Getting coefficients from a string - c++

I have a project to write a program that receives a polynomial string from the user up to the 5th power (ex. x^3+6x^2+9x+24) and prints out all the real and imaginary roots. The coefficients should be stored in a dynamic array.
The problem is getting these coefficients from the string. One of the coefficients can be a 0 (ex. 2x^2-18) so I can't store the coefficients from left to right by using an increment, because in this case a=2, b=-18, and c has no value, which is wrong.
Another problem is if the coefficient is 1, because in this case nothing will be written beside the x for the program to read (ex. x^2-x+14). Another problem is if the user adds a space, several, or none (ex. x ^3 +4x^ 2- 12 x + 1 3).
I have been thinking of pseudocode for a long time now, but nothing is coming to mind. I thought of detecting numbers from left to right and reading numbers and stopping at x, but the first and second problems occur. I thought of finding each x and then checking the numbers before it, but the second problem occurs, and also I don't know how big the number the user inputs.

Here is another Regex that you can use to get your coefficients after deleting whitespace characters:
(\d*)(x?\^?)(\d*)
It uses groups (indicated by the brackets). Every match has 3 groups:
Your coefficient
x^n, x or nothing
The exponent
If (1) is null (e.g. does not exist), it means your coefficient is 1.
If (2) and (3) are null, you have the last single number without x.
If only (3) is null, you have a single x without ^n.
You can try some examples on online regex sites like this one, where you can see the results on the right.
There are many tutorials online how to use Regex with C++.

You should normalize your input string, for example, remove all space then parse coefficients.
Let see my example. Please change it for your case.
#include <iostream>
#include <regex>
#include <iterator>
#include <string>
#include <vector>
#include <algorithm>
int main(int argc, char *argv[]) {
std::string input {argv[1]};
input.erase(remove_if(input.begin(), input.end(), isspace), input.end());
std::cout << input << std::endl;
std::vector<int> coeffs;
std::regex poly_regex(R"(\s*\+?\-?\s*\d*\s*x*\^*\s*\d*)");
auto coeff_begin = std::sregex_iterator(input.begin(), input.end(), poly_regex);
auto coeff_end = std::sregex_iterator();
for (std::sregex_iterator i = coeff_begin; i != coeff_end; ++i) {
std::smatch match = *i;
std::string match_str = match.str();
// std::cout << " " << match_str << "\n";
std::size_t plus_pos = match_str.find('+');
std::size_t minus_pos = match_str.find('-');
std::size_t x_pos = match_str.find('x');
if (x_pos == std::string::npos) {
std::cout << match_str.substr(plus_pos + 1) << std::endl;
} else if (x_pos == 0) {
std::cout << 1 << std::endl;
} else if (minus_pos != std::string::npos) {
if (x_pos - minus_pos == 1) std::cout << -1 << std::endl;
else std::cout << match_str.substr(minus_pos, x_pos - minus_pos) << std::endl;
}
else {
std::cout << match_str.substr(plus_pos + 1, x_pos - plus_pos - 1) << std::endl;
}
}
for (auto i: coeffs) std::cout << i << " ";
return 0;
}

Related

std::setprecision sets the number of significant figures. How do I use iomanip to set the precision?

