how to convert unsigned char* to integer fast, - c++

In my project i wrote such function for conversion:
// len should be > 0
uint32_t stringToInt(unsigned char const* buffer, int len) {
uint32_t result = buffer[0] - '0';
for (int i = 1; i < len; i++) {
result *= 10;
result += buffer[i] - '0';
}
return result;
}
Are there any stl / boost methods that can do the same with the same speed? If not so then probaly you can further improve my version?
I can not use atoi because it doesn't allow to provide len. I also don't want to create temp buffer just for atoi call.

From atoi implementation in C, with minor modification to account for length.
int my_atoi(char *p,int len) {
int k = 0;
for (int i=0;i<len;i++) {
k = (k<<3)+(k<<1)+(*p)-'0';
p++;
}
return k;
}
If you want the fastest possible atoi implementation, one way to go would be to check gcc source for their implementation of atoi and modify it to match your additional requirement for length...

Saw this one years ago. Its probably slower but more fun:
int stringToInt(char *buffer, int len) {
if (len > 0) {
return myStoi(buffer + len - 1, len);
}
return 0;
}
int myStoi(char *curr, int len) {
if (--len) {
return myStoi(curr-1, len) * 10 + (*curr - '0');
}
return (*curr - '0');
}
This, like the original, assumes the buffer contents are valid digit characters. (i.e. no leading spaces or dashes or plus-signs, no internal dots or alpha chars)

Related

Add comma in a sprintf function

I was looking at topic: How to format a number from 1123456789 to 1,123,456,789 in C?
So I mounted my code based on an existing one.
void printfcomma(char *buf, const char* text int n) {
if (n < 1000) {
sprintf(buf, "%s %d", text, n);
return;
}
printfcomma(buf, n / 1000);
sprintf(buf, "%s ,%03d", text, n %1000);
return;
}
sprintf is only returning the final 3 digits. Example: ,536
Does anyone have any idea why they are not showing the other numbers
You are overwriting.
You should do sprintf(s+ strlen(s),"abcde");
void printfcomma(char *buf,int n) {
if (n < 1000) {
sprintf(buf+strlen(buf), "%d", n);
return;
}
printfcomma(buf, n / 1000);
sprintf(buf+strlen(buf), ",%03d", n %1000);
return;
}
In calling function
memset(s,0,sizeof(s));// s is the char array.
printfcomma(s,100000536);
Output
100,000,536
As answered by #coderredoc, code is over-writing buf.
An alternative to calling strlen() is to take advantage of the return value of sprintf().
The sprintf function returns the number of characters written in the array, not counting the terminating null character, or a negative value if an encoding error occurred. C11dr §7.21.6.6 3
Further: code should handle negative numbers too.
const char* text use is unclear. Example code below does not use it.
int printfcomma(char *buf, int n) {
if (n > -1000 && n < 1000) { // avoid (abs(n) < 1000) here. abs(INT_MIN) is a problem
return sprintf(buf, "%d", n);
}
int len = printfcomma(buf, n / 1000);
if (len > 0) {
len += sprintf(buf + len, ",%03d", text, abs(n % 1000)); // do not print `-`
}
return len;
}
Usage
char s[sizeof(int)*CHAR_BIT]; // Somehow, insure buffer size is sufficient.
printfcomma(s, INT_MIN);
puts(s); --> -2,147,483,648

