I am trying to understand how to use cin.get(). I would like to only accept 20 characters of input, but any method I've tried using to detect further input stop
I am accepting input one character at a time and inputting it into an array like so:
void getInput(char arr[], int size) {
char number = '0';
for (int i = 0; i < size + 1; i++) {
char number;
cin.get(number);
if ((number >= '0') && (number <= '9')) {
arr[i] = number;
}
if ((number == ' ') || (number == '\n')) {
break;
}
}
}
I've tried doing something like this, but you can see that my understanding is not excellent:
if (number == '\n') {
done == true;
}
and wrapping the whole thing in a while (!done) for example, but that doesn't work in situations where I input 20 characters or less.
You could initialise a variable int count= 1;. With each iteration you can increment it with count++;and put the check
if (count==20)
break;
I hope this helps.
You could round down the passed size parameter to 20 if it's larger than 20. Then the for-loop will break at most after 20 iterations.
#include <iostream>
using namespace std;
void getInput(char arr[], size_t size) {
const size_t MAX_SIZE = 20;
if (MAX_SIZE < size) {
// 'round down' size if it's too big
size = MAX_SIZE;
}
char number = '0';
for (unsigned int i = 0; i < size; i++) {
cin.get(number);
if ((number >= '0') && (number <= '9')) {
arr[i] = number;
} else if ((number == ' ') || (number == '\n')) {
break;
}
}
}
int main() {
char arr[32] = {};
// even though arr could fit 32 chars
// only MAX_SIZE=20 numbers at most will be accepted
getInput(arr, 32);
cout << arr << endl;
return 0;
}
I'm assuming size is the length of arr (can't yet comment so could not ask this), if so then watch out for doing i < size + 1; that let's i take values from 0 to size inclusive, and then the for-loop eventually reaches arr[size] which is off the end of the array. Use i < size instead.
Related
Wrote a function that calculates the average length of words in a sentence.
Why does the program print 0 instead of the average?
Please help me fix my mistake.
If you know how to make an implementation in one function, please write.
#include <iostream>
#include <string>
using namespace std;
int CountWordsAndLetters(char* str, int& words, int& letters)
{
words = 0;
int i = 0;
letters = 0;
while (str[i] == ' ')
i++;
for (; str[i]; i++) {
if (((str[i] >= 'a') && (str[i] <= 'z'))
|| ((str[i] >= 'A') && (str[i] <= 'Z')))
letters++;
if (str[i] == ' ') {
words++;
while (1)
if (str[i] == ' ')
i++;
else {
i--;
break;
}
}
}
words = words + 1;
return (words);
}
float AverageLetters(float words, float letters)
{
float a = (double)(letters / words);
return a;
}
int main()
{
char array[255];
int words = 0;
int letters = 0;
cout << "Enter the string\n\n";
gets_s(array);
int size;
for (size = 0; array[size]; size++)
;
char* str = new char[size];
CountWordsAndLetters(str, words, letters);
cout << "\nAverage number of letters per word: "
<< AverageLetters(words, letters);
return 0;
}
If you know how to make an implementation in one function, please write.
Here, you are allocating an uninitialized array of char:
char* str = new char[size];
You put nothing in it.
You then pass it to CountWordsAndLetters:
// here -------v
CountWordsAndLetters(str, words, letters);
You should consider simply sending array instead:
CountWordsAndLetters(array, words, letters);
Here's a live example of your code working.
Our teacher gave us this exercise:
"Given a string like '-5,14' write a function that returns the float value of -5,14
I used double here just to test the precision, but it also didn't work with floats.
[also i'm from Europe, we use the comma instead of the dot. Oh also we aren't allowed to use the type string and bool, we have to "make" them like in C]
This is what i came up with, and it seems to work a little bit. Positive numbers are similar, but wrong, and given a negative number, the result is similar to 10 times the positive of the given number.