I have always found iomanip confusing and counter intuitive. I need help.
A quick internet search finds (https://www.vedantu.com/maths/precision) "We thus consider precision as the maximum number of significant digits after the decimal point in a decimal number" (the emphasis is mine). That matches my understanding too. However I wrote a test program and:
stm << std::setprecision(3) << 5.12345678;
std::cout << "5.12345678: " << stm.str() << std::endl;
stm.str("");
stm << std::setprecision(3) << 25.12345678;
std::cout << "25.12345678: " << stm.str() << std::endl;
stm.str("");
stm << std::setprecision(3) << 5.1;
std::cout << "5.1: " << stm.str() << std::endl;
stm.str("");
outputs:
5.12345678: 5.12
25.12345678: 25.1
5.1: 5.1
If the precision is 3 then the output should be:
5.12345678: 5.123
25.12345678: 25.123
5.1: 5.1
Clearly the C++ standard has a different interpretation of the meaning of "precision" as relates to floating point numbers.
If I do:
stm.setf(std::ios::fixed, std::ios::floatfield);
then the first two values are formatted correctly, but the last comes out as 5.100.
How do I set the precision without padding?
You can try using this workaround:
decltype(std::setprecision(1)) setp(double number, int p) {
int e = static_cast<int>(std::abs(number));
e = e != 0? static_cast<int>(std::log10(e)) + 1 + p : p;
while(number != 0.0 && static_cast<int>(number*=10) == 0 && e > 1)
e--; // for numbers like 0.001: those zeros are not treated as digits by setprecision.
return std::setprecision(e);
}
And then:
auto v = 5.12345678;
stm << setp(v, 3) << v;
Another more verbose and elegant solution is to create a struct like this:
struct __setp {
double number;
bool fixed = false;
int prec;
};
std::ostream& operator<<(std::ostream& os, const __setp& obj)
{
if(obj.fixed)
os << std::fixed;
else os << std::defaultfloat;
os.precision(obj.prec);
os << obj.number; // comment this if you do not want to print immediately the number
return os;
}
__setp setp(double number, int p) {
__setp setter;
setter.number = number;
int e = static_cast<int>(std::abs(number));
e = e != 0? static_cast<int>(std::log10(e)) + 1 + p : p;
while(number != 0.0 && static_cast<int>(number*=10) == 0)
e--; // for numbers like 0.001: those zeros are not treated as digits by setprecision.
if(e <= 0) {
setter.fixed = true;
setter.prec = 1;
} else
setter.prec = e;
return setter;
}
Using it like this:
auto v = 5.12345678;
stm << setp(v, 3);
I don't think you can do it nicely. There are two candidate formats: defaultfloat and fixed. For the former, "precision" is the maximum number of digits, where both sides of the decimal separator count. For the latter "precision" is the exact number of digits after the decimal separator.
So your solution, I think, is to use fixed format and then manually clear trailing zeros:
#include <iostream>
#include <iomanip>
#include <sstream>
void print(const double number)
{
std::ostringstream stream;
stream << std::fixed << std::setprecision(3) << number;
auto string=stream.str();
while(string.back()=='0')
string.pop_back();
if(string.back()=='.') // in case number is integral; beware of localization issues
string.pop_back();
std::cout << string << "\n";
}
int main()
{
print(5.12345678);
print(25.12345678);
print(5.1);
}
The fixed format gives almost what you want except that it preserves trailing zeros. There is no built-in way to avoid that but you can easily remove those zeros manually. For example, in C++20 you can do the following using std::format:
std::string format_fixed(double d) {
auto s = fmt::format("{:.3f}", d);
auto end = s.find_last_not_of('0');
return end != std::string::npos ? std::string(s.c_str(), end + 1) : s;
}
std::cout << "5.12345678: " << format_fixed(5.12345678) << "\n";
std::cout << "25.12345678: " << format_fixed(25.12345678) << "\n";
std::cout << "5.1: " << format_fixed(5.1) << "\n";
Output:
5.12345678: 5.123
25.12345678: 25.123
5.1: 5.1
The same example with the {fmt} library, std::format is based on: godbolt.
Disclaimer: I'm the author of {fmt} and C++20 std::format.

