This is a snippet of my code:
#include <iostream>
#include <fstream>
#include <string>
#include <stdlib.h> // atoi()
int main() {
std::string line;
std::ifstream numbers("numbertest.txt");
if (numbers.is_open()) {
while (std::getline(numbers, line)) {
for (int i = 0; i < line.length() - 4; i++) {
for (int n = 0; n < 5; n++) {
std::cout << atoi((line.substr(i, 5)[n]).c_str());
}
I want to operate with numbers in groups of 5, from a file. Why is atoi() not working here? It says "expression must have class type" under the second parentheses on the atoi line.
line.substr(i, 5) creates a temporary std::string containing 5 characters in line from position i
std::string foo = "hello world";
int i = 2;
std::cout << foo.substr(2, 5) << '\n';
would print "llo wo".
The [n] operator returns the nth character of the substring, which is of type char, you are then calling .c_str() on that character rather than on the substring.
You can avoid the .c_str() entirely by using std::stoi, e.g.
std::cout << "line.substr(i, 5) = " << line.substr(i, 5) << '\n';
std::cout << std::stoi(line.substr(i, 5));
aoti and stoi both take a string representation of a number as their input and return the numeric value. For example:
std::string input = "123a";
// std::cout << input * 10 << '\n'; // illegal: input is a string, not a number.
int i = std::stoi(input); // converts to integer representation, i.e. 123
std::cout << i * 10 << '\n'; // outputs 1230
----- EDIT -----
You're actually asking all the wrong questions. What you want to do is take an input pattern and output all of the patterns of 5 characters in it.
Example input: "1020304050"
Example output: 10203 02030 20304 03040 30405 04050
You don't need to convert these to numbers to output them, you can just output the characters. The problem with your original code wasn't the conversion it was the incorrect sequence of operators.
std::substring is expensive, it has to create a new, temporary string, copy characters from the original into it, and then return it, and it does it for every call.
The following should achieve what you're trying to do:
while (std::getline(numbers, line)) {
for (size_t i = 0; i < line.length() - 4; i++) {
for (size_t n = 0; n < 5; n++) {
std::cout << line[i + n];
}
std::cout << '\n';
}
}
If you really want to invoke substr, you could also implement this as
while (std::getline(numbers, line)) {
for (size_t i = 0; i < line.length() - 4; i++) {
std::cout << line.substr(i, 5) << '\n';
}
}
Here's a working demonstration: http://ideone.com/mXv2z5
Try atoi( line.substr(i,5).c_str() )
Or if you want for each character
std::cout << ((line.substr(i, 5)[n]) - '0');
Or even better
std::cout << (line[i+n]) - '0');
Note that: atoi is not ascii to integer. It converts a ctype string to number. For a single character, this conversion should be done using arithmetic or lookup table.
Also there is no point converting characters to integer and then print it (back to chacters). You should better print digit character itself.
Moreover in C++, I would prefer to use stringstream instead or atoi. On C++11 there are even more advanced solutions like sto*.
Related
I am trying to make a program that turns a string into encryption by going ten letters ahead of each letter. https://gyazo.com/86f9d708c2f02cf2d70dbc1cd9fa9a06 I am doing part 2. When I input "helloworld" something like 0x45 something comes up. Please help! This is due soon!
I am tried messing around with the for loops but it didn't help.
#include <iostream>
using namespace std;
int main()
{
//Input Message
cout << "Enter a message" << endl;
string message;
getline(cin, message);
//Convert Message to Numbers
int numMess[message.length()];
for (int i = 0; i<message.length(); i++) {
numMess[i] = (int)message[i];
}
cout << numMess << endl;
//Encrypt Number Message by adding ten to each one
int encryptNumMess[message.length()];
for (int a = 0; a < message.length(); a++){
encryptNumMess[a] = numMess[a] + 10;
if (encryptNumMess[a] > 122) {
encryptNumMess[a] = 97;
}
}
cout << encryptNumMess << endl;
//Convert Encrypted Number Message to letters
string encryption[message.length()];
for (int b = 0; b<message.length(); b++) {
encryption[b] = (char)encryptNumMess[b];
}
cout << encryption << endl;
return 0;
}
I expect when I type "helloworld" the final product will be "rovvygybvn"
If you are willing to scrap the hand-coded loops, you can use the STL algorithms such as std::transform to accomplish this:
But first, there are a few things you should do:
Don't use magic numbers such as 122, 97, etc. Instead use the actual character constants, i.e a, b, etc. However if we assume ASCII, where the alphabetic character codes are contiguous, your particular program could simply use a constant string to denote the alphabet, and then use simple indexing to pick out the character.
