define a
const char* foo = "0123456789ABC...";
where ... are the rest of the characters that you want to consider.
Then and foo[value] will evaluate to a particular char. For example foo[0] will be '0', and foo[10] will be 'A'.
If you assume a particular encoding (such as the common but by no means ubiquitous ASCII) then your code is not strictly portable.
Characters use an encoding (typically ASCII) to map numbers to a particular character. The codes for the characters '0' to '9' are consecutive, so for values less than 10 you add the value to the character constant '0'. For values 10 or more, you add the value minus 10 to the character constant 'A':
char result;
if (value >= 10) {
result = 'A' + value - 10;
} else {
result = '0' + value;
}
Converting Int to Char
I take it that OP wants more that just a 1 digit conversion as radix was supplied.
To convert an int into a string, (not just 1 char) there is the sprintf(buf, "%d", value) approach.
To do so to any radix, string management becomes an issue as well as dealing the corner case of INT_MIN
The following C99 solution returns a char* whose lifetime is valid to the end of the block. It does so by providing a compound literal via the macro.
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
// Maximum buffer size needed
#define ITOA_BASE_N (sizeof(unsigned)*CHAR_BIT + 2)
char *itoa_base(char *s, int x, int base) {
s += ITOA_BASE_N - 1;
*s = '\0';
if (base >= 2 && base <= 36) {
int x0 = x;
do {
*(--s) = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"[abs(x % base)];
x /= base;
} while (x);
if (x0 < 0) {
*(--s) = '-';
}
}
return s;
}
#define TO_BASE(x,b) itoa_base((char [ITOA_BASE_N]){0} , (x), (b))
Sample usage and tests
void test(int x) {
printf("base10:% 11d base2:%35s base36:%7s ", x, TO_BASE(x, 2), TO_BASE(x, 36));
printf("%ld\n", strtol(TO_BASE(x, 36), NULL, 36));
}
int main(void) {
test(0);
test(-1);
test(42);
test(INT_MAX);
test(-INT_MAX);
test(INT_MIN);
}
Output
base10: 0 base2: 0 base36: 0 0
base10: -1 base2: -1 base36: -1 -1
base10: 42 base2: 101010 base36: 16 42
base10: 2147483647 base2: 1111111111111111111111111111111 base36: ZIK0ZJ 2147483647
base10:-2147483647 base2: -1111111111111111111111111111111 base36:-ZIK0ZJ -2147483647
base10:-2147483648 base2: -10000000000000000000000000000000 base36:-ZIK0ZK -2147483648
Ref How to use compound literals to fprintf() multiple formatted numbers with arbitrary bases?
Check out the ascii table
The values stored in a char are interpreted as the characters corresponding to that table. The value of 10 is a newline
So characters in C are based on ASCII (or UTF-8 which is backwards-compatible with ascii codes). This means that under the hood, "A" is actually the number "65" (except in binary rather than decimal). All a "char" is in C is an integer with enough bytes to represent every ASCII character. If you want to convert an int to a char, you'll need to instruct the computer to interpret the bytes of an int as ASCII values - and it's been a while since I've done C, but I believe the compiler will complain since char holds fewer bytes than int. This means we need a function, as you've written. Thus,
if(value < 10) return '0'+value;
return 'A'+value-10;
will be what you want to return from your function. Keep your bounds checks with "radix" as you've done, imho that is good practice in C.
1. Converting int to char by type casting
Source File charConvertByCasting.c
#include <stdio.h>
int main(){
int i = 66; // ~~Type Casting Syntax~~
printf("%c", (char) i); // (type_name) expression
return 0;
}
Executable charConvertByCasting.exe command line output:
C:\Users\boqsc\Desktop\tcc>tcc -run charconvert.c
B
Additional resources:
https://www.tutorialspoint.com/cprogramming/c_type_casting.htm
https://www.tutorialspoint.com/cprogramming/c_data_types.htm
2. Convert int to char by assignment
Source File charConvertByAssignment.c
#include <stdio.h>
int main(){
int i = 66;
char c = i;
printf("%c", c);
return 0;
}
Executable charConvertByAssignment.exe command line output:
C:\Users\boqsc\Desktop\tcc>tcc -run charconvert.c
B
You can do
char a;
a = '0' + 5;
You will get character representation of that number.