C++ code - problems with сode execution time

there is code.
#include "pch.h"
#include <algorithm>
#include <iostream>
#include <vector>
#include <stdlib.h>
using namespace std;
vector<int> SearchInt(vector<int> vec, int num) {
vector<int> temp(2);
sort(begin(vec), end(vec));
int j = 0;
for (int i : vec) {
if (i > num) {
temp[0] = i;
temp[1] = j;
return { temp };
}
//cout << i << " !>= " << num << endl ;
j++;
}
cout << "NO";
exit(0);
}
int main()
{
int n;
cin >> n;
vector<int> nums(n, 0);
vector<int> NewNums(n, 0);
for (int i = 0; i < n; i++) {
cin >> nums[i];
}
if (n != nums.size()) {
cout << "://";
return 0;
}
sort(begin(nums), end(nums));
NewNums[1] = nums[nums.size() - 1];
nums.erase(nums.begin() + nums.size() - 1);
NewNums[0] = nums[nums.size() - 1];
nums.erase(nums.begin() + nums.size() - 1);
for (int j = 2; j <= NewNums.size() - 1; j++) {
NewNums[j] = SearchInt(nums, NewNums[j-1]- NewNums[j-2])[0];
nums.erase(nums.begin() + SearchInt(nums, NewNums[j] - NewNums[j - 1])[1]);
}
if (NewNums[NewNums.size()-1] < NewNums[NewNums.size() - 2] + NewNums[0]) {
cout << "YES" << endl;
for (int i : NewNums) {
cout << i << " ";
}
return 0;
}
else {
cout << "NO";
return 0;
}
}
His task is to check whether it is possible from the given Each number is less than the sum of the two adjacent ones.
(each number is less than both of two adjacent ones)
But there is a problem - with a large number of numbers, the code takes too long. Please help me to optimize it, or just give some advice.
numbers cаn not be null.
time limit: 3.0 s
n <= 500000
You are given n numbers a1, a2,…, an. Is it possible to arrange them in a circle so that each number is strictly less than the sum of its neighbors?
For example, for the array [1,4,5,6,7,8], the left array satisfies the condition, while the right array does not, since 5≥4 + 1 and 8> 1 + 6.
Input data
The first line contains one integer n (3≤n≤105) - the number of numbers.
The second line contains n integers a1, a2,…, an (1≤ai≤109) - the numbers themselves. The given numbers are not necessarily different.
Output
If there is no solution, print "NO" on the first line.
If it exists, print "YES" on the first line. After that, on the second line print n numbers - the elements of the array in the order in which they will stand on the circle. The first and last elements you print are considered neighbors on the circle. If there are multiple solutions, output any of them. You can print a circle starting with any of the numbers.
First I'll only briefly analyze technical shortcomings of your code - without analyzing its meaning. After that I'll write my solution of the problem you defined.
Performance problems of your code are due to some strange decisions:
(1) passing std::vector<int> by value and not by reference to SearchInt function - this implies allocating and copying of the whole array on each function invocation,
(2) call SearchInt two times per loop iteration in function main instead of only one,
(3) sort array within each invocation of SearchInt - it is already sorted before the loop.
To be honest your code feels ridiculously time-consuming. I'm only wondering if that was your intention to make it as slow as you possibly can...
I will not analyze correctness of your code according to problem description. To be honest even after fixing technical shortcomings your code seems to me utterly sub-optimal and quite incomprehensible - so it is just easier to solve the problem from scratch to me.
The answer to the problem as defined is YES if the biggest number is smaller than the sum of the second big and the third big and NO otherwise - this follows from the fact that all numbers are positive (in range 1 - 109 according to newly found problem description). If the answer is YES then to make a circle that satisfies the problem description you just need in a sorted sequence of input numbers switch places of the biggest number and the next big one - that's all.
Here is my code for that (for slightly relaxed input format - I'm not checking if number of items is on a separate line and that all items are on the same line - but all correct inputs will be parsed just fine):
#include <set>
#include <iostream>
int main()
{
std::multiset<unsigned> input_set;
unsigned n;
if( !( std::cin >> n ) )
{
std::cerr << "Input error - failed to read number of items." << std::endl;
return 2;
}
if( n - 3U > 105U - 3U )
{
std::cerr << "Wrong number of items value - " << n << " (must be 3 to 105)" << std::endl;
return 2;
}
for( unsigned j = 0; j < n; ++j )
{
unsigned x;
if( !( std::cin >> x ) )
{
std::cerr << "Input error - failed to read item #" << j << std::endl;
return 2;
}
if( x - 1U > 109U - 1U )
{
std::cerr << "Wrong item #" << j << " value - " << x << " (must be 1 to 109)" << std::endl;
return 2;
}
input_set.insert(x);
}
std::multiset<unsigned>::const_reverse_iterator it = input_set.rbegin();
std::multiset<unsigned>::const_reverse_iterator it0 = it;
std::multiset<unsigned>::const_reverse_iterator it1 = ++it;
if( *it0 >= *it1 + *++it )
{
std::cout << "NO (the biggest number is bigger than the sum of the second big and the third big numbers)" << std::endl;
return 1;
}
std::cout << "YES" << std::endl;
std::cout << "Circle: " << *it1 << ' ' << *it0;
do
{
std::cout << ' ' << *it;
}
while( ++it != input_set.rend() );
std::cout << std::endl;
return 0;
}