const char *alphabet = "abcdefghijklmnopqrstuvwxyz";
Then to get the letter a, a simple subtraction is all that's required to get the index:
char ch = 'b';
int index = ch - 'a'; // same as 'b' - 'a' == 98 - 97 == 1
std::cout << alphabet[index]; // will print 'b'
Given this, the next thing is to figure out what character is reached if you add 10 to the value, and if greater than 26, wrap around to the beginning of the alphabet. This can be done using modulus (remainder after division)
char ch = 'x';
int index = (ch - 'a' + 10) % 26; // Same as ('x' - 'a' + 10) % 26 == (120 - 97 + 10) % 26 == 33 % 26 == 7
std::cout << alphabet[index]; // will print 'h'
The next thing is to figure out the opposite, where given an encrypted character, you have to find the unencrypted character by subtracting 10. Here this wraps the opposite way, so a little more work needs to be done (not shown, but code sample reflects what is done).
Putting this all together, and using std::transform and lambdas, we get the following small program:
#include <iostream>
#include <algorithm>
#include <string>
#include <iterator>
#include <cmath>
int main()
{
//Input Message
const char *alphabet="abcdefghijklmnopqrstuvwxyz";
std::string message = "helloworld";
std::string result;
// set the encrypted string using the formula above and std::transform
std::transform(message.begin(), message.end(), std::back_inserter(result),
[&](char ch) { return alphabet[(ch - 'a' + 10) % 26]; });
std::cout << "Encrypted: " << result << '\n';
// convert back to unencrypted using the above formula and std::transform
std::string result2;
std::transform(result.begin(), result.end(), std::back_inserter(result2),
[&](char ch)
{ int index = ch - 'a' - 10; index = index < 0?26 - (abs(index) % 26):index % 26; return alphabet[index];});
std::cout << "Unencrypted: " << result2;
}
Output:
Encrypted: rovvygybvn
Unencrypted: helloworld
This code works for encrypt, if you want to decrypt you should chande newAlphabet and oldAlphabet
I comment in the code that which newAlphabet and oldAlphabet are for encrypt and which are for decrypt
#include <windows.h>
#include <stdio.h>
#include <string>
#include <iostream>
using namespace std;
int main()
{
// For Encrypt
string newAlphabet = "abcdefghijklmnopqrstuvwxyz";
string oldAlphabet = "klmnopqrstuvwxyzabcdefghij";
// For Decrypt
//string newAlphabet = "klmnopqrstuvwxyzabcdefghij";
//string oldAlphabet = "abcdefghijklmnopqrstuvwxyz";
string input = "";
string output = "";
getline(cin, input);
int inputLen = input.size();
if (oldAlphabet.size() != newAlphabet.size())
return false;
for (int i = 0; i < inputLen; ++i)
{
int oldCharIndex = oldAlphabet.find(tolower(input[i]));
if (oldCharIndex >= 0)
output += isupper(input[i]) ? toupper(newAlphabet[oldCharIndex]) : newAlphabet[oldCharIndex];
else
output += input[i];
}
cout << output << endl;
return 0;
}
As others have already mentioned int numMess[message.length()]; is not valid c++.
If it works for you, you're using compiler extension which you really shouldn't rely on. The correct way would be:
std::vector <int> numMess(message.length());
Look up the std::vector reference for more info.
Next, int encryptNumMess[100]; creates a C array style array. encryptNumMess is the base pointer to the array. when you try std::cout << encryptNumMess it'll output the pointer value, NOT the array. You'll need a for loop for doing that, like so :
for(int i = 0; i < 100; ++i)
std::cout << encryptNumMess[i] << " ";
std::cout << endl;
The above also works when you convert this to a vector like we did with numMess whereas in that case, std::cout << encryptNumMess wouldn't even compile.
Thirdly, string encryption[100] creates an array of 100 strings! Not a string of size 100. To do that:
std::string foo(message.length(), '\0');
We have to specify what character to fill the string with. Thus us '\0'.
And now, for the string, to output it, you may use std::cout << foo.
