After calculating a number, I need to get several digits of this one. I cast it to string, and print a digit, but in one piece of code, it works correct (prints only one digit), but after I equate it to another variable, the program prints 2 digits.
#include "stdafx.h"
#include <iostream>
#include <chrono>
#include <ctime>
#include <sstream>
using namespace std;
int main()
{
time_t seconds;
int res = 0;
seconds = time(NULL);
double nump = seconds;
cout.precision(45);
for (int i = 1; i <= 100; i++) {
nump = nump /10;
}
std::ostringstream strs;
strs.precision(55);
strs << nump;
std::string str = strs.str();
cout << str[str.size() - 9] << endl; // here we get a gidit from the string (e.g. 5)
res = str[str.size() - 9];
cout << res << endl; // here we get a number (e.g. 49)
system("pause");
return 0;
}
I cant understand whats going on. Please help!
That's because here
res = str[str.size() - 9];
You are storing the value of a char in an int. Printing the same value as int can have different results when you send it to std::cout than when you print it as char. For integers this operator is called, while for chars this operator is called instead.
In your example, you probably have a value of '1' (which is 49 in ASCII). When you print it as a char it prints 1, and when you print it as an int, it prints 49.
One way to solve this problem is to make int res be a char instead.
Related
I have a homework assignment. The input is a three-digit number. Print the arithmetic mean of its digits. I am new to C++ and cannot write the code so that it takes 1 number as input to a string. I succeed, only in a column.
#include <iostream>
int main()
{
int a,b,c;
std::cin >> a >> b >> c;
std::cout << (a+b+c)/3. << std::endl;
return 0;
}
If you write it in Python it looks like this. But I don't know how to write the same thing in C ++ :(
number = int(input())
digital3 = number % 10
digital2 = (number//10)%10
digital1 = number//100
summ = (digital1+digital2+digital3)/3
print(summ)
The most direct translation from Python differs mostly in punctuation and the addition of types:
#include <iostream>
int main()
{
int number;
std::cin >> number;
int digital3 = number % 10;
int digital2 = (number/10)%10;
int digital1 = number/100;
int summ = (digital1+digital2+digital3)/3;
std::cout << summ << std::endl;
}
In your code, you use three different numbers and take the mean of their sum (not the sum of three-digits number). The right way is:
#include <iostream>
int main()
{
int a;
std::cin >> a;
std::cout << ((a/100) + ((a/10)%10) + (a%10))/3.<< std::endl;
return 0;
}
EDIT: This answer is incorrect. I thought the goal was to average three numbers. Not three DIGITS. Bad reading on my part
*Old answer *
I'm not sure I'm interpreting the question correctly. I ran your code
and confirmed it does what I expected it to...
Are you receiving three digit chars (0-9) and finding the average of
them? If so, I'd trying using a
for loop using getChar()
Here is a range of functions that may be of use to you.
Regex strip
Convert string to int: int myInt = stoi(myStr.c_str())
Convert int to string: std::string myStr = myInt.to_string()
If you need to improve your printing format
Using printf
If using cout, you can kindve hack your way through it!
The input is a three-digit number.
If it means, you'll be given a number that will always have 3 digits, then you can try the following approach.
Separate each digit
Find all digits sum
Divide the sum by 3
If you're given the number as a string, all you've to do is convert that string into int. Rest of the approach is the same as abve.
Sample code:
int main()
{
int a;
std::cin >> a;
int sum = (a % 10); // adding 3rd digit
a /= 10;
sum += (a % 10); // adding 2nd digit
a /= 10;
sum += (a % 10); // adding 1st digit
std::cout << (double)sum / 3.0 << std::endl;
return 0;
}
Here's a possible solution using std::string:
EDIT added digits check
#include <iostream>
#include <string>
#include <cctype>
int main()
{
std::string s;
std::cin >> s;
if(s.length() == 3 && isdigit(s[0]) && isdigit(s[1]) && isdigit(s[2]))
{
std::cout<<double(s[0] + s[1] + s[2])/3 - '0'<<std::endl;
}
else
{
std::cout<<"Wrong input"<<std::endl;
}
return 0;
}
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'm working on this code that takes a numeric string and fills an array with each "digit" of the string. The issue I'm having is trying to convert an integer to a string. I tried using to_string to no avail.
Here is the code (note this is pulled from a larger program with other functions):
#include <cstdlib>
#include <stdlib.h>
#include <iostream>
#include <time.h>
#include <typeinfo>
int fillarr(int &length) {
int arr[length];
string test = "10010"; //test is an example of a numeric string
int x = 25 + ( std::rand() % ( 10000 - 100 + 1 ) );
std::string xstr = std::to_string(x); //unable to resolve identifier to_string
cout << xstr << endl;
cout << typeid (xstr).name() << endl; //just used to verify type change
length = test.length(); //using var test to play with the function
int size = (int) length;
for (unsigned int i = 0; i < test.size(); i++) {
char c = test[i];
cout << c << endl;
arr[int(i)] = atoi(&c);
}
return *arr;
}
How can I convert int x to a string? I have this error: unable to resolve identifier to_string.
As mentioned by user 4581301, you need an #include <string> to use string functions.
The following, though is wrong:
arr[int(i)] = atoi(&c);
The atoi() will possibly crash because c by itself is not a string and that mean there will be no null terminator.
You would have to use a buffer of 2 characters and make sure the second one is '\0'. Something like that:
char buf[2];
buf[1] = '\0';
for(...)
{
buf[0] = test[i];
...
}
That being said, if your string is decimal (which is what std::to_string() generates) then you do not need atoi(). Instead you can calculate the digit value using a subtraction (much faster):
arr[int(i)] = c - '0';
Okay I modified my code a bit per suggestion from everyone and ended up handling the conversion like this:
string String = static_cast<ostringstream*>( &(ostringstream() << x) )->str();
cout << String << endl;
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;
}