It should work like this:
I read the string into an array of characters;
I check if the first character is a minus. if so, subtract 1 from the number of integer figures because i will count them later starting from index 0;
I count the number of integer figures with a loop from the start of the array to the ',' character;
I count the number of decimal figures with a loop from after the ',' to the end of the string;
[Keep in mind for the next step that, following the ASCII table, the code for the character of a number is that number + 48]
I add to the result variable every integer figure multiplied by ten to the power of whatever place in the number it has.
I do the same for the deicmal values but with the negative exponent.
if the number was negative, i multiply the result with -1.
But for some reason it's not working properly. The lower the number is, the less accurate it is (given 4,5 the result is 9, but given 345,543 the result is 350,43)
#include <iostream>
#define EOS '\0'
#define DIM 100
#define TRUE 1
#define FALSE 0
void leggiN(char* c)
{
std::cout << "Insert a number: ";
std::cin >> c;
}
double stof(char* str)
{
double Result = 0;
double ascii_to_int = 48;
int i = 0;
int j = 0;
int IntegerDigits = 0;
int DecimalDigits = 0;
int CommaIndex;
int isNegative = FALSE;
if (str[0] == '-')
{
IntegerDigits = -1;
isNegative = TRUE;
}
while (str[i] != ',')
{
++IntegerDigits;
++i;
}
CommaIndex = i;
++i;
while (str[i] != EOS)
{
++DecimalDigits;
++i;
}
for (i = (CommaIndex - 1); i >= 0; --i)
{
Result += (str[i] - ascii_to_int) * (std::pow(10, j));
++j;
}
j = 0;
for (i = (CommaIndex + 1); str[i] != EOS; ++i)
{
Result += (str[i] - ascii_to_int) * (std::pow(10, -j));
++j;
}
if (isNegative == 1)
Result = Result * -1;
return Result;
}
int main()
{
char str[DIM];
leggiN(str);
std::cout << stof(str);
}
use j = 1 to start your second for loop. You are trying to raise 10 to the power of -0
j = 1;
for (i = (CommaIndex + 1); str[i] != EOS; ++i)
{
Result += (str[i] - ascii_to_int) * (std::pow(10, -j));
++j;
}
If your code return 9.0 when you enter "4,5", your problem has nothing to do with imprecision.
There are other problems in your code, I've tried to un it and got a SEGFAULT...
#include <iostream>
#define EOS '\0' // 0 being such a special value, there is no need to
// define a named constant for it.
#define DIM 100
#define TRUE 1 // the language defines boolean values, avoid defining
#define FALSE 0 // unnecessary named constants for something that already
// exists.
void leggiN(char* c)
{
std::cout << "Insert a number: ";
std::cin >> c; // Inserting from cin to a char* is a BIG no-no.
// some compilers won't even allow it, for good reasons
// i.e.: what is the length of the array pointed to?
}
double stof(char* str) // you are indicating that you may modify str?
{
double Result = 0;
double ascii_to_int = 48; // this is a terrible name.
int i = 0;
int j = 0;
int IntegerDigits = 0;
int DecimalDigits = 0;
int CommaIndex;
int isNegative = FALSE;
if (str[0] == '-') // is str a valid pointer? what happens if NULL ??
{
IntegerDigits = -1;
isNegative = TRUE;
// you fail to skip the sing character, should have ++i here.
}
while (str[i] != ',') // what happens if there is no ',' in the string?
{ // you should check for str[i] == 0.
++IntegerDigits;
++i;
}
CommaIndex = i;
++i;
while (str[i] != EOS)
{
++DecimalDigits; // why do you count decimal digits?
++i; // you do not use this result anyway...
}
for (i = (CommaIndex - 1); i >= 0; --i)
{
// what happens if you have non-digit characters? they participate
// in the conversion??
// you call std::pow(), but do not include <cmath> at the top of the file.
// isn't str[i] - '0' clearer ?
Result += (str[i] - ascii_to_int) * (std::pow(10, j));
++j;
}
j = 0;
for (i = (CommaIndex + 1); str[i] != EOS; ++i)
{
Result += (str[i] - ascii_to_int) * (std::pow(10, -j));
++j;
}
if (isNegative == 1) // you had defined constants fot this, but don't use them.