Borrowing the idea from the existing answers, i.e. making use of array index.
Here is a "just works" simple demo for "integer to char[]" conversion in base 10, without any of <stdio.h>'s printf family interfaces.
Test:
$ cc -o testint2str testint2str.c && ./testint2str
Result: 234789
Code:
#include <stdio.h>
#include <string.h>
static char digits[] = "0123456789";
void int2str (char *buf, size_t sz, int num);
/*
Test:
cc -o testint2str testint2str.c && ./testint2str
*/
int
main ()
{
int num = 234789;
char buf[1024] = { 0 };
int2str (buf, sizeof buf, num);
printf ("Result: %s\n", buf);
}
void
int2str (char *buf, size_t sz, int num)
{
/*
Convert integer type to char*, in base-10 form.
*/
char *bufp = buf;
int i = 0;
// NOTE-1
void __reverse (char *__buf, int __start, int __end)
{
char __bufclone[__end - __start];
int i = 0;
int __nchars = sizeof __bufclone;
for (i = 0; i < __nchars; i++)
{
__bufclone[i] = __buf[__end - 1 - i];
}
memmove (__buf, __bufclone, __nchars);
}
while (num > 0)
{
bufp[i++] = digits[num % 10]; // NOTE-2
num /= 10;
}
__reverse (buf, 0, i);
// NOTE-3
bufp[i] = '\0';
}
// NOTE-1:
// "Nested function" is GNU's C Extension. Put it outside if not
// compiled by GCC.
// NOTE-2:
// 10 can be replaced by any radix, like 16 for hexidecimal outputs.
//
// NOTE-3:
// Make sure inserting trailing "null-terminator" after all things
// done.
NOTE-1:
"Nested function" is GNU's C Extension. Put it outside if not
compiled by GCC.
NOTE-2:
10 can be replaced by any radix, like 16 for hexidecimal outputs.
NOTE-3:
Make sure inserting trailing "null-terminator" after all things
done.

How to pack a hexadecimal value in an unsigned char variable in a C program?

I have a hexadecimal value "F69CF355B6231FDBD91EB1E22B61EA1F" in a string and I am using this value in my program by hardcoding the value in an unsigned char variable like this:
unsigned char a[] = { 0xF6 ,0x9C ,0xF3 ,0x55 ,0xB6 ,0x23 ,0x1F ,0xDB ,0xD9 ,0x1E ,0xB1 ,0xE2 ,0x2B ,0x61 ,0xEA ,0x1F};
Is there any function or any other method by which I can take the value from a string and put it into an unsigned variable in the hexadecimal format by packing it?
#include <stdio.h>
#include <ctype.h>
int hctoi(const char h){
if(isdigit(h))
return h - '0';
else
return toupper(h) - 'A' + 10;
}
int main(void){
const char cdata[]="F69CF355B6231FDBD91EB1E22B61EA1F";
unsigned char udata[(sizeof(cdata)-1)/2];
const char *p;
unsigned char *up;
for(p=cdata,up=udata;*p;p+=2,++up){
*up = hctoi(p[0])*16 + hctoi(p[1]);
}
{ //check code
int i;
for(i=0;i<sizeof(udata);++i)
printf("%02X", udata[i]);
}
return 0;
}
You can translate a hexadecimal value in a string into a value using sscanf. If you want an array of values, then you could write a function to split up the input string into two character segments and use sscanf to convert each piece. (I haven't done C for an eternity so I don't know if that's a good way to do it.)
