How can I add letters in a sentence? [closed] - c++

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I've asked to write code that gets a char array(sentence), if the there is an 'i' in the sentence I need to add the letter 'b' the letter 'i' again like this example:
pig -> pibig
I tried to use string.h functions but I didn't succeed to make it right.

Use std::string in string header file, and std::string::insert whenever you need to insert a char in string:
std::string my_string = "my satringa";
for (size_t i = 0; i < my_string.length(); ++i)
{
if (my_string.at(i) == 'a')
{
my_string.insert(i + 1, "b");
}
}
std::clog << my_string << std::endl;
Output:
> my sabtringab
If you are forced to use C-style strings, don't worry do all of your operations on std::string and then take the underlying stored string with std::string::c_str() as a C-style string (and don't forget to take a copy).

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how to write correctly into multidimentional char array unknown amount of values but fixed amound of chars [closed]

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I'm sick and tired of solving why my ch[0] is of value "Thomas EdisonÇ#", when it should be "Thomas Edison"
int main(){
using namespace std;
ifstream in("U2.txt");
int n;
in>>n; //n=rows, so in every line there will be "name surname", time, money
char ch[n][21]; //I'm trying to get Name+Surname which must be 20 char long
in.read(ch[0], 20);
cout << ch[0]; //but getting Thomas EdisonÇ#
return 0;}
It works on one dimentional ch[21], but there's gonna be lots of values so I want to use ch[n][21]
Any other out of my box solution is welcome, I'm tired
You are forgetting that C strings need to be nul terminated
in.read(ch[0], 20);
ch[0][20] = '\0'; // add the nul terminator
cout << ch[0]; // now correct output

Making a password type program that needs to accept letter and number combinations? [closed]

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I am making a shopping list program. For this program, I need to be able to type in a user input that accepts both number (1564, 121,1, etc) and word (hello, goodbye, etc) combinations. The program reads numbers just fine, but it cannot process words. Thank you in advance. The part of the code I am stuck with is below:
int code, option, count = 0;
double quantity, price, cost;
string description;
cin >> code;
while ((code != 123456789) && (count < 2))
{
cout << "Incorrect code, try again \n";
cin >> code;
count++;
if (count == 2)
{
cout << "max # of tries reached. Goodbye. \n";
system("pause");
}
}
Your code variable is now an int. If you wanted that to be a string, declare it so: std::string code;. Note that you might need to #include <string> in the very beginning. Also, if you want to compare it with numbers, either you call something like atoi() (string has .cstr()), or better yet, you might just compare it with "123456789". HTH.

Deleting undesired characters at the end of a char [closed]

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With this code below I dont know how to delete the undesired characters appearing at the end of the message array. It is compulsory for me to use char, can't use strings, because of the rest of my code.
recvbuf is also a char* recvbuf=new char
char* message=new char[140];
for (int i=1; i<141; i++){
message[i-1]=recvbuf[i];
}
printf("Message: %s\n", message);
delete[]recvbuf;
Though it is recommended you use strings to implement this code, the problem can be fixed by manually appending a null character \0 at the end of your char array.
You can introduce it as:
char* message=new char[141];
for (int i=1; i<141; i++){
message[i-1]=recvbuf[i];
}
message[140] = '\0'; //newly introduced line.
printf("Message: %s\n", message);
delete[]recvbuf;
NOTE 1: The size of the array was increased from 140 to 141 during initialization to make room for the \0 character at the end.
Cheers!

Modifying specific characters in text input (C++) [closed]

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I receive text with special characters (such as á) so I have to manually search and replace each one with code (in this case "á")
I would like to have code to search and replace such instances automatically after user input. Since I'm a noob, I'll show you the code I have so far - however meager it may be.
// Text fixer
#include <iostream>
#include <fstream>
#include <string>
int main(){
string input;
cout << "Input text";
cin >> input;
// this is where I'm at a loss. How should I manipulate the variable?
cout << input;
return 0;
}
Thank you!
An easy method is to use an array of substitution strings:
std::string replacement_text[???];
The idea is that you use the incoming character as the index into the array and extract the replacement text.
For example:
replacement_text[' '] = " ";
// ...
std::string new_string = replacement_text[input_character];
Another method is to use switch and case to convert the character.
Alternative techniques are a lookup table and std::map.
The lookup table could be an array of mapping structures:
struct Entry
{
char key;
std::string replacement_text;
}
Search the table using the key field to match the incoming character. Use the replacement_text to get the replacement text.

How to use regex in C++? [closed]

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I have a string like below:
std::string myString = "This is string\r\nIKO\r\n. I don't exp\r\nO091\r\nect some characters.";
Now I want to get rid of the characters between \r\n including \r\n.
So the string must look like below:
std::string myString = "This is string. I don't expect some characters.";
I am not sure, how many \r\n's going to appear.
And I have no idea what characters are coming between \r\n.
How could I use regex in this string?
Personally, I'd do a simple loop with find. I don't see how using regular expressions helps much with this task. Something along these lines:
string final;
size_t cur = 0;
for (;;) {
size_t pos = myString.find("\r\n", cur);
final.append(myString, cur, pos - cur);
if (pos == string::npos) {
break;
}
pos = myString.find("\r\n", pos + 2);
if (pos == string::npos) {
// Odd number of delimiters; handle as needed.
break;
}
cur = pos + 2;
}
Regular expressions are "greedy" by default in most regex libaries.
Just make your regex say "\r\n.*\r\n" and you should be fine.
EDIT
Then split your input using the given regex. That should yield two strings you can combine into the desired result.