print fibo big numbers in c++ or c language

I write this code for show fibonacci series using recursion.But It not show correctly for n>43 (ex: for n=100 show:-980107325).
#include<stdio.h>
#include<conio.h>
void fibonacciSeries(int);
void fibonacciSeries(int n)
{
static long d = 0, e = 1;
long c;
if (n>1)
{
c = d + e;
d = e;
e = c;
printf("%d \n", c);
fibonacciSeries(n - 1);
}
}
int main()
{
long a, n;
long long i = 0, j = 1, f;
printf("How many number you want to print in the fibonnaci series :\n");
scanf("%d", &n);
printf("\nFibonacci Series: ");
printf("%d", 0);
fibonacciSeries(n);
_getch();
return 0;
}
The value of fib(100) is so large that it will overflow even a 64 bit number. To operate on such large values, you need to do arbitrary-precision arithmetic. Arbitrary-precision arithmetic is not provided by C nor C++ standard libraries, so you'll need to either implement it yourself or use a library written by someone else.
For smaller values that do fit your long long, your problem is that you use the wrong printf format specifier. To print a long long, you need to use %lld.
Code overflows the range of the integer used long.
Could use long long, but even that may not handle Fib(100) which needs at least 69 bits.
Code could use long double if 1.0/LDBL_EPSILON > 3.6e20
Various libraries exist to handle very large integers.
For this task, all that is needed is a way to add two large integers. Consider using a string. An inefficient but simply string addition follows. No contingencies for buffer overflow.
#include <stdio.h>
#include <string.h>
#include <assert.h>
char *str_revese_inplace(char *s) {
char *left = s;
char *right = s + strlen(s);
while (right > left) {
right--;
char t = *right;
*right = *left;
*left = t;
left++;
}
return s;
}
char *str_add(char *ssum, const char *sa, const char *sb) {
const char *pa = sa + strlen(sa);
const char *pb = sb + strlen(sb);
char *psum = ssum;
int carry = 0;
while (pa > sa || pb > sb || carry) {
int sum = carry;
if (pa > sa) sum += *(--pa) - '0';
if (pb > sb) sum += *(--pb) - '0';
*psum++ = sum % 10 + '0';
carry = sum / 10;
}
*psum = '\0';
return str_revese_inplace(ssum);
}
int main(void) {
char fib[3][300];
strcpy(fib[0], "0");
strcpy(fib[1], "1");
int i;
for (i = 2; i <= 1000; i++) {
printf("Fib(%3d) %s.\n", i, str_add(fib[2], fib[1], fib[0]));
strcpy(fib[0], fib[1]);
strcpy(fib[1], fib[2]);
}
return 0;
}
Output
Fib( 2) 1.
Fib( 3) 2.
Fib( 4) 3.
Fib( 5) 5.
Fib( 6) 8.
...
Fib(100) 3542248xxxxxxxxxx5075. // Some xx left in for a bit of mystery.
Fib(1000) --> 43466...about 200 more digits...8875
You can print some large Fibonacci numbers using only char, int and <stdio.h> in C.
There is some headers :
#include <stdio.h>
#define B_SIZE 10000 // max number of digits
typedef int positive_number;
struct buffer {
size_t index;
char data[B_SIZE];
};
Also some functions :
void init_buffer(struct buffer *buffer, positive_number n) {
for (buffer->index = B_SIZE; n; buffer->data[--buffer->index] = (char) (n % 10), n /= 10);
}
void print_buffer(const struct buffer *buffer) {
for (size_t i = buffer->index; i < B_SIZE; ++i) putchar('0' + buffer->data[i]);
}
void fly_add_buffer(struct buffer *buffer, const struct buffer *client) {
positive_number a = 0;
size_t i = (B_SIZE - 1);
for (; i >= client->index; --i) {
buffer->data[i] = (char) (buffer->data[i] + client->data[i] + a);
buffer->data[i] = (char) (buffer->data[i] - (a = buffer->data[i] > 9) * 10);
}
for (; a; buffer->data[i] = (char) (buffer->data[i] + a), a = buffer->data[i] > 9, buffer->data[i] = (char) (buffer->data[i] - a * 10), --i);
if (++i < buffer->index) buffer->index = i;
}
Example usage :
int main() {
struct buffer number_1, number_2, number_3;
init_buffer(&number_1, 0);
init_buffer(&number_2, 1);
for (int i = 0; i < 2500; ++i) {
number_3 = number_1;
fly_add_buffer(&number_1, &number_2);
number_2 = number_3;
}
print_buffer(&number_1);
}
// print 131709051675194962952276308712 ... 935714056959634778700594751875
Best C type is still char ? The given code is printing f(2500), a 523 digits number.
Info : f(2e5) has 41,798 digits, see also Factorial(10000) and Fibonacci(1000000).
Well, you could want to try implementing BigInt in C++ or C.
Useful Material:
How to implement big int in C++
For this purporse you need implement BigInteger. There is no such build-in support in current c++. You can view few advises on stack overflow
Or you also can use some libs like GMP
Also here is some implementation:
E-maxx - on Russian language description.
Or find some open implementation on GitHub
Try to use a different format and printf, use unsigned to get wider range of digits.
If you use unsigned long long you should get until 18 446 744 073 709 551 615 so until the 93th number for fibonacci serie 12200160415121876738 but after this one you will get incorrect result because the 94th number 19740274219868223167 is too big for unsigned long long.
Keep in mind that the n-th fibonacci number is (approximately) ((1 + sqrt(5))/2)^n.
This allows you to get the value for n that allows the result to fit in 32 /64 unsigned integers. For signed remember that you lose one bit.