Determine whether a value is within the maximum range for that data type in c++

What is the correct way to determine if a number (in my case it is a value of power of two calculated by pow(2,n)) is within the limits of values that one variable type can take? I'm doing it like this: if(pow (2,128)>std::numeric_limits<float>::max()), but this is evaluated as true although it is expected that float's maximum value is 2^128 or something more. Is there any better way to do this comparison?
For these kinds of limit checking, you can move the terms around to stay within the limits of the type.
In this case, pow(2,n) == exp(ln(2)*n) mathematically, so, rearranging terms, you can use n > ln(maxval)/ln(2)
You can take the base 2 logarithm of the maximum limit for the type of variable and compare it to n. For example: if(n > std::log2(std::numeric_limits<float>::max()). You probably don't want n to be exactly on the limit though, since I think stuff like floating point error might cause some problems.
First of all can you answer what is the result of pow(2, 128)?
The real question is what is the type for this expression?
The second question is do you know how floating point numbers work?
Take a look on this code to give you a hints:
#include <cmath>
#include <iostream>
#include <limits>
template<class T>
void printInfo(const std::string& desc, T x)
{
std::cout << desc << ' ' << typeid(x).name() << ' ' << x << std::endl;
}
int main()
{
printInfo("A", std::pow(2, 128));
printInfo("B", std::pow(2.0f, 128));
printInfo("A", std::pow(2, 128.0f));
auto c = std::pow(2.0f, 128.0f);
printInfo("C", c);
std::cout << (c > std::numeric_limits<float>::max()) << std::endl;
std::cout << (c == std::numeric_limits<float>::infinity()) << std::endl;
return 0;
}
https://wandbox.org/permlink/bHdKqToDKdC0hSvW
I recommend review documentation of numeric_limits.
And analyze this code:
#include <cmath>
#include <iostream>
#include <limits>
template<class T>
void print2exp()
{
std::cout << typeid(T).name() << '\n';
std::cout << "Radix = " << std::numeric_limits<T>::radix << '\n';
auto maxExp = std::numeric_limits<T>::max_exponent;
std::cout << "Max exp = " << maxExp << '\n';
std::cout << "2^maxExp = " << std::pow(static_cast<T>(2), static_cast<T>(maxExp)) << '\n';
std::cout << "2^(maxExp - 1) = " << std::pow(static_cast<T>(2), static_cast<T>(maxExp - 1)) << '\n';
}
int main()
{
print2exp<float>();
print2exp<double>();
print2exp<long double>();
return 0;
}
https://wandbox.org/permlink/J0hACKUKvKlV8lYK
So proper approach to this is (assuming that radix is 2):
if (x < std::numeric_limits<T>::max_exponent) {
return std::pow(static_cast<T>(2), static_cast<T>(x));
} else {
throw invalid_argument("x is to big to be use as 2^x");
}

double to string conversion with fixed width

I would like to print a double value, into a string of no more than 8 characters. The printed number should have as many digits as possible, e.g.
5.259675
48920568
8.514e-6
-9.4e-12
I tried C++ iostreams, and printf-style, and neither respects the provided size in the way I would like it to:
cout << setw(8) << 1.0 / 17777.0 << endl;
printf( "%8g\n", 1.0 / 17777.0 );
gives:
5.62525e-005
5.62525e-005
I know I can specify a precision, but I would have to provide a very small precision here, in order to cover the worst case. Any ideas how to enforce an exact field width without sacrificing too much precision? I need this for printing matrices. Do I really have to come up with my own conversion function?
A similar question has been asked 5 years ago: Convert double to String with fixed width , without a satisfying answer. I sure hope there has been some progress in the meantime.
This seems not too difficult, actually, although you can't do it in a single function call. The number of character places used by the exponent is really quite easy to predict:
const char* format;
if (value > 0) {
if (value < 10e-100) format = "%.1e";
else if (value < 10e-10) format = "%.2e";
else if (value < 1e-5) format = "%.3e";
}
and so on.
Only, the C standard, where the behavior of printf is defined, insists on at least two digits for the exponent, so it wastes some there. See c++ how to get "one digit exponent" with printf
Incorporating those fixes is going to make the code fairly complex, although still not as bad as doing the conversion yourself.
If you want to convert to fixed decimal numbers (e.g. drop the +/-"E" part), then it makes it a lot easier to accomplish:
#include <stdio.h>
#include <cstring> // strcpy
#include <iostream> // std::cout, std::fixed
#include <iomanip> // std::setprecision
#include <new>
char *ToDecimal(double val, int maxChars)
{
std::ostringstream buffer;
buffer << std::fixed << std::setprecision(maxChars-2) << val;
std::string result = buffer.str();
size_t i = result.find_last_not_of('\0');
if (i > maxChars) i = maxChars;
if (result[i] != '.') ++i;
result.erase(i);
char *doubleStr = new char[result.length() + 1];
strcpy(doubleStr, (const char*)result.c_str());
return doubleStr;
}
int main()
{
std::cout << ToDecimal(1.26743237e+015, 8) << std::endl;
std::cout << ToDecimal(-1.0, 8) << std::endl;
std::cout << ToDecimal(3.40282347e+38, 8) << std::endl;
std::cout << ToDecimal(1.17549435e-38, 8) << std::endl;
std::cout << ToDecimal(-1E4, 8) << std::endl;
std::cout << ToDecimal(12.78e-2, 8) << std::endl;
}
Output:
12674323
-1
34028234
0.000000
-10000
0.127800

