Before you read ahead or try to help, this question is regarding my homework so the requirements to this question will be very specific.
I am writing a code that takes a user input between 0 and 511 and converts it into a binary number. Then the program will replace all the 1's in the binary number with T and all the 0's in the number as H. Afterwards it will print out the results (the binary number with the H and T replacement) as a 3*3 matrix.
This is the desired output (not what I have but what I want):
Enter a number between 0 and 511: 299
The binary number is: 100101011
The matrix is:
THH
THT
HTT
The problem with my code is that I am unsure of how to replace an array that consists of all integers to have certain parts of the index to be either characters or strings. For sure the part with the binary number conversion works but the replacement of the 0's and 1's of the array is where the trouble is at. I am also unsure of how to print out the matrix result. I assume it goes either of 2 ways: 1. The program creates a new array for the previous array's elements stored and prints out the matrix array instead. 2. There is a way to only print the array 3 lines at a time. The only way I can think of is to somehow cut the for loop short and add a line break after every 3 values. I am aware that there are a few pointable errors in my code but I do not know how to fix them.
Although this is in the C++ language, what I have learned is the C style syntax (no std:: kinds of code or stuff like that because I haven't learned it yet and I will not understand it) So far I have learned basic arrays, loops, and functions.
#include <iostream>
using namespace std;
int main(){
int arr[10];
int input, i;
cout<<"Enter a number between 0 and 511: ";
cin>> input;
for(i = 0; input > 0; i++){
arr[i] = (input % 2);
input = input / 2;
}
cout<<"The binary number is: ";
for(i = i - 1; i >= 0; i--){
cout<<arr[i];
}
string newArr[10] = arr[10]; //the error here states that the array initializer must be an initializer list
for(int i = 0; i < sizeof(arr)/sizeof(arr[10]); i++){
if(arr[i] == 1){
arr[i] = "T"; //the error here mentions that a string/ character cannot be assigned with a integer array
}
else{
arr[i] = "H";
}
}
for(int i = 0; i < sizeof(arr)/sizeof(arr[10]); i++){
cout<<arr[i]<< " ";
}
}
This would be sufficient:
#include <iostream>
using namespace std;
int main()
{
// you never actually checked if the input is valid
// so you may or may not want this loop:
int input;
do
{
cout << "Enter a number between 0 and 511: ";
cin >> input;
} while ((input < 0) || (input > 511));
// space for matrix, new lines and null
// to construct a null terminated string
char buffer[3 * (3 + 1) + 1];
int i = 0;
// since the bits are read from left to right
// I will use a mask instead of bit shifting the input
int bit = 1 << 9;// 2^9 == 512
for (int r = 0; r < 3; r++)// rows
{
for (int c = 0; c < 3; c++)// columns
{
// this could come after the check
// and then bit would start at 256
bit >>= 1;
// perform the check and add the corresponding letter
buffer[i++] = (bit & input) ? 'T' : 'H';
}
// add new lines
buffer[i++] = '\n';
}
// if you don't want the last '\n'
// this could be { buffer[--i] = '\0'; }
buffer[i++] = '\0';
cout << buffer;
}
I have got some doubts while solving - Name That Number.
It goes like this -
Among the large Wisconsin cattle ranchers, it is customary to brand cows with serial numbers to please the Accounting Department. The cowhands don't appreciate the advantage of this filing system, though, and wish to call the members of their herd by a pleasing name rather than saying, "C'mon, #4734, get along."
Help the poor cowhands out by writing a program that will translate the brand serial number of a cow into possible names uniquely associated with that serial number. Since the cowhands all have cellular saddle phones these days, use the standard Touch-Tone(R) telephone keypad mapping to get from numbers to letters (except for "Q" and "Z"):
2: A,B,C 5: J,K,L 8: T,U,V
3: D,E,F 6: M,N,O 9: W,X,Y
4: G,H,I 7: P,R,S
Acceptable names for cattle are provided to you in a file named "dict.txt", which contains a list of fewer than 5,000 acceptable cattle names (all letters capitalized). Take a cow's brand number and report which of all the possible words to which that number maps are in the given dictionary which is supplied as dict.txt in the grading environment (and is sorted into ascending order).
