How can i do a Bitwise OR on strings?
A:
10001
01010
------
11011
Why on strings?
The Bits can have length of 40-50.Maybe this could be problematic on int ?
Any Ideas ?
I would say std::bitset is more than enough for your situation, but for more flexibility you can use boost::dynamic_bitset. Here is an example on std::bitset:
const size_t N = 64;
string a_str = "10001", b_str = "01010";
bitset<N> a(a_str), b(b_str);
bitset<N> c = a | b;
cout << c;
You should take a look at the C++ std::bitset class, which does exactly what you want.
For each char:
char result = (a - '0') | (b - '0') + '0';
Where a and b are two chars with ascii character 0 or 1 in them.
Why not just use a vector of int values? Doesn't the bitset still use a byte per bit?
You can also use a vector of bool values, but this is also implementation specific.
Depending on whether you need storage efficiency or speed (or the utility of container methods that a couple of these approaches lack) you might profile to decide which approach to use.
This is similar to Andreas Brinck's answer, only it returns a full output string and can compare strings of different (arbitrary) lengths.
Example in C# (not near c++ compiler right now), but it should be simple to convert it to a language of your choice.
public static string BitwiseOr(string input1, string input2)
{
char[] inarr1 = (char[])input1.ToCharArray().Reverse().ToArray();
char[] inarr2 = (char[])input2.ToCharArray().Reverse().ToArray();
char[] outarr = new char[input1.Length > input2.Length ? input1.Length : input2.Length];
for (int i = 0; i < outarr.Length ; i++)
{
char c1 = i < input1.Length ? inarr1[i] : '0';
char c2 = i < input2.Length ? inarr2[i] : '0';
outarr[i] = (char)((c1 - '0') | (c2 - '0') + '0');
}
return new string((char[])outarr.Reverse().ToArray());
}
Of course this is only valid if you really need it to be in a string, if not you should (as suggested in other answers) use a vector or similar data type.
Related
Well currently I am re creating my own version of enigma as a little project but if you understand how the enigma machine works it has rotors which connect a character to a completely different character for example A might be connected to F or U may be connected to C and this is done three times. Currently I am getting the char for the rotor by using this function:
char getRotorOne(char i) {
if(i == 'a') {
return 'g';
}if(i == 'b') {
return 'A';
}if(i == 'c') {
return 'o';
}
The main problem with this is it takes a long time to write and it seems inefficient and I think there must be a better way. The other problem with this is on the original enigma machine there were only the 26 letters of the alphabet on this there are the 94 tapeable characters of ascii (32-126) is there any other simpler way that this can be done? If you think this question is unclear or you don't understand please tell me instead of just flagging my post, then you can help me improve my question.
Use tables! Conveniently, C string literals are arrays of characters. So you can do this:
// abc
const char* lower_mapping = "gAo";
// ABC
const char* upper_mapping = "xyz";
char getRotorOne(char i) {
if (i >= 'a' && i <= 'z') return lower_mapping[i - 'a'];
if (i >= 'A' && i <= 'Z') return upper_mapping[i - 'A'];
assert(false && "Unknown character cannot be mapped!");
}
Since chars are really just small integers, and ASCII guarantees contiguous ranges for a-z and A-Z (and 0-9) you can subtract from a given character the first one in its range (so, 'a' or 'A') to get an index into that range. That index can then be used to look up the corresponding character via a table, which in this case is just a simple hardcoded string literal.
This is an improvement on Cameron's answer. You should use a simple char array for each rotor, but as you said you want to process ASCII characters in the range 32-126, you should build each mapping as an array of 95 characters:
char rotor1[95] ="aXc;-0..."; // the 95 non control ascii characters in arbitrary order
Then you write your rotor function that way:
char getRotorOne(char i) {
if ((i < 32) || (i > 126)) return i; // do not change non processed characters
return rotor1[i - 32]; // i - 32 is in range 0 - 94: rotor1[i - 32] is defined
}
This question already has answers here:
C++ - how to find the length of an integer
(17 answers)
Closed 7 years ago.
In Java, I use
int length = String.valueOf(input).length();
to find the length of an integer.
My question is: Are there any similar ways to do so in C++?
I have already tried the for loops and while loops such as:
while (input > 0){
input/=10;
count++;
So, apart from the loops are there anything else available in C++. Thank you for your answer.
If you want an exact counterpart of what you have written in Java, you can use:
int length = to_string(input).length();
Note that to_string is a C++11 feature. Also, be careful with negative numbers.