Base 64 Encoding Losing data

This is my fourth attempt at doing base64 encoding. My first tries work but it isn't standard. It's also extremely slow!!! I used vectors and push_back and erase a lot.
So I decided to re-write it and this is much much faster! Except that it loses data. -__-
I need as much speed as I can possibly get because I'm compressing a pixel buffer and base64 encoding the compressed string. I'm using ZLib. The images are 1366 x 768 so yeah.
I do not want to copy any code I find online because... Well, I like to write things myself and I don't like worrying about copyright stuff or having to put a ton of credits from different sources all over my code..
Anyway, my code is as follows below. It's very short and simple.
const static std::string Base64Chars = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
inline bool IsBase64(std::uint8_t C)
{
return (isalnum(C) || (C == '+') || (C == '/'));
}
std::string Copy(std::string Str, int FirstChar, int Count)
{
if (FirstChar <= 0)
FirstChar = 0;
else
FirstChar -= 1;
return Str.substr(FirstChar, Count);
}
std::string DecToBinStr(int Num, int Padding)
{
int Bin = 0, Pos = 1;
std::stringstream SS;
while (Num > 0)
{
Bin += (Num % 2) * Pos;
Num /= 2;
Pos *= 10;
}
SS.fill('0');
SS.width(Padding);
SS << Bin;
return SS.str();
}
int DecToBinStr(std::string DecNumber)
{
int Bin = 0, Pos = 1;
int Dec = strtol(DecNumber.c_str(), NULL, 10);
while (Dec > 0)
{
Bin += (Dec % 2) * Pos;
Dec /= 2;
Pos *= 10;
}
return Bin;
}
int BinToDecStr(std::string BinNumber)
{
int Dec = 0;
int Bin = strtol(BinNumber.c_str(), NULL, 10);
for (int I = 0; Bin > 0; ++I)
{
if(Bin % 10 == 1)
{
Dec += (1 << I);
}
Bin /= 10;
}
return Dec;
}
std::string EncodeBase64(std::string Data)
{
std::string Binary = std::string();
std::string Result = std::string();
for (std::size_t I = 0; I < Data.size(); ++I)
{
Binary += DecToBinStr(Data[I], 8);
}
for (std::size_t I = 0; I < Binary.size(); I += 6)
{
Result += Base64Chars[BinToDecStr(Copy(Binary, I, 6))];
if (I == 0) ++I;
}
int PaddingAmount = ((-Result.size() * 3) & 3);
for (int I = 0; I < PaddingAmount; ++I)
Result += '=';
return Result;
}
std::string DecodeBase64(std::string Data)
{
std::string Binary = std::string();
std::string Result = std::string();
for (std::size_t I = Data.size(); I > 0; --I)
{
if (Data[I - 1] != '=')
{
std::string Characters = Copy(Data, 0, I);
for (std::size_t J = 0; J < Characters.size(); ++J)
Binary += DecToBinStr(Base64Chars.find(Characters[J]), 6);
break;
}
}
for (std::size_t I = 0; I < Binary.size(); I += 8)
{
Result += (char)BinToDecStr(Copy(Binary, I, 8));
if (I == 0) ++I;
}
return Result;
}
I've been using the above like this:
int main()
{
std::string Data = EncodeBase64("IMG." + ::ToString(677) + "*" + ::ToString(604)); //IMG.677*604
std::cout<<DecodeBase64(Data); //Prints IMG.677*601
}
As you can see in the above, it prints the wrong string. It's fairly close but for some reason, the 4 is turned into a 1!
Now if I do:
int main()
{
std::string Data = EncodeBase64("IMG." + ::ToString(1366) + "*" + ::ToString(768)); //IMG.1366*768
std::cout<<DecodeBase64(Data); //Prints IMG.1366*768
}
It prints correctly.. I'm not sure what is going on at all or where to begin looking.
Just in-case anyone is curious and want to see my other attempts (the slow ones): http://pastebin.com/Xcv03KwE
I'm really hoping someone could shed some light on speeding things up or at least figuring out what's wrong with my code :l
The main encoding issue is that you are not accounting for data that is not a multiple of 6 bits. In this case, the final 4 you have is being converted into 0100 instead of 010000 because there are no more bits to read. You are supposed to pad with 0s.
After changing your Copy like this, the final encoded character is Q, instead of the original E.
std::string data = Str.substr(FirstChar, Count);
while(data.size() < Count) data += '0';
return data;
Also, it appears that your logic for adding padding = is off because it is adding one too many = in this case.
As far as comments on speed, I'd focus primarily on trying to reduce your usage of std::string. The way you are currently converting the data into a string with 0 and 1 is pretty inefficent considering that the source could be read directly with bitwise operators.
I'm not sure whether I could easily come up with a slower method of doing Base-64 conversions.
The code requires 4 headers (on Mac OS X 10.7.5 with G++ 4.7.1) and the compiler option -std=c++11 to make the #include <cstdint> acceptable:
#include <string>
#include <iostream>
#include <sstream>
#include <cstdint>
It also requires a function ToString() that was not defined; I created:
std::string ToString(int value)
{
std::stringstream ss;
ss << value;
return ss.str();
}
The code in your main() — which is what uses the ToString() function — is a little odd: why do you need to build a string from pieces instead of simply using "IMG.677*604"?