For instance, brand number 4734 produces all the following names:
GPDG GPDH GPDI GPEG GPEH GPEI GPFG GPFH GPFI GRDG GRDH GRDI
GREG GREH GREI GRFG GRFH GRFI GSDG GSDH GSDI GSEG GSEH GSEI
GSFG GSFH GSFI HPDG HPDH HPDI HPEG HPEH HPEI HPFG HPFH HPFI
HRDG HRDH HRDI HREG HREH HREI HRFG HRFH HRFI HSDG HSDH HSDI
HSEG HSEH HSEI HSFG HSFH HSFI IPDG IPDH IPDI IPEG IPEH IPEI
IPFG IPFH IPFI IRDG IRDH IRDI IREG IREH IREI IRFG IRFH IRFI
ISDG ISDH ISDI ISEG ISEH ISEI ISFG ISFH ISFI
As it happens, the only one of these 81 names that is in the list of valid names is "GREG".
Write a program that is given the brand number of a cow and prints all the valid names that can be generated from that brand number or ``NONE'' if there are no valid names. Serial numbers can be as many as a dozen digits long.
Here is what I tried to solve this problem. Just go through all the names in the list and check which is satisfying the constraints given.
int numForChar(char c){
if (c=='A'||c=='B'||c=='C') return 2;
else if(c=='D'||c=='E'||c=='F') return 3;
else if(c=='G'||c=='H'||c=='I') return 4;
else if(c=='J'||c=='K'||c=='L') return 5;
else if(c=='M'||c=='N'||c=='O') return 6;
else if(c=='P'||c=='R'||c=='S') return 7;
else if(c=='T'||c=='U'||c=='V') return 8;
else if(c=='W'||c=='X'||c=='Y') return 9;
else return 0;
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
freopen("namenum.in","r",stdin);
freopen("namenum.out","w",stdout);
string S; cin >> S;
int len = S.length();
freopen("dict.txt","r",stdin);
string x;
while(cin >> x){
string currName = x;
if(currName.length() != S.length()) continue;
string newString = x;
for(int i=0;i<len;i++){
//now encode the name as a number according to the rules
int num = numForChar(currName[i]);
currName[i] = (char)num;
}
if(currName == S){
cout << newString << "\n";
}
}
return 0;
}
Unfortunately, when I submit it to the judge, for some reason, it says no output produced that is my program created an empty output file. What's possibly going wrong?
Any help would be much appreciated. Thank You.
UPDATE: I tried what Some Programmer Dude suggested by adding a statement else return 0; at the end of the numOfChar function in case of a different alphabet. Unfortunately, it didn't work.
So after looking further at the question and exploring the information for Name That Number. I realized that it is not a current contest, and just a practice challenge. Thus, I updated my answer and also giving you my version of a successful submission. Nonetheless, that is a spoiler and will be posted after why your code was not working.
First, you forgot a } after the declaration of your number function. Secondary, you did not implement anything to check whether if the input fail to yield a valid name. Third, when you use numForChar() on the character of currName, the function yielded an integer value. That is not a problem, the problem is that it is not the ASCII code but is a raw number. You then compare that against a character of the input string. Of which, is an ASCII's value of a digit. Thus, your code can't never find a match. To fix that you can just add 48 to the return value of the numForChar() function or xor the numForChar() return's value to 48.
You are on the right track with your method. But there is a few hints. If you are bored you can always skip to the spoiler. You don't need to use the numForChar() function to actually get a digit value from a character. You can just use a constant array. A constant array is faster than that many if loop.
For example, you know that A, B, C will yield two and A's ASCII code is 65, B's is 66, and C's equal to 67. For that 3, you can have an array of 3 indexes, 0, 1, 2 and all of them stores a 2. Thus, if you get B, you subtract B's ASCII code 65 will yield 1. That that is the index to get the value from.
For getting a number to a character you can have a matrix array of char instead. Skip the first 2 index, 0 and 1. Each first level index, contain 3 arrays of 3 characters that are appropriate to their position.
For dictionary comparing, it is right that we don't need to actually look at the word if the length are unequal. However, besides that, since their dictionary words are sorted, if the word's first letter is lower than the range of the input first letter, we can skip that. On the other hand, if words' first letter are now higher than the highest of the input first letter, there isn't a point in continue searching. Take note that my English for code commenting are almost always bad unless I extensively document it.