Anything like substr but instead of stopping at the byte you specified, it stops at a specific string [duplicate]

This question already has answers here:
How do you search a std::string for a substring in C++?
(6 answers)
Closed 8 years ago.
I have a client for a pre-existing server. Let's say I get some packets "MC123, 456!##".
I store these packets in a char called message. To print out a specific part of them, in this case the numbers part of them, I would do something like "cout << message.substr(3, 7) << endl;".
But what if I receive another message "MC123, 456, 789!##". "cout << message.substr(3,7)" would only print out "123, 456", whereas I want "123, 456, 789". How would I do this assuming I know that every message ends with "!##".
First - Sketch out the indexing.
std::string packet1 = "MC123, 456!##";
// 0123456789012345678
// ^------^ desired text
std::string packet2 = "MC123, 456, 789!##";
// 0123456789012345678
// ^-----------^ desired text
The others answers are ok. If you wish to use std::string find,
consider rfind and find_first_not_of, as in the following code:
// forward
void messageShow(std::string packet,
size_t startIndx = 2);
// /////////////////////////////////////////////////////////////////////////////
int main (int, char** )
{
// 012345678901234567
// |
messageShow("MC123, 456!##");
messageShow("MC123, 456, 789!##");
messageShow("MC123, 456, 789, 987, 654!##");
// error test cases
messageShow("MC123, 456, 789##!"); // missing !##
messageShow("MC123x 456, 789!##"); // extraneous char in packet
return(0);
}
void messageShow(std::string packet,
size_t startIndx) // default value 2
{
static size_t seq = 0;
seq += 1;
std::cout << packet.size() << " packet" << seq << ": '"
<< packet << "'" << std::endl;
do
{
size_t bangAtPound_Indx = packet.rfind("!##");
if(bangAtPound_Indx == std::string::npos){ // not found, can't do anything more
std::cerr << " '!##' not found in packet " << seq << std::endl;
break;
}
size_t printLength = bangAtPound_Indx - startIndx;
const std::string DIGIT_SPACE = "0123456789, ";
size_t allDigitSpace = packet.find_first_not_of(DIGIT_SPACE, startIndx);
if(allDigitSpace != bangAtPound_Indx) {
std::cerr << " extraneous char found in packet " << seq << std::endl;
break; // something extraneous in string
}
std::cout << bangAtPound_Indx << " message" << seq << ": '"
<< packet.substr(startIndx, printLength) << "'" << std::endl;
}while(0);
std::cout << std::endl;
}
This outputs
13 packet1: 'MC123, 456!##'
10 message1: '123, 456'
18 packet2: 'MC123, 456, 789!##'
15 message2: '123, 456, 789'
28 packet3: 'MC123, 456, 789, 987, 654!##'
25 message3: '123, 456, 789, 987, 654'
18 packet4: 'MC123, 456, 789##!'
'!##' not found in packet 4
18 packet5: 'MC123x 456, 789!##'
extraneous char found in packet 5
Note: String indexes start at 0. The index of the digit '1' is 2.
The correct approach is to look for existence / location of the "known termination" string, then take the substring up to (but not including) that substring.
Something like
str::string termination = "!#$";
std::size_t position = inputstring.find(termination);
std::string importantBit = message.substr(0, position);
You could check the front of the string separately as well. Combining these, you could use regular expressions to make your code more robust, using a regex like
MC([0-9,]+)!#\$
This will return the bit between MC and !#$ but only if it consists entirely of numbers and commas. Obviously you can adapt this as needed.
UPDATE you asked in your comment how to use the regular expression. Here is a very simple program. Note - this is using C++11: you need to make sure our compiler supports it.
#include <iostream>
#include <regex>
int main(void) {
std::string s ("ABC123,456,789!#$");
std::smatch m;
std::regex e ("ABC([0-9,]+)!#\\$"); // matches the kind of pattern you are looking for
if (std::regex_search (s,m,e)) {