If it's for 32 single hex-values (16 bytes, 128bit) only then you might take look at the methods provided by libuuid.
libuuid is part of the e2fsprogs package. Anyhow some linux distros, Debian for example, ship libuuid as a separate package. To use the Debian package for your developement you also need to look here.
Check this answer for doing this stuff in c++ using sscanf().
For c, it would be something like this:
char *str = "F69CF355B6231FDBD91EB1E22B61EA1F";
char substr[3] = "__";
unsigned char *a = NULL;
len = strlen(str);
a = malloc(sizeof(unsigned char)*(len/2)+1);
for ( i = 0; i < len/2; i++) {
substr[0] = str[i*2];
substr[1] = str[i*2 + 1];
sscanf( substr, "%hx", &a[i] );
}
free(a);
Introduce auxiliary functions data_length and data_get to easily iterate over your data. The following program dumps unpacked unsigned chars on stdout, one per line:
#include <stdlib.h>
#include <stdio.h>
#include <ctype.h>
#include <string.h>
/* htoi(H)
Return the value associated to the hexadecimal digit H. */
int
htoi(char h)
{
int a = -1;
if(isdigit(h))
{
a = h - '0';
}
else
{
a = toupper(h) - 'A' + 10;
}
return a;
}
/* data_length(D)
The length of the data stored at D. */
int
data_length(const char* d)
{
return strlen(d) / 2;
}
/* data_get(D, K)
Return the K-th unsigned char located encoded in d. */
unsigned char
data_get(const char *d, int k)
{
return htoi(d[2*k]) * 0x10 +
htoi((d+1)[2*k]);
}
int
main()
{
const char cdata[]="F69CF355B6231FDBD91EB1E22B61EA1F";
for(int i = 0; i < data_length(cdata); ++i)
{
printf("0x%02hhx\n", data_get(cdata, i));
}
return EXIT_SUCCESS;
}

C Library function for converting a string of hex digits to ints?

I have a variable length string where each character represents a hex digit. I could iterate through the characters and use a case statement to convert it to hex but I feel like there has to be a standard library function that will handle this. Is there any such thing?
Example of what I want to do. "17bf59c" -> int intarray[7] = { 1, 7, 0xb, 0xf, 5, 9, 0xc}
No, there's no such function, probably because (and now I'm guessing, I'm not a C standard library architect by a long stretch) it's something that's quite easy to put together from existing functions. Here's one way of doing it decently:
int * string_to_int_array(const char *string, size_t length)
{
int *out = malloc(length * sizeof *out);
if(out != NULL)
{
size_t i;
for(i = 0; i < length; i++)
{
const char here = tolower(string[i]);
out[i] = (here <= '9') ? (here - '\0') : (10 + (here - 'a'));
}
}
return out;
}
Note: the above is untested.
Also note things that maybe aren't obvious, but still subtly important (in my opinion):
Use const for pointer arguments that are treated as "read only" by the function.
Don't repeat the type that out is pointing at, use sizeof *out.
Don't cast the return value of malloc() in C.
Check that malloc() succeeded before using the memory.
Don't hard-code ASCII values, use character constants.
The above still assumes an encoding where 'a'..'f' are contigous, and would likely break on e.g. EBCDIC. You get what you pay for, sometimes. :)
using strtol
void to_int_array (int *dst, const char *hexs)
{
char buf[2] = {0};
char c;
while ((c = *hexs++)) {
buf[0] = c;
*dst++ = strtol(buf,NULL,16);
}
}
Here's another version that allows you to pass in the output array. Most of the time, you don't need to malloc, and that's expensive. A stack variable is typically fine, and you know the output is never going to be bigger than your input. You can still pass in an allocated array, if it's too big, or you need to pass it back up.
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
/* str of length len is parsed to individual ints into output
* length of output needs to be at least len.
* returns number of parsed elements. Maybe shorter if there
* are invalid characters in str.