Your Code(fixed):
#include <iostream>
#include <fstream>
#include <string>
using namespace std;
int numForChar(char c){
if (c=='A'||c=='B'||c=='C') return 2;
else if(c=='D'||c=='E'||c=='F') return 3;
else if(c=='G'||c=='H'||c=='I') return 4;
else if(c=='J'||c=='K'||c=='L') return 5;
else if(c=='M'||c=='N'||c=='O') return 6;
else if(c=='P'||c=='R'||c=='S') return 7;
else if(c=='T'||c=='U'||c=='V') return 8;
else if(c=='W'||c=='X'||c=='Y') return 9;
else return 0;
}
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
ifstream fin("namenum.in");
ifstream dict("dict.txt");
ofstream fout("namenum.out");
string S;
fin >> S;
int len = S.length();
bool match = false;
string x;
while(dict >> x){
string currName = x;
if(currName.length() != S.length()) continue;
string newString = x;
for(int i=0;i<len;i++){
//now encode the name as a number according to the rules
int num = numForChar(currName[i]) ^ 48;
currName[i] = (char)num;
}
if(currName == S){
fout << newString << "\n";
match = true;
}
}
if ( match == false ){
fout << "NONE" << endl;
}
return 0;
}
Spoiler Code(Improved):
#include <fstream>
#include <string>
using namespace std;
// A = 65
// 65 - 0 = 65
const char wToN[] = {
// A ,B ,C ,D ,E ,F ,G ,H ,I ,
'2','2','2','3','3','3','4','4','4',
// J ,K ,L ,M ,N ,O ,P ,Q ,R ,S
'5','5','5','6','6','6','7','7','7','7',
// T ,U ,V ,W ,X ,Y ,Z
'8','8','8','9','9','9','9'
};
// 2 = {A, B, C} = 2[0] = A, 2[1] = B, 2[2] C
const char nToW[10][3] = {
{}, // 0 skip
{}, // 1
{'A','B','C'},
{'D','E','F'},
{'G','H','I'},
{'J','K','L'},
{'M','N','O'},
{'P','R','S'},
{'T','U','V'},
{'W','X','Y'}
};
int main(){
ifstream fin("namenum.in");
ifstream dict("dict.txt");
ofstream fout("namenum.out");
string S;
fin >> S;
// Since this will not change
// make this a const to make it
// run faster.
const int len = S.length();
// lastlen is last Index of length
// We calculate this value here,
// So we do not have to calculate
// it for every loop.
const int lastLen = len - 1;
int i = 0;
unsigned char digits[len];
unsigned char firstLetter[3];
// If not match print None
bool match = false;
for ( ; i < len; i++ ){
// No need to check upper bound
// constrain did not call for check.
if ( S[i] < '2' ) {
fout << "NONE" << endl;
return 0;
}
}
const char digit1 = S[0] ^ 48;
// There are 3 set of first letter.
// We get them by converting digits[0]'s
// value using the nToW array.
firstLetter[0] = nToW[digit1][0];
firstLetter[1] = nToW[digit1][1];
firstLetter[2] = nToW[digit1][2];
string dictStr;
while(dict >> dictStr){
// For some reason, when keeping the i = 0 here
// it seem to work faster. That could be because of compiler xor.
i = 0;
// If it is higher than our range
// then there is no point contineuing.
if ( dictStr[0] > firstLetter[2] ) break;
// Skip if first character is lower
// than our range. or If they are not equal in length
if ( dictStr[0] < firstLetter[0] || dictStr.length() != len ) continue;
// If we are in the letter range
// we always check the second letter
// not the first, since we skip the first
i = 1;
for ( int j = 1; j < len; j++ ){
// We convert each letter in the word
// to the corresponding int value
// by subtracting the word ASCII value
// to 65 and use it again our wToN array.
// if it does not match the digits at
// this current position we end the loop.
if ( wToN[dictStr[i] - 65] != S[j] ) break;
// if we get here and there isn't an unmatch then it is a match.
if ( j == lastLen ) {
match = true;
fout << dictStr << endl;
break;
}
i++;
}
}
// No match print none.
if ( match == false ){
fout << "NONE" << endl;
}
return 0;
}
I suggest you use c++ file handling. Overwriting stdin and stdout doesn't seem appropriate.