Lastly, since arithmetic is allowed on char, the entire program may be shortened to just this
#include <iostream>
int main()
{
// Input Message
std::cout << "Enter a message" << std::endl;
std::string message, encryption;
getline(std::cin, message);
// Resize encryption string to the desired length
encryption.resize(message.length());
// Do the encryption
for(size_t i = 0; i < message.length(); ++i) {
encryption[i] = message[i] + 10;
if (encryption[i] > 122) {
encryption[i] = 97;
}
}
// Output the string
std::cout << encryption << std::endl;
return 0;
}
Of course, your encryption algorithm is still not correct as per instructions, but I'll leave that for you to figure out. I believe #PaulMcKenzie has already told you most of how to fix it, and also to not use magic numbers.
The string str_hex contains the hex-values for the letters A-J, which corresponds to the decimal values 65-74. I'm trying to cast each hex-value to its decimal value following this example. It works nice for the std::cout case inside the for-loop, but the output-std::string still has the ascii-values. Why does this not work or is there a nicer/more proper way to build my output string?
#include <string>
#include <iostream>
#include <stdint.h>
int main()
{
std::string str_hex("\x41\x42\x43\x44\x45\x46\x47\x48\x49\x4a\x4b", 10);
std::string str_output = "";
for (int i = 0; i < 10; ++i)
{
uint8_t tmp = str_hex[i];
str_output.append(1, (unsigned)tmp);
std::cout << "cout+for: " << (unsigned)tmp << std::endl;
if(i<9)
str_output.append(1, '-');
}
std::cout << std::endl << "cout+str_append: " << str_output << std::endl;
return 0;
}
Compiling and running the program gives the following output:
cout+for: 65
cout+for: 66
cout+for: 67
...
cout+str_append: A-B-C-D-E-F-G-H-I-J
The desired output is:
cout+str_append: 65-66-67-68-...
The method string::append accepts, among the various overload, a size_t and a char, see reference.
string& append (size_t n, char c);
Therefore, in your code line
str_output.append(1, (unsigned)tmp);
you are implicitly converting the unsigned tmp to a char, i.e., to a single letter. To obtain the output you want, you have to convert tmp to a string containing the number, and then append it to str_output. You can do that by using
str_output+=std::to_string(tmp);
instead of str_output.append(1, (unsigned)tmp);.
You have to change your string append to for the change from a number to its "string":
str_output.append(std::to_string(tmp));
It's not one character that you want to add, but a string representing the number.
I am writing a code where I take user user text input, convert it to binary, store each binary character in an element in an array and then print A or T for 0 and G or C for 1 at random. But the ATGC seem to not follow this rule and they come at random for every digit; 0 and 1. So If the binary is 0010101 I need output as ATGACTG. Also when I store the binary in an int variable, the zero in front of it vanishes. Is there a way to keep it?
#include <iostream>
#include <cstdlib>
#include <bitset>
#include <string>
#include <ctime>
int main()
{
using namespace std;
int p, i=0, a[100000];
int s;
string myString;
int binary;
cout << "Type your text: ";
std::getline (std::cin,myString);
for (std::size_t k=0; k < myString.size(); ++k)
{
std::bitset<8> y(myString[k]);
std::string dna = y.to_string();
binary = atoi(dna.c_str());
cout << binary;
while (binary != 0)
{
a[i] = binary % 10;
binary = binary / 10;
i++;
}
}
std::cout << std::endl;
srand(time(0));
for (int j = (i-1); j>-1; j--)
{
if (a[j] == 0)
{
p = rand() %2;
if (p==0)
cout<< "A";
else
cout<< "T";
}
if (a[j] == 1)
{
s = rand() %2;
if (s == 0)
cout<< "G";
else
cout<< "C";
}
else
{
cout << "";
}
}
}
I don't know why exactly you wrote so much wrong code, but I've managed to extract (and change) the code that actually does the job.
#include <iostream>
#include <string>
#include <bitset>
#include <ctime>
int main()
{
int i = 0, a[8];
std::string myString;
std::cout << "Type your text: " << std::endl;
std::getline(std::cin, myString);
for(auto x : std::bitset<8>(myString).to_string())
a[i++] = x == '1';
std::cout << std::endl;
srand(time(0));
for(int j = 0; j < i; ++j)
if(a[j] == 0)
std::cout << (rand() % 2 ? "T" : "A");
else if(a[j] == 1)
std::cout << (rand() % 2 ? "C" : "G");
std::cout << std::endl;
}
And here's neater version of main:
int main()
{
std::vector<int> a; // using std::vector
std::bitset<8> bs;
std::cout << "Type your text: " << std::endl;
std::cin >> bs; // std::bitset can be read from stream via operator>>
for(auto x : bs.to_string())
a.push_back(x == '1');
std::cout << std::endl;
srand(time(0));
for(auto x : a)
if(x == 0)
std::cout << (rand() % 2 ? "T" : "A");
else if(x == 1)
std::cout << (rand() % 2 ? "C" : "G");
std::cout << std::endl;
}
Just ask if you want an explanation on some specific part.