*/
int string_to_array(const char *str, int *output)
{
int *out = output;
for (; *str; str++) {
if (isxdigit(*str & 0xff)) {
char ch = tolower(*str & 0xff);
*out++ = (ch >= 'a' && ch <= 'z') ? ch - 'a' + 10 : ch - '0';
}
}
return out - output;
}
int main(void)
{
int values[10];
int len = string_to_array("17bzzf59c", values);
int i = 0;
for (i = 0; i < len; i++)
printf("%x ", values[i]);
printf("\n");
return EXIT_SUCCESS;
}
#include <stdio.h>
int main(){
char data[] = "17bf59c";
const int len = sizeof(data)/sizeof(char)-1;
int i,value[sizeof(data)/sizeof(char)-1];
for(i=0;i<len;++i)
sscanf(data+i, "%1x",value + i);
for(i=0;i<len;++i)
printf("0x%x\n", value[i]);
return 0;
}

Grab all integers from irregular strings in C

I am looking for a (relatively) simple way to parse a random string and extract all of the integers from it and put them into an Array - this differs from some of the other questions which are similar because my strings have no standard format.
Example:
pt112parah salin10n m5:isstupid::42$%&%^*%7first3
I would need to eventually get an array with these contents:
112 10 5 42 7 3
And I would like a method more efficient then going character by character through a string.
Thanks for your help
A quick solution. I'm assuming that there are no numbers that exceed the range of long, and that there are no minus signs to worry about. If those are problems, then you need to do a lot more work analyzing the results of strtol() and you need to detect '-' followed by a digit.
The code does loop over all characters; I don't think you can avoid that. But it does use strtol() to process each sequence of digits (once the first digit is found), and resumes where strtol() left off (and strtol() is kind enough to tell us exactly where it stopped its conversion).
#include <stdlib.h>
#include <stdio.h>
#include <ctype.h>
int main(void)
{
const char data[] = "pt112parah salin10n m5:isstupid::42$%&%^*%7first3";
long results[100];
int nresult = 0;
const char *s = data;
char c;
while ((c = *s++) != '\0')
{
if (isdigit(c))
{
char *end;
results[nresult++] = strtol(s-1, &end, 10);
s = end;
}
}
for (int i = 0; i < nresult; i++)
printf("%d: %ld\n", i, results[i]);
return 0;
}
Output:
0: 112
1: 10
2: 5
3: 42
4: 7
5: 3
More efficient than going through character by character?
Not possible, because you must look at every character to know that it is not an integer.
Now, given that you have to go though the string character by character, I would recommend simply casting each character as an int and checking that:
//string tmp = ""; declared outside of loop.
//pseudocode for inner loop:
int intVal = (int)c;
if(intVal >=48 && intVal <= 57){ //0-9 are 48-57 when char casted to int.
tmp += c;
}
else if(tmp.length > 0){
array[?] = (int)tmp; // ? is where to add the int to the array.
tmp = "";
}
array will contain your solution.
Just because I've been writing Python all day and I want a break. Declaring an array will be tricky. Either you have to run it twice to work out how many numbers you have (and then allocate the array) or just use the numbers one by one as in this example.
NB the ASCII characters for '0' to '9' are 48 to 57 (i.e. consecutive).
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
int main(int argc, char **argv)
{
char *input = "pt112par0ah salin10n m5:isstupid::42$%&%^*%7first3";
int length = strlen(input);
int value = 0;
int i;
bool gotnumber = false;
for (i = 0; i < length; i++)
{
if (input[i] >= '0' && input[i] <= '9')
{
gotnumber = true;
value = value * 10; // shift up a column
value += input[i] - '0'; // casting the char to an int
}
else if (gotnumber) // we hit this the first time we encounter a non-number after we've had numbers
{
printf("Value: %d \n", value);
value = 0;
gotnumber = false;
}
}
return 0;
}
EDIT: the previous verison didn't deal with 0
Another solution is to use the strtok function
/* strtok example */
#include <stdio.h>
#include <string.h>
int main ()
{
char str[] = "pt112parah salin10n m5:isstupid::42$%&%^*%7first3";
char * pch;
printf ("Splitting string \"%s\" into tokens:\n",str);
pch = strtok (str," abcdefghijklmnopqrstuvwxyz:$%&^*");
while (pch != NULL)
{
printf ("%s\n",pch);
pch = strtok (NULL, " abcdefghijklmnopqrstuvwxyz:$%&^*");
}
return 0;
}
Gives:
112
10
5
42
7
3
Perhaps not the best solution for this task, since you need to specify all characters that will be treated as a token. But it is an alternative to the other solutions.