std::cout << "match[0] = " << m[0] << std::endl;
std::cout << "match[1] = " << m[1] << std::endl;
}
}
On my Mac, I can compile the above program with
clang++ -std=c++0x -stdlib=libc++ match.cpp -o match
If instead of just digits and commas you want "anything" in your expression (but it's still got fixed characters in front and behind) you can simply do
std::regex e ("ABC(.*)!#\\$");
Here, .+ means "zero or more of 'anything'" - but followed by !#$. The double backslash has to be there to "escape" the dollar sign, which has special meaning in regular expressions (it means "the end of the string").
The more accurately your regular expression reflects exactly what you expect, the better you will be able to trap any errors. This is usually a very good thing in programming. "Always check your inputs".
One more thing - I just noticed you mentioned that you might have "more stuff" in your string. This is where using regular expressions quickly becomes the best. You mentioned a string
MC123, 456!##*USRChester.
and wanted to extract 123, 456 and Chester. That is - stuff between MC and !#$, and more stuff after USR (if that is even there). Here is the code that shows how that is done:
#include <iostream>
#include <regex>
int main(void) {
std::string s1 ("MC123, 456!#$");
std::string s2 ("MC123, 456!#$USRChester");
std::smatch m;
std::regex e ("MC([0-9, ]+)!#\\$(?:USR)?(.*)$"); // matches the kind of pattern you are looking for
if (std::regex_search (s1,m,e)) {
std::cout << "match[0] = " << m[0] << std::endl;
std::cout << "match[1] = " << m[1] << std::endl;
std::cout << "match[2] = " << m[2] << std::endl;
}
if (std::regex_search (s2,m,e)) {
std::cout << "match[0] = " << m[0] << std::endl;
std::cout << "match[1] = " << m[1] << std::endl;
std::cout << "match[2] = " << m[2] << std::endl;
if (match[2].length() > 0) {
std::cout << m[2] << ": " << m[1] << std::endl;
}
}
}
Output:
match[0] = MC123, 456!#$
match[1] = 123, 456
match[2] =
match[0] = MC123, 456!#$USRChester
match[1] = 123, 456
match[2] = Chester
Chester: 123, 456
The matches are:
match[0] : "everything in the input string that was consumed by the Regex"
match[1] : "the thing in the first set of parentheses"
match[2] : "The thing in the second set of parentheses"
Note the use of the slightly tricky (?:USR)? expression. This says "This might (that's the ()? ) be followed by the characters USR. If it is, skip them (that's the ?: part) and match what follows.
As you can see, simply testing whether m[2] is empty will tell you whether you have just numbers, or number plus "the thing after the USR". I hope this gives you an inkling of the power of regular expressions for chomping through strings like yours.
If you are sure about the ending of the message, message.substr(3, message.size()-6) will do the trick.
However, it is good practice to check everything, just to avoid surprises.
Something like this:
if (message.size() < 6)
throw error;
if (message.substr(0,3) != "MCX") //the exact numbers do not match in your example, but you get the point...
throw error;
if (message.substr(message.size()-3) != "!##")
throw error;
string data = message.substr(3, message.size()-6);
Just calculate the offset first.
string str = ...;
size_t start = 3;
size_t end = str.find("!##");
assert(end != string::npos);
return str.substr(start, end - start);
You can get the index of "!##" by using:
message.find("!##")
Then use that answer instead of 7. You should also check for it equalling std::string::npos which indicates that the substring was not found, and take some different action.
string msg = "MC4,512,541,3123!##";
for (int i = 2; i < msg.length() - 3; i++) {
if (msg[i] != '!' && msg[i + 1] != '#' && msg[i + 2] != '#')
cout << msg[i];
}
or use char[]
char msg[] = "MC4,123,54!##";
sizeof(msg -1 ); //instead of msg.length()
// -1 for the null byte at the end (each char takes 1 byte so the size -1 == number of chars)