I told you not to convert the string to an integer. You didn't listen. This is why leading 0 vanishes.
Your output seams to be completely random because you reverse the order of characters in the sequence when reading the information from a.
Here is how I'd solve your problem: run online
#include <iostream>
#include <string>
#include <bitset>
#include <ctime>
#include <cstdlib>
int main()
{
std::cout << "Type your text: " << std::endl;
std::string in_str;
std::getline(std::cin, in_str);
std::string binary_str;
for(int i = 0; i < in_str.size(); ++i)
{
char c = in_str.at(i);
binary_str.append(std::bitset<8>(c).to_string());
}
std::cout << binary_str << std::endl;
srand(time(0));
for(int i = 0; i < binary_str.size(); ++i)
{
char c = binary_str.at(i);
if(c == '0')
std::cout << (rand() % 2 ? "T" : "A");
else
std::cout << (rand() % 2 ? "C" : "G");
}
std::cout << std::endl;
}
If you have any questions, ask me in the comments.
Edit: the OP asked me to explain all mistakes in his program.
Where did all those zeros gone?
To answer this question I'll have to explain all things your program does line-by-line.
Here you convert a symbol to a bitset:
std::bitset<8> y(myString[k])
For example: if k is 'a', then the y would be 01100001.
Here you convert the bitset to a string:
std::string dna = y.to_string();
In our example the dna would be "01100001".
Here you convert the string to an integer:
binary = atoi(dna.c_str());
A very simplified version of what atoi does:
binary = 0;
for(int i = 0; i < dna.size(); ++i)
binary = binary * 10 + (dna.at(i) - '0')
In our example the binary would be 1100001.
Note: that's NOT where you loose zeros. At this point you are still able to extract them because you know that you need to extract 8 digits. So you can append leading zeros to up it's length to 8.
The next line is where you actually loose zeros the first time because cout doesn't know that you want to print 8 digits.
cout << binary;
In our example it would print 1100001.
And here you loose zeros again because you stop extracting digits as soon as binary == 0 even if you extracted less than 8 digits. Also note that you are actually reversing what the function atoi just did with the only difference that you don't get your leading zeros back and the reverse order of bits (see the next paragraph):
while (binary != 0)
{
a[i] = binary % 10;
binary = binary / 10;
i++;
}
Why the output is "random"?
Here you are iterating through myString in the standard order
for (std::size_t k=0; k < myString.size(); ++k)
e.g. if myString is "abc" than
in the first iteration myString[k] would be 'a'
in the second iteration myString[k] would be 'b'
in the third iteration myString[k] would be 'c':
But in this loop you extract digits in reverse order:
while (binary != 0)
{
a[i] = binary % 10;
binary = binary / 10;
i++;
}
eg if binary is 1100001
in the 1st iteration you extract 1 and binary becomes 110000
in the 2nd iteration you extract 0 and binary becomes 11000
in the 3rd iteration you extract 0 and binary becomes 1100
in the 4th iteration you extract 0 and binary becomes 110
in the 5th iteration you extract 0 and binary becomes 11
in the 6th iteration you extract 1 and binary becomes 1
in the 7th iteration you extract 1 and binary becomes 0
Now you end up with an array where bits inside a character code are reversed, but different characters are stored in the array in the normal order.
e.g. If the input string was "abc", then a would become:
1,0,0,0,0,1,1, 0,1,0,0,0,1,1, 1,1,0,0,0,1,1
reversed 'a' reversed 'b' reversed 'c'
If you iterate through a in normal order, the order of bits inside character codes would be reversed. If you iterate through a in reverse order, you get the reversed order of characters.
As a rule of thumb: don't program by guessing, program by thinking.
Further reading
The Zen of Python. Most of this aphorisms are applicable to every programming language with the exception of Brainfuck
Raw C arrays are evil
I have the following code :
string a = "wwwwaaadexxxxxx";
Intended Output : "w4a3d1e1x6";
somewhere in my code I have int count = 1; ... count++;
Also, somewhere in my code I have to print this count as a[i] but as a number only .. like 1,2,3 and not the character equivalent of 1,2,3.