And if you don't mind using C++ instead of C (usually there isn't a good reason why not), then you can reduce your solution to just two lines of code (using AXE parser generator):
vector<int> numbers;
auto number_rule = *(*(axe::r_any() - axe::r_num())
& *axe::r_num() >> axe::e_push_back(numbers));
now test it:
std::string str = "pt112parah salin10n m5:isstupid::42$%&%^*%7first3";
number_rule(str.begin(), str.end());
std::for_each(numbers.begin(), numbers.end(), [](int i) { std::cout << "\ni=" << i; });
and sure enough, you got your numbers back.
And as a bonus, you don't need to change anything when parsing unicode wide strings:
std::wstring str = L"pt112parah salin10n m5:isstupid::42$%&%^*%7first3";
number_rule(str.begin(), str.end());
std::for_each(numbers.begin(), numbers.end(), [](int i) { std::cout << "\ni=" << i; });
and sure enough, you got the same numbers back.
#include <stdio.h>
#include <string.h>
#include <math.h>
int main(void)
{
char *input = "pt112par0ah salin10n m5:isstupid::42$%&%^*%7first3";
char *pos = input;
int integers[strlen(input) / 2]; // The maximum possible number of integers is half the length of the string, due to the smallest number of digits possible per integer being 1 and the smallest number of characters between two different integers also being 1
unsigned int numInts= 0;
while ((pos = strpbrk(pos, "0123456789")) != NULL) // strpbrk() prototype in string.h
{
sscanf(pos, "%u", &(integers[numInts]));
if (integers[numInts] == 0)
pos++;
else
pos += (int) log10(integers[numInts]) + 1; // requires math.h
numInts++;
}
for (int i = 0; i < numInts; i++)
printf("%d ", integers[i]);
return 0;
}
Finding the integers is accomplished via repeated calls to strpbrk() on the offset pointer, with the pointer being offset again by an amount equaling the number of digits in the integer, calculated by finding the base-10 logarithm of the integer and adding 1 (with a special case for when the integer is 0). No need to use abs() on the integer when calculating the logarithm, as you stated the integers will be non-negative. If you wanted to be more space-efficient, you could use unsigned char integers[] rather than int integers[], as you stated the integers will all be <256, but that isn't a necessity.

K&R Exercise 2-3 "Hex to int converter" Problem

The program I wrote works in demographics consisting of only single Hexadecimal values. (Probably not the most elegant solution, but I'm a new programmer) My question is, how would I go about handling of multiple hexadecimal digits, such as 0xAF, or 0xFF, etc? I'm not exactly sure, and I've seemed confuse myself greatly, in the attempt. I'm not asking for someone to hold my hand, but to give me a tip where I've gone wrong in this code and thoughts on how to fix it.
Thanks :)
/* Exercise 2-3. Write the function htoi(s), which converts a string of
* hexadecimal digits (including an optional 0x or 0X) into it's equivalent
* integer value. The allowable digits are 0...9 - A...F and a...f.