Result = Result * -1;
return Result;
}
int main()
{
char str[DIM];
leggiN(str);
std::cout << stof(str);
}
Here is one way to achieve what you want.
#include <iostream>
#include <string>
const char DECIMAL_POINT = ','; // we'll use a named constant here....
// usually, we'd have to check the locale
// for regional specific information.
// works like atod(), conversion stops at end of string of first illegal character.
double stof(const char* str) {
// check input, must be not null, not empty
if (!str || str[0] == 0)
return 0;
int i = 0;
bool isNegative = false;
// take care of leading sign
if (str[0] == '-' || str[0] == '+') {
isNegative = (str[0] == '-');
++i;
}
// convert integer part.
double result = 0;
while ('0' <= str[i] && str[i] <= '9') {
result = (result * 10) + (str[i] - '0');
++i;
}
// only do decimals if they are there.
if (str[i] != DECIMAL_POINT)
return (isNegative) ? -result : result;
++i; // skip decimal point
double decimals = 0;
double multiplier = .1;
while ('0' <= str[i] && str[i] <= '9') {
decimals += (str[i] - '0') * multiplier;
++i;
multiplier *= .1;
}
result += decimals;
return (isNegative) ? -result : result;
}
int main() {
// always use std::string to read strings from cin.
std::string str;
std::cout << "Insert a number: ";
std::cin >> str;
std::cout << "in: " << str << " out: " << stof(str.c_str()) << '\n';
return 0;
}
I have an assignment to make a program that should convert a number from it's integer value to a binary value. For some reason my array is always filled with zeroes and won't add "1"'s from my if statements. I know there are probably solutions to this assignment on internet but I would like to understand what is problem with my code. Any help is appreciated.
Here is what I tried:
#include <iostream>
/*Write a code that will enable input of one real number in order to write out it's binary equivalent.*/
int main() {
int number;
int binaryNumber[32] = { 0 };
std::cout << "Enter your number: ";
std::cin >> number;
while (number > 0) {
int i = 0;
if ((number / 10) % 2 == 0) {
binaryNumber[i] = 0;
}
if ((number / 10) % 2 != 0) {
binaryNumber[i] = 1;
}
number = number / 10;
i++;
}
for (int i = 31; i >= 0; i--) {
std::cout << binaryNumber[i];
}
return 0;
}
You need to remove number/10 in both the if statements. Instead, just use number. you need the last digit every time to get the ith bit.
Moreover, you need to just half the number in every iteration rather than doing it /10.
// Updated Code
int main() {
int number;
int binaryNumber[32] = { 0 };
std::cout << "Enter your number: ";
std::cin >> number;
int i = 0;
while (number > 0) {
if (number % 2 == 0) {
binaryNumber[i] = 0;
}
if (number % 2 != 0) {
binaryNumber[i] = 1;
}
number = number / 2;
i++;
}
for (int i = 31; i >= 0; i--) {
std::cout << binaryNumber[i];
}
return 0;
}
The first thing is the variable 'i' in the while loop. Consider it more precisely: every time you iterate over it, 'i' is recreated again and assigned the value of zero. It's the basics of the language itself.
The most relevant mistake is logic of your program. Each iteration we must take the remainder of division by 2, and then divide our number by 2.
The correct code is:
#include <iostream>
int main()
{
int x = 8;
bool repr[32]{};
int p = 0;
while(x)
{
repr[p] = x % 2;
++p;
x /= 2;
}
for(int i = 31; i >= 0; --i)
std::cout << repr[i];
return 0;
}
... is always filled with zeroes ... I would like to understand what is problem with my code
int i = 0; must be before the while, having it inside you only set the index 0 of the array in your loop because i always values 0.