The number of digits can be calculated without converting to a string first by using the number's logarithm:
std::size_t intlen(int i) {
if (i == 0) return 1;
else if (i < 0) return 2 + static_cast<std::size_t>(std::log10(-i));
else if (i > 0) return 1 + static_cast<std::size_t>(std::log10(i));
}
The logartihm is only defined for positive numbers, so negatives and zero have to be handled separately, counting the - sign as an additional character. Replace log10 by log2 to obtain the number of binary digits (this is possible for any base).
Note however that converting to strings first (e.g. by using std::to_string) is a locale-dependent operation and can thus yield different results for different language settings - some locales insert a thousands separator (e.g. 100,000) which will not show up using the above formula.
unsigned int number_of_digits = 0;
do {
++number_of_digits;
n /= base; } while (n);
// n is your base number.
Talking about pre-C++11, you can use the same approach, but with sprintf.
Convert integer to a char array, and then get its length:
char buffer[30];
int length = sprintf(buffer, "%d", input);
Here is the working IDEOne example.
Apart from the loops there is recursion. For example, for positive integers you can do:
unsigned int len(unsigned int n)
{
return n ? len(n/10)+1 : 0;
}
The question is simple. I'm looking for an easy and efficient way of joining an array of strings (or arrays of any other type, for that matter, since strings are an alias for char[]), with an optional separator.
In JavaScript, this functionality would already exist with the join method. Being new to D, I failed to find something as easy as that in the standard library. It would be too bad if I had to implement a utility function myself.
So instead of something like this:
string merge (const string arr[] , const string separator) {
if (arr.length == 0) return "";
string r = arr[0];
for (int i = 1 ; i < arr.length ; i++) {
r ~= separator ~ arr[i];
}
return r;
}
What would an experienced D programmer do?
I'm not a D programmer, but I'll take a crack at it, the library reference has a join method.
From the docs:
const string[] arr = ["apple", "banana"];
assert(arr.join(",") == "apple,banana");
assert(arr.join() == "applebanana");
See also std.algorithm.joiner for a version that is lazy and doesn't allocate any memory.
PROBLEM SOLVED: thanks everyone!
I am almost entirely new to C++ so I apologise in advance if the question seems trivial.
I am trying to convert a string of letters to a set of 2 digit numbers where a = 10, b = 11, ..., Y = 34, Z = 35 so that (for example) "abc def" goes to "101112131415". How would I go about doing this? Any help would really be appreciated. Also, I don't mind whether capitalization results in the same number or a different number. Thank you very much in advance. I probably won't need it for a few days but if anyone is feeling particularly nice how would I go about reversing this process? i.e. "101112131415" --> "abcdef" Thanks.
EDIT: This isn't homework, I'm entirely self taught. I have completed this project before in a different language and decided to try C++ to compare the differences and try to learn C++ in the process :)
EDIT: I have roughly what I want, I just need a little bit of help converting this so that it applies to strings, thanks guys.
#include <iostream>
#include <sstream>
#include <string>
int returnVal (char x)
{
return (int) x - 87;
}
int main()
{
char x = 'g';
std::cout << returnVal(x);
}
A portable method is to use a table lookup:
const unsigned int letter_to_value[] =
{10, 11, 12, /*...*/, 35};
// ...
letter = toupper(letter);
const unsigned int index = letter - 'A';
value = letter_to_value[index];
cout << index;
Each character has it's ASCII values. Try converting your characters into ASCII and then manipulate the difference.
Example:
int x = 'a';
cout << x;
will print 97; and
int x = 'a';
cout << x - 87;
will print 10.
Hence, you could write a function like this:
int returnVal(char x)
{
return (int)x - 87;
}
to get the required output.
And your main program could look like:
int main()
{
string s = "abcdef"
for (unsigned int i = 0; i < s.length(); i++)
{
cout << returnVal(s[i]);
}
return 0;
}
This is a simple way to do it, if not messy.
map<char, int> vals; // maps a character to an integer
int g = 1; // if a needs to be 10 then set g = 10
string alphabet = "abcdefghijklmnopqrstuvwxyz";
for(char c : alphabet) { // kooky krazy for loop
vals[c] = g;
g++;
}
What Daniel said, try it out for yourself.
As a starting point though, casting:
int i = (int)string[0] + offset;
will get you your number from character, and: stringstream will be useful too.
How would I go about doing this?
By trying to do something first, and looking for help only if you feel you cannot advance.