*
*/
#include <stdio.h>
#include <string.h>
#define NL '\n'
#define MAX 24
int htoi(char *hexd);
int
main(void)
{
char str[MAX] = {0};
char hex[] = "0123456789ABCDEFabcdef\0";
int c;
int i;
int x = 0;
while((c = getchar()) != EOF) {
for(i = 0; hex[i] != '\0'; i++) {
if(c == hex[i])
str[x++] = c;
}
if(c == NL) {
printf("%d\n", htoi(str));
x = 0, i = x;
}
}
return 0;
}
int
htoi(char *hexd)
{
int i;
int n = 0;
for(i = 0; isdigit(hexd[i]); i++)
n = (16 * i) + (hexd[i] - '0');
for(i = 0; isupper(hexd[i]); i++) /* Let's just deal with lowercase characters */
hexd[i] = hexd[i] + 'a' - 'A';
for(i = 0; islower(hexd[i]); i++) {
hexd[i] = hexd[i] - 'a';
n = (16 + i) + hexd[i] + 10;
n = hexd[i] + 10;
}
return n;
}
Someone has alredy asked this (hex to int, k&r 2.3).
Take a look, there are many good answers, but you have to fill in the blanks.
Hex to Decimal conversion [K&R exercise]
Edit:
in
char hex[] = "0123456789ABCDEFabcdef\0";
The \0 is not necesary. hex is alredy nul terminated. Is len (0...f) + 1 = 17 bytes long.
I'll pick on one loop, and leave it to you to rethink your implementation. Specifically this:
for(i = 0; isdigit(hexd[i]); i++)
n = (16 * i) + (hexd[i] - '0');
doesn't do what you probably think it does...
It only processes the first span of characters where isdigit() is TRUE.
It stops on the first character where isdigit() is FALSE.
It doesn't run past the end because isdigit('\0') is known to be FALSE. I'm concerned that might be accidentally correct, though.
It does correctly convert a hex number that can be expressed solely with digits 0-9.
Things to think about for the whole program:
Generally, prefer to not modify input strings unless the modification is a valuable side effect. In your example code, you are forcing the string to lower case in-place. Modifying the input string in-place means that a user writing htoi("1234") is invoking undefined behavior. You really don't want to do that.
Only one of the loops over digits is going to process a non-zero number of digits.
What happens if I send 0123456789ABCDEF0123456789ABCDEF to stdin?
What do you expect to get for 80000000? What did you get? Are you surprised?
Personally, I wouldn't use NL for '\n'. C usage pretty much expects to see \n in a lot of contexts where the macro is not convenient, so it is better to just get used to it now...
I think that the MAX size of string should be either 10 or 18 instead of 24. (If you have already checked the int on your machine and followed the reasoning bellow, it would be beneficial to include it as a comment in your code.)
10 : since htoi() returns an int , int is usually 4 bytes (check your system's too), so the hexadecimal number can be atmost 8 digits in length (4bits to 1 hex digit, 8 bits to a byte), and we want to allow for the optional 0x or 0X.
18 : would be better if htoi() returned a long and its 8 bytes (again, check your system's), so the hexadecimal number can be atmost 16 digits in length, and we want to allow for the optional 0x or 0X.
Please note that that sizes of int and long are machine dependent, and please look at exercise 2.1 in the K&R book to find them.
Here is my version of a classic htoi() function to convert multiple hexadecimal values into decimal integers. It's a full working program compile it and run.
#include <stdio.h>
#include <ctype.h>
#include <string.h>
#include <stdlib.h>
int htoi(const char*);
int getRawInt(char);
int main(int argc, char **argv) {
char hex[] = " ";
printf("Enter a hexadecimal number (i.e 33A)\n");
scanf("%s", hex);
printf("Hexedecimal %s in decimal is %d\n", hex, htoi(hex)); // result will be 826
return 0;
}
int htoi(const char *hex) {
const int LEN = strlen(hex) -1;
int power = 1;
int dec = 0;
for(int i = LEN; i >= 0; --i) {
dec += getRawInt(hex[i]) * power;
power *= 16;
}
return dec;
}
int getRawInt(char c) {
if(isalpha(c)) {
return toupper(c) - 'A' + 10;
} return c-'0';
}

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