Also, it is worth printing out the intermediate result:
int main()
{
std::string Data = EncodeBase64("IMG." + ::ToString(677) + "*" + ::ToString(604));
std::cout << Data << std::endl;
std::cout << DecodeBase64(Data) << std::endl; //Prints IMG.677*601
}
This yields:
SU1HLjY3Nyo2MDE===
IMG.677*601
The output string (SU1HLjY3Nyo2MDE===) is 18 bytes long; that has to be wrong as a valid Base-64 encoded string has to be a multiple of 4 bytes long (as three 8-bit bytes are encoded into four bytes each containing 6 bits of the original data). This immediately tells us there are problems. You should only get zero, one or two pad (=) characters; never three. This also confirms that there are problems.
Removing two of the pad characters leaves a valid Base-64 string. When I use my own home-brew Base-64 encoding and decoding functions to decode your (truncated) output, it gives me:
Base64:
0x0000: SU1HLjY3Nyo2MDE=
Binary:
0x0000: 49 4D 47 2E 36 37 37 2A 36 30 31 00 IMG.677*601.
Thus it appears you have encode the null terminating the string. When I encode IMG.677*604, the output I get is:
Binary:
0x0000: 49 4D 47 2E 36 37 37 2A 36 30 34 IMG.677*604
Base64: SU1HLjY3Nyo2MDQ=
You say you want to speed up your code. Quite apart from fixing it so that it encodes correctly (I've not really studied the decoding), you will want to avoid all the string manipulation you do. It should be a bit manipulation exercise, not a string manipulation exercise.
I have 3 small encoding routines in my code, to encode triplets, doublets and singlets:
/* Encode 3 bytes of data into 4 */
static void encode_triplet(const char *triplet, char *quad)
{
quad[0] = base_64_map[(triplet[0] >> 2) & 0x3F];
quad[1] = base_64_map[((triplet[0] & 0x03) << 4) | ((triplet[1] >> 4) & 0x0F)];
quad[2] = base_64_map[((triplet[1] & 0x0F) << 2) | ((triplet[2] >> 6) & 0x03)];
quad[3] = base_64_map[triplet[2] & 0x3F];
}
/* Encode 2 bytes of data into 4 */
static void encode_doublet(const char *doublet, char *quad, char pad)
{
quad[0] = base_64_map[(doublet[0] >> 2) & 0x3F];
quad[1] = base_64_map[((doublet[0] & 0x03) << 4) | ((doublet[1] >> 4) & 0x0F)];
quad[2] = base_64_map[((doublet[1] & 0x0F) << 2)];
quad[3] = pad;
}
/* Encode 1 byte of data into 4 */
static void encode_singlet(const char *singlet, char *quad, char pad)
{
quad[0] = base_64_map[(singlet[0] >> 2) & 0x3F];
quad[1] = base_64_map[((singlet[0] & 0x03) << 4)];
quad[2] = pad;
quad[3] = pad;
}
This is written as C code rather than using native C++ idioms, but the code shown should compile with C++ (unlike the C99 initializers elsewhere in the source). The base_64_map[] array corresponds to your Base64Chars string. The pad character passed in is normally '=', but can be '\0' since the system I work with has eccentric ideas about not needing padding (pre-dating my involvement in the code, and it uses a non-standard alphabet to boot) and the code handles both the non-standard and the RFC 3548 standard.
The driving code is:
/* Encode input data as Base-64 string. Output length returned, or negative error */
static int base64_encode_internal(const char *data, size_t datalen, char *buffer, size_t buflen, char pad)
{
size_t outlen = BASE64_ENCLENGTH(datalen);
const char *bin_data = (const void *)data;
char *b64_data = (void *)buffer;
if (outlen > buflen)
return(B64_ERR_OUTPUT_BUFFER_TOO_SMALL);
while (datalen >= 3)
{
encode_triplet(bin_data, b64_data);
bin_data += 3;
b64_data += 4;
datalen -= 3;
}
b64_data[0] = '\0';
if (datalen == 2)
encode_doublet(bin_data, b64_data, pad);
else if (datalen == 1)
encode_singlet(bin_data, b64_data, pad);
b64_data[4] = '\0';
return((b64_data - buffer) + strlen(b64_data));
}
/* Encode input data as Base-64 string. Output length returned, or negative error */
int base64_encode(const char *data, size_t datalen, char *buffer, size_t buflen)
{
return(base64_encode_internal(data, datalen, buffer, buflen, base64_pad));
}
The base64_pad constant is the '='; there's also a base64_encode_nopad() function that supplies '\0' instead. The errors are somewhat arbitrary but relevant to the code.
The main point to take away from this is that you should be doing bit manipulation and building up a string that is an exact multiple of 4 bytes for a given input.
std::string EncodeBase64(std::string Data)
{
std::string Binary = std::string();
std::string Result = std::string();
for (std::size_t I = 0; I < Data.size(); ++I)
{
Binary += DecToBinStr(Data[I], 8);
}
if (Binary.size() % 6)
{
Binary.resize(Binary.size() + 6 - Binary.size() % 6, '0');
}
for (std::size_t I = 0; I < Binary.size(); I += 6)
{
Result += Base64Chars[BinToDecStr(Copy(Binary, I, 6))];
if (I == 0) ++I;
}
if (Result.size() % 4)
{
Result.resize(Result.size() + 4 - Result.size() % 4, '=');
}
return Result;
}