But there are several other problems in your code :
using int binaryNumber[32] you suppose your int are on 32bits. Do not use 32 but sizeof(int)*CHAR_BIT, and the same for your last loop in case you want to also write 0 on the left of the first 1
you look at the value of (number / 10) % 2, you must look at the value of number % 2
it is useless to do the test then its reverse, just use else, or better remove the two ifs and just do binaryNumber[i] = number & 1;
number = number / 10; is the right way when you want to produce the value in decimal, in binary you have to divide by 2
in for (int i = 31; i >= 0; i--) { except for numbers needing 32 bits you will write useless 0 on the left, why not using the value of i from the while ?
There are some logical errors in your code.
You have taken (number/10) % 2, instead, you have to take (number %2 ) as you want the remainder.
Instead of taking i = 31, you should use this logic so you can print the following binary in reverse order:
for (int j = i - 1; j >= 0; j--)
{
cout << BinaryNumb[j];
}
Here is the code to convert an integer to its binary equivalent:
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
// function to convert integer to binary
void DecBinary(int n)
{
// Array to store binary number
int BinaryNumb[32];
int i = 0;
while (n > 0)
{
// Storing remainder in array
BinaryNumb[i] = n % 2;
n = n / 2;
i++;
}
// Printing array in reverse order
for (int j = i - 1; j >= 0; j--)
{
cout << BinaryNumb[j];
}
}
// Main Program
int main()
{
int testcase;
//Loop is optional
for(int i = 0; i < testcase; i++)
{
cin >> n;
DecToBinary(n);
}
return 0;
}
int function2 (const char * string1) returning the number of unique digits appearing in the string, e.g. function2 ("ab512af6kc1") -> 3.
int function2(const char* string1) {
int zero = 0, one = 0, two = 0, three = 0, four = 0, five = 0, six = 0,
seven = 0, eight = 0, nine = 0, counter = 0;
for (int i = 0; i < strlen(string1); i++) {
if (string1[i] == '0') {
zero++;
}
if (string1[i] == '1') {
one++;
}
if (string1[i] == '2') {
two++;
}
if (string1[i] == '3') {
three++;
}
if (string1[i] == '4') {
four++;
}
if (string1[i] == '5') {
five++;
}
if (string1[i] == '6') {
six++;
}
if (string1[i] == '7') {
seven++;
}
if (string1[i] == '8') {
eight++;
}
if (string1[i] == '9') {
nine++;
}
}
if (zero == 1) {
counter++;
}
if (one == 1) {
counter++;
}
if (two == 1) {
counter++;
}
if (three == 1) {
counter++;
}
if (four == 1) {
counter++;
}
if (five == 1) {
counter++;
}
if (six == 1) {
counter++;
}
if (seven == 1) {
counter++;
}
if (eight == 1) {
counter++;
}
if (nine == 1) {
counter++;
}
return counter;
}
It's every correct in this code, but it's long a bit. Could someone help me and write SHORTER code? It's the only way that I can measure up to this exercise.
You can use an array instead of 10 variables. Calculate the index in the array by converting the character to an integer.
int function2(const char *in) {
// Array to hold digits occurence counts.
unsigned digits[10]{};
// Iterate over the characters in input.
// Better (auto i : std::string_view(in)) in C++17.
for (auto i = in; *i; ++i) {
if (isdigit(*i)) {
// Increment the proper digit index.
digits[*i - '0']++;
}
}
int count = 0;
// Go through digit occurences.
for (auto i : digits) {
// If the digit occurred only once.
if (i == 1) {
// Increment the count.
count++;
}
}
return count;
}
To shorten your code, use an array instead of 10 individual variables:
int digits[10] = {0}; // instead of int zero = 0, one = 0, ...
To check whether a char is a representation of a digit, use isdigit:
if (isdigit(string1[i])) // instead of if (string1[i] == '0'), if (string1[i] == '1'), ...
The only non-trivial part is to convert a char to the corresponding int:
string1[i] - '0'
This code subtracts the character code of 0 (usually 48) from the character code of a digit (usually 49 for 1, 50 for 2, ..., 57 for 9). The result is an index to your array of counters.