That being said, the most obvious solution that comes to mind is based on the fact that characters (i.e. 'a', 'G') are really numbers. Suppose you have the following:
char c = 'a';
You can get the number associated with c by doing:
int n = static_cast<int>(c);
Then, add some offset to 'n':
n += 10;
...and cast it back to a char:
c = static_cast<char>(n);
Note: The above assumes that characters are consecutive, i.e. the number corresponding to 'a' is equal to the one corresponding to 'z' minus the amount of letters between the two. This usually holds, though.
This can work
int Number = 123; // number to be converted to a string
string Result; // string which will contain the result
ostringstream convert; // stream used for the conversion
convert << Number; // insert the textual representation of 'Number' in the characters in the stream
Result = convert.str(); // set 'Result' to the contents of the stream
you should add this headers
#include <sstream>
#include <string>
Many answers will tell you that characters are encoded in ASCII and that you can convert a letter to an index by subtracting 'a'.
This is not proper C++. It is acceptable when your program requirements include a specification that ASCII is in use. However, the C++ standard alone does not require this. There are C++ implementations with other character sets.
In the absence of knowledge that ASCII is in use, you can use translation tables:
#include <limits.h>
// Define a table to translate from characters to desired codes:
static unsigned int Translate[UCHAR_MAX] =
{
['a'] = 10,
['b'] = 11,
…
};
Then you may translate characters to numbers by looking them up in the table:
unsigned char x = something;
int result = Translate[x];
Once you have the translation, you could print it as two digits using printf("%02d", result);.
Translating in the other direction requires reading two characters, converting them to a number (interpreting them as decimal), and performing a similar translation. You might have a different translation table set up for this reverse translation.
Just do this !
(s[i] - 'A' + 1)
Basically we are converting a char to number by subtracting it by A and then adding 1 to match the number and letters
We have this example:
struct X {
int e0 : 6;
int e1 : 6;
int e2 : 6;
...
int e10 : 6;
};
struct X c;
How can I access the members "automatically", something like that:
c.e{0-10} ?
Say if I want to read c.e0, then c.e1 ...
If my struct would have 1000 elements, I do not think that I should write so much code, right ?
Can you help me with a workaround, an idea ?
I mention that I already read other posts related somehow to this problem, but I did not find a solution.
Thank you very much !
As others have said, you cannot do exactly what you want with bit fields. It looks like you want to store a large number of 6 bit integers with maximum space efficiency. I will not argue whether this is a good idea or not. Instead I will present an old-school C like way of doing exactly that, using C++ features for encapsulation (untested). The idea is that 4 6 bit integers require 24 bits, or 3 characters.
// In each group of 3 chars store 4 6 bit ints
const int nbr_elements = 1000;
struct X
{
// 1,2,3 or 4 elements require 3 chars, 5,6,7,8 require 6 chars etc.
char[ 3*((nbr_elements-1)/4) + 3 ] storage;
int get( int idx );
};
int X::get( int idx )
{
int dat;
int offset = 3*(idx/4); // eg idx=0,1,2,3 -> 0 idx=4,5,6,7 -> 3 etc.
char a = storage[offset++];
char b = storage[offset++];
char c = storage[offset];
switch( idx%4) // bits lie like this; 00000011:11112222:22333333
{
case 0: dat = (a>>2)&0x3f; break;
case 1: dat = ((a<<4)&0x30) + ((b>>4)&0x0f); break;
case 2: dat = ((b<<2)&0x3c) + ((c>>6)&0x03); break;
case 3: dat = c&0x3f; break;
}
return dat;
}
I will leave the companion put() function as an exercise.
It sounds like a struct isn't the right tool for what you're trying to do. You need either an array or a vector. Arrays are used for storing a number of the same type of data. Vectors are array wrappers that manage the addition and removal of items automatically.
If you need a list of the same data, and some other data (say a string) you can make an array or a vector part of your struct.
struct X {
int[10] numbs;
string name;
};
X c;
You can't. To do this would require some form of reflection, which is not supported in either C or C++.
Since your bitfields are of the same size, you could encapsulate std::bitset (or vector<bool>, gulp...) and provide your own iterators (each increment moving the bookmark six bits) and operator[] (etc) to allow your code to be more simple to write.
I am sure performance would suck compared to the bitfields though.
How about something like this:
char getByte(char *startPos, int index) {
int i = (index*6) / 8;
if (index % 4 == 0)
return 0b11111100 & startPos[i] >> 2;
else if (index % 4 == 3)
return 0b00111111 & startPos[i];
else if (index % 4 == 2)
return (0b00001111 & startPos[i] << 2) | (0b11000000 & startPos[i+1] >> 6);
else
return (0b00000011 & startPos[i] << 4) | (0b11110000 & startPos[i+1] >> 4);
}