Add these,
std::ifstream dict ("dict.txt");
std::ofstream fout ("namenum.out");
std::ifstream fin ("namenum.in");
Accordingly change,
cin >> S --to--> fin >> S;
cin >> x --to--> dict >> x
cout << newString --to--> fout << newString
I have to convert a decimal value into a string that shows the binary value, e.g. given 8, I need to print a string "1000". I have the conversion from decimal to binary, but when I print the values directly form the char array, I get little question marks instead of numbers. I know it has something to do with the way char arrays read values, but I can't figure out how to correct the issue.
void dec2Bin(int value, char binaryString[]) {
int remainder = 0;
int binDigit = 0;
int i = 0;
while (value != 0) {
binDigit = value % 2;
value /= 2;
binaryString[i] = char(binDigit);
i++;
}
for (int k = i - 1; k > 0; k--) {
cout << binaryString[k];
}
}
int main()
{
cout << "Enter a decimal number: ";
int num;
cin >> num;
char binaryString[20] = "";
dec2Bin(num, binaryString);
return 0;
}
When you do
binaryString[i] = char(binDigit);
you are assigning the decimal value 0 or 1 to binaryString[i]. That's okay, a char is basically nothing more than a small integer.
The problems comes when you want to print the value, as the only overloaded << operator to handle char treats the characters as a character, and in most encodings the values 0 and 1 are not printable.
There are two solutions:
Either you convert the character you want to print into a larger integer which won't be treated as a character:
cout << static_cast<int>(binaryString[k]);
Or you make the array contain actual printable characters instead:
binaryString[i] = binDigit + '0';
So, I am trying to figure out the best/simplest way to do this. For my algorithms class we are supposed read in a string (containing up to 40 characters) from a file and use the first character of the string (data[1]...we are starting the array at 1 and wanting to use data[0] as something else later) as the number of rotations(up to 26) to rotate letters that follow (it's a Caesar cipher, basically).
An example of what we are trying to do is read in from a file something like : 2ABCD and output CDEF.
I've definitely made attempts, but I am just not sure how to compare the first letter in the array char[] to see which number, up to 26, it is. This is how I had it implemented (not the entire code, just the part that I'm having issues with):
int rotation = 0;
char data[41];
for(int i = 0; i < 41; i++)
{
data[i] = 0;
}
int j = 0;
while(!infile.eof())
{
infile >> data[j+1];
j++;
}
for(int i = 1; i < 27; i++)
{
if( i == data[1])
{
rotation = i;
cout << rotation;
}
}
My output is always 0 for rotation.
I'm sure the problem lies in the fact that I am trying to compare a char to a number and will probably have to convert to ascii? But I just wanted to ask and see if there was a better approach and get some pointers in the right direction, as I am pretty new to C++ syntax.
Thanks, as always.
Instead of formatted input, use unformatted input. Use
data[j+1] = infile.get();
instead of
infile >> data[j+1];
Also, the comparison of i to data[1] needs to be different.
for(int i = 1; i < 27; i++)
{
if( i == data[1]-'0')
// ^^^ need this to get the number 2 from the character '2'.
{
rotation = i;
std::cout << "Rotation: " << rotation << std::endl;
}
}
You can do this using modulo math, since characters can be treated as numbers.
Let's assume only uppercase letters (which makes the concept easier to understand).
Given:
static const char letters[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
const std::string original_text = "MY DOG EATS HOMEWORK";
std::string encrypted_text;
The loop:
for (unsigned int i = 0; i < original_text.size(); ++i)
{
Let's convert the character in the string to a number:
char c = original_text[i];
unsigned int cypher_index = c - 'A';
The cypher_index now contains the alphabetic offset of the letter, e.g. 'A' has index of 0.
Next, we rotate the cypher_index by adding an offset and using modulo arithmetic to "circle around":
cypher_index += (rotation_character - 'A'); // Add in the offset.
cypher_index = cypher_index % sizeof(letters); // Wrap around.
Finally, the new, shifted, letter is created by looking up in the letters array and append to the encrypted string:
encrypted_text += letters[cypher_index];
} // End of for loop.
The modulo operation, using the % operator, is great for when a "wrap around" of indices is needed.
With some more arithmetic and arrays, the process can be expanded to handle all letters and also some symbols.
First of all you have to cast the data chars to int before comparing them, just put (int) before the element of the char array and you will be okay.