C++ Bytes To Bits Conversion And Then Print

Code Taken From: Bytes to Binary in C Credit: BSchlinker
The following code I modified to take more than 1 Byte at a time. I modified it, and got it half working and then got really confused on my loops. :( Ive spent the last day and a half trying to figure it out... but my C++ skills are not really that good (still learning!)
#include <iostream>
using namespace std;
char show_binary(unsigned char u, unsigned char *result,int len);
int main()
{
unsigned char p40[3] = {0x40, 0x00, 0x0a};
unsigned char bits[8*(sizeof(p40))];
int c;
c=sizeof(p40);
show_binary(*p40, bits, 3);
cout << "\n\n";
cout << "BIN = ";
do{
for (int i = 0; i < 8; i++)
printf("%d",bits[i+(8*c)]);
c++;
}while(c < 3);
cout << "\n";
int a;
cin >> a;
return 0;
}
char show_binary(unsigned char u, unsigned char *result, int len)
{
unsigned char mask = 1;
unsigned char bits[8*sizeof(result)];
int a,b,c;
a=0;
b=0;
c=len;
do{
for (int i = 0; i < 8; i++)
bits[i+(8*a)] = (u[&a] & (mask << i)) != 0;
a++;
}while(a < len);
//Need to reverse it?
do{
for (int i = 8; i != -1; i--)
result[i+(8*c)] = bits[i+(8*c)];
b++;
c--;
}while(b < len);
return *result;
}
After I spit out:
cout << "BIN = ";
do{
for (int i = 0; i < 8; i++)
printf("%d",bits[i+(8*c)]);
c++;
}while(c < 3);
Id like to take bit[11] ~ bit[the end] and compute a BYTE every 8 bits. If that makes sense. But first the function should work. Any pro tips on how this should be done? And of course, rip my code apart. I like to learn.
Man, there is a lot going on in this code, so it's hard to know where to start. Suffice to say, you're trying a bit too hard. It sounds like you are trying to 1) pass in a byte array; 2) turn those bytes into a string representation of the binary; and 3) turn that string representation back into a value?
It just so happens I recently did something similar to this in C, which should still work using a C++ compiler.
#include <stdio.h>
#include <string.h>
/* A macro to get a substring */
#define substr(dest, src, dest_size, startPos, strLen) snprintf(dest, dest_size, "%.*s", strLen, src+startPos)
/* Pass in char* array of bytes, get binary representation as string in bitStr */
void str2bs(const char *bytes, size_t len, char *bitStr) {
size_t i;
char buffer[9] = "";
for(i = 0; i < len; i++) {
sprintf(buffer,
"%c%c%c%c%c%c%c%c",
(bytes[i] & 0x80) ? '1':'0',
(bytes[i] & 0x40) ? '1':'0',
(bytes[i] & 0x20) ? '1':'0',
(bytes[i] & 0x10) ? '1':'0',
(bytes[i] & 0x08) ? '1':'0',
(bytes[i] & 0x04) ? '1':'0',
(bytes[i] & 0x02) ? '1':'0',
(bytes[i] & 0x01) ? '1':'0');
strncat(bitStr, buffer, 8);
buffer[0] = '\0';
}
}
To get the string of binary back into a value it can by done with bit shifting:
unsigned char bs2uc(char *bitStr) {
unsigned char val = 0;
int toShift = 0;
int i;
for(i = strlen(bitStr)-1; i >= 0; i--) {
if(bitStr[i] == '1') {
val = (1 << toShift) | val;
}
toShift++;
}
return val;
}
Once you had a binary string you could then take substrings of any arbitrary 8 bits (or less, I guess) and turn them back into bytes.
char *bitStr; /* Let's pretend this is populated with a valid string */
char byte[9] = "";
substr(byte, bitStr, 9, 4, 8);
/* This would create a substring of length 8 starting from index 4 of bitStr */
unsigned char b = bs2uc(byte);
I've actually created a whole suite of value -> binary string -> value functions if you'd like to take a look at them. GitHub - binstr