I am trying the following : printf("%c%d",a[i],count);
i also read something like :
stringstream ss;
ss << 100
What is the correct way to do so in CPP?
EDIT :
so i Modified the code to add a number at index i in a string as :
stringstream newSS;
newSS << count;
char t = newSS.str().at(0);
a[i] = t;
You can use a stringstream to concatenate the string and the count,
stringstream newSS;
newSS << a[i] << count;
and then finally convert it to string and then print it or return (if this is done inside a function)
string output = newSS.str();
cout << output << endl;
But if your objective is only to print the output, then using the printf is fine.
If you need to update in place, then you can use two pointers. Let them be i,j.
You use j to set the new value and i to count the count. This is the standard runLength Encoding Problem.
There is no "correct" way. You could use snprintf, stringstream, etc. Or you could roll your algorithm. Assuming this is a base 10 number, you want the number in base 10.
#include <iostream>
#include <string>
#include <algorithm>
int main(void)
{
int a = 1194;
int rem = 0;
std::string output;
do {
rem = a % 10;
a = a / 10;
output.append(1, rem + '0');
} while(a != 0);
std::reverse(output.begin(), output.end());
std::cout << output << std::endl;
return 0;
}
Hello i want to convert two characters at a time in a string to binary? how can i do that by applying simple arithmetic (that is by making my own function?)
For example: our string is = hello world:
Desired output (two characters at a time):
he // need binaryform of 0's and 1's (16 bits for 2 characters 'h' and 'e'
ll // similarly
o(space) // single space also counts as a character with 8 zero bit in binary.
wo
rl
d(space) // space equals a character again with 8 zero bits
how to go about with it. i dont want any ascii in between. directly from character to binary...is that possible?
If you're looking for a way to textually represent the binary representation of characters, then here's a small example of how you can do it:
A small function that prints out the binary representation of c to std::cout (will only work for standard ASCII letters):
void printBinary(char c) {
for (int i = 7; i >= 0; --i) {
std::cout << ((c & (1 << i))? '1' : '0');
}
}
Use it like this (will only print out pairs of characters):
std::string s = "hello "; // Some string.
for (int i = 0; i < s.size(); i += 2) {
printBinary(s[i]);
std::cout << " - ";
printBinary(s[i + 1]);
std::cout << " - ";
}
Outputs:
01101000 - 01100101 - 01101100 - 01101100 - 01101111 - 00100000 -
Edit:
Actually, using std::bitset this is all that is needed:
std::string s = "hello "; // Some string.
for (int i = 0; i < s.size(); i += 2) {
std::cout << std::bitset<8>(s[i]) << " ";
std::cout << std::bitset<8>(s[i + 1]) << " ";
}
Outputs:
01101000 01100101 01101100 01101100 01101111 00100000
If you want to store the binary numbers of the character pairs in a std::vector, as mentioned in a comment, then this will do it:
std::vector<std::string> bitvec;
std::string bits;
for (int i = 0; i < s.size(); i += 2) {
bits = std::bitset<8>(s[i]).to_string() + std::bitset<8>(s[i + 1]).to_string();
bitvec.push_back(bits);
}
This can be accomplished quickly and easily using the bitset class in the C++ STL.
Below is a function that you can use:
#include <string>
#include <bitset>
string two_char_to_binary(string s) // s is a string of 2 characters of the input string
{
bitset<8> a (s[0]); // bitset constructors only take integers or string that consists of 1s and 0s e.g. "00110011"
bitset<8> b (s[1]); // The number 8 represents the bit depth
bitset<16> ans (a.to_string() + b.to_string()); // We take advantage of the bitset constructor that takes a string of 1s and 0s and the concatenation operator of the C++ string class
return ans.to_string();
}
Sample Usage:
using namespace std;
int main(int argc, char** argv)
{
string s = "hello world";
if(s.length() % 2 != 0) // Ensure string is even in length
s += " ";
for(int i=0; i<s.length(); i += 2)
{
cout << two_char_to_binary(s.substr(i, 2)) << endl;
}
return 0;
}
I guess the thing you're looking for is casting. Try like this:
char *string = "hello world ";
short *tab = (short*)string;
for(int i = 0; i < 6; i++)
std::cout << tab[i] << std::endl;