Second, keep in mind that the ASCII table doesn't start with letters. There are some funny symbols up until 60-so element. So when you make i to be equal to data[1] you are practically giving it a number way higher than 27 so the loop stops.
The ASCII integer value of uppercase letters ranges from 65 to 90. In C and its descendents, you can just use 'A' through 'Z' in your for loop:
change
for(int i = 1; i < 27; i++)
to
for(int i = 'A'; i <= 'Z'; i++)
and you'll be comparing uppercase values. The statement
cout << rotation;
will print the ASCII values read from infile.
How much of the standard library are you permitted to use? Something like this would likely work better:
#include <iostream>
#include <string>
#include <sstream>
int main()
{
int rotation = 0;
std::string data;
std::stringstream ss( "2ABCD" );
ss >> rotation;
ss >> data;
for ( int i = 0; i < data.length(); i++ ) {
data[i] += rotation;
}
// C++11
// for ( auto& c : data ) {
// c += rotation;
// }
std::cout << data;
}
Live demo
I used a stringstream instead of a file stream for this example, so just replace ss with your infile. Also note that I didn't handle the wrap-around case (i.e., Z += 1 isn't going to give you A; you'll need to do some extra handling here), because I wanted to leave that to you :)
The reason your rotation is always 0 is because i is never == data[1]. ASCII character digits do not have the same underlying numeric value as their integer representations. For example, if data[1] is '5', it's integer value is actually 49. Hint: you'll need to know these values when handle the wrap-around case. Do a quick google for "ANSI character set" and you'll see all the different values.
Your determination of the rotation is also flawed in that you're only checking data[1]. What happens if you have a two-digit number, like 10?
This little exercise is meant to get a string from the user that could be decimal, hexadecimal, or octal. 1st I need to identify which kind of number the string is. 2nd I need to convert that number to int and display the number in its proper format, eg:
cout <<(dec,hex,oct, etc)<< number;
Here's what I came up with. I'd like a simpler, cleaner way to write this.
string number = "";
cin >> number;
string prefix = "dec";
char zero = '0';
char hex_prefix = 'x';
string temp = "";
int value = 0;
for(int i =0; i<number.size();++i)
{
if(number[0] == zero)//must be octal or hex
{
if (number[0] == zero && number[1] == hex_prefix ) //is hex
{
prefix = "hex";
for(int i = 0; i < (number.size() - 2); ++i)
{
temp[i] = number[i+2];
}
value = atoi(temp.c_str());
}
//... code continues to deal with octal and decimal
You are checking number[0] twice, that's the first most obvious problem.
The inner if already checks both number[0] and number[1], I don't see the point of the outer one.
The outermost loop is also hard to understand, do you expect non-hex data before the number, or what? Your question could be clearer on how the expected input string looks.
I think the cleanest would be to ignore this, and push it into existing (library) code that can parse integers in any base. In C I would recommend strtoul(), you can of course use that in C++ too.
You have two inner loop with same value integer this could be a conflict problem in your code. I suggest you look at the isdigit and islower methods in the c++ library and take advantage of those methods to accomplish your task. isdigit & islower
Good Luck
This is prints the number after deleting the hex prefix, otherwise return 0:
#include<iostream>
#include<cmath>
#include<stdlib.h>
using namespace std;
int main(){
string number = "";
cin >> number;
string prefix = "dec";
char zero = '0';
char hex_prefix = 'x';
string temp = "";
int value = 0;
if (number.size()>=2 && number[0] == zero && number[1] == hex_prefix ) //is hex
{
prefix = "hex";
for(int i = 0; i < (number.size() - 2); ++i)
{
temp[i] = number[i+2];
}
value = atoi(temp.c_str());
}
cout<<value;
return 0;
}
This partial solution that I found is as clean as possible, but it doesn't report the format of the integer:
int string_to_int(std::string str)
{
std::istringstream stream;
stream.unsetf(std::ios_base::dec);
int result;
if (stream >> result)
return result;
else
throw std::runtime_error("blah");
}
...
cout << string_to_int("55") << '\n'; // prints 55
cout << string_to_int("0x37") << '\n'; // prints 55
The point here is stream.unsetf(std::ios_base::dec) - it unsets the "decimal" flag that is set by default. This format flag tells iostreams to expect a decimal integer. If it is not set, iostreams expect the integer in any base.