Convert char array to single int?

Anyone know how to convert a char array to a single int?
char hello[5];
hello = "12345";
int myNumber = convert_char_to_int(hello);
Printf("My number is: %d", myNumber);
There are mulitple ways of converting a string to an int.
Solution 1: Using Legacy C functionality
int main()
{
//char hello[5];
//hello = "12345"; --->This wont compile
char hello[] = "12345";
Printf("My number is: %d", atoi(hello));
return 0;
}
Solution 2: Using lexical_cast(Most Appropriate & simplest)
int x = boost::lexical_cast<int>("12345");
Solution 3: Using C++ Streams
std::string hello("123");
std::stringstream str(hello);
int x;
str >> x;
if (!str)
{
// The conversion failed.
}
If you are using C++11, you should probably use stoi because it can distinguish between an error and parsing "0".
try {
int number = std::stoi("1234abc");
} catch (std::exception const &e) {
// This could not be parsed into a number so an exception is thrown.
// atoi() would return 0, which is less helpful if it could be a valid value.
}
It should be noted that "1234abc" is implicitly converted from a char[] to a std:string before being passed to stoi().
I use :
int convertToInt(char a[1000]){
int i = 0;
int num = 0;
while (a[i] != 0)
{
num = (a[i] - '0') + (num * 10);
i++;
}
return num;;
}
Use sscanf
/* sscanf example */
#include <stdio.h>
int main ()
{
char sentence []="Rudolph is 12 years old";
char str [20];
int i;
sscanf (sentence,"%s %*s %d",str,&i);
printf ("%s -> %d\n",str,i);
return 0;
}
I'll just leave this here for people interested in an implementation with no dependencies.
inline int
stringLength (char *String)
{
int Count = 0;
while (*String ++) ++ Count;
return Count;
}
inline int
stringToInt (char *String)
{
int Integer = 0;
int Length = stringLength(String);
for (int Caret = Length - 1, Digit = 1; Caret >= 0; -- Caret, Digit *= 10)
{
if (String[Caret] == '-') return Integer * -1;
Integer += (String[Caret] - '0') * Digit;
}
return Integer;
}
Works with negative values, but can't handle non-numeric characters mixed in between (should be easy to add though). Integers only.
For example, "mcc" is a char array and "mcc_int" is the integer you want to get.
char mcc[] = "1234";
int mcc_int;
sscanf(mcc, "%d", &mcc_int);
With cstring and cmath:
int charsToInt (char* chars) {
int res{ 0 };
int len = strlen(chars);
bool sig = *chars == '-';
if (sig) {
chars++;
len--;
}
for (int i{ 0 }; i < len; i++) {
int dig = *(chars + i) - '0';
res += dig * (pow(10, len - i - 1));
}
res *= sig ? -1 : 1;
return res;
}
Ascii string to integer conversion is done by the atoi() function.
Long story short you have to use atoi()
ed:
If you are interested in doing this the right way :
char szNos[] = "12345";
char *pNext;
long output;
output = strtol (szNos, &pNext, 10); // input, ptr to next char in szNos (null here), base