So, to increase the proper array element, use the following code:
digit = string1[i] - '0';
digits[digit]++; // instead of zero++, one++, ...
After the code goes over the input string, count the number of digits which appeared once:
int counter = 0;
for (digit = 0; digit < 10; ++digit)
{
if (digits[digit] == 1)
++counter;
}
Use a hash table collection class to keep track of unique digits. In this case, unordered_set will do just fine. Don't even bother converting the char to integer. You're just looking for unique chars between '0' and '9'.
#include <string>
#include <unordered_set>
size_t getUniqueDigits(const std::string& string1)
{
std::unordered_set<char> table;
for (char c : string1)
{
if ((c >= '0') && (c <= '9'))
{
table.insert(c);
}
}
return table.size();
}
A more traditional "C" based solution that doesn't use any std:: collections or objects is to use an array to be that "set"
int getUniqueDigits(const char* string1)
{
int table[10] = {0};
int count = 0;
const size_t len = (string1 != nullptr) ? strlen(string1) : 0;
for(size_t i = 0; i < len; i++)
{
char c = string1[i];
if ((c >= '0') && (c <= '9'))
{
table[c - '0'] = 1;
}
}
for (char j = '0'; j <= '9'; j++)
{
count += table[j];
}
return count;
}
Just use an ordinary array as for example in this demonstrative program
#include <iostream>
size_t unique_digits( const char *s )
{
unsigned char digits[10] = { 0 };
for ( ; *s; ++s )
{
if ( '0' <= *s && *s <= '9' )
{
if ( digits[*s - '0'] != 2 ) ++digits[*s - '0'];
}
}
size_t count = 0;
for ( unsigned char c : digits ) count += c == 1;
return count;
}
int main()
{
std::cout << unique_digits( "ab512af6kc1" ) << '\n';
return 0;
}
The program output is
3
Or you can declare the array of the element type size_t. In this case the function will look the following way
#include <iostream>
size_t unique_digits( const char *s )
{
size_t digits[10] = { 0 };
for ( ; *s; ++s )
{
if ( '0' <= *s && *s <= '9' )
{
++digits[*s - '0'];
}
}
size_t count = 0;
for ( unsigned char c : digits ) count += c == 1;
return count;
}
int main()
{
std::cout << unique_digits( "ab512af6kc1" ) << '\n';
return 0;
}
I think you already have many good solutions.
Here is mine version anyway
int function2(const char* string1) {
int count[10] = {0};
int counter = 0;
int i;
for(i = 0; i < strlen(string1); i++)
int a = (++count[string1[i]-'0']);
if(a == 1)counter++;
if(a == 2)counter--;
return counter;
}
I haven't tried it. Hope there is no error
Edit:
I tried it. It seems to work fine now.
I am trying to write a code that takes a binary number input as a string and will only accept 1's or 0's if not there should be an error message displayed. Then it should go through a loop digit by digit to convert the binary number as a string to decimal. I cant seem to get it right I have the fact that it will only accept 1's or 0's correct. But then when it gets into the calculations something messes up and I cant seem to get it correct. Currently this is the closest I believe I have to getting it working. could anyone give me a hint or help me with what i am doing wrong?
#include <iostream>
#include <string>
using namespace std;
string a;
int input();
int main()
{
input();
int decimal, x= 0, length, total = 0;
length = a.length();
// atempting to make it put the digits through a formula backwords.
for (int i = length; i >= 0; i--)
{
// Trying to make it only add the 2^x if the number is 1
if (a[i] = '1')
{
//should make total equal to the old total plus 2^x if a[i] = 1
total = total + pow(x,2);
}
//trying to let the power start at 0 and go up each run of the loop
x++;
}
cout << endl << total;
int stop;
cin >> stop;
return 0;
}
int input()
{
int x, x2, count, repeat = 0;
while (repeat == 0)
{
cout << "Enter a string representing a binary number => ";
cin >> a;
count = a.length();
for (x = 0; x < count; x++)
{
if (a[x] != '0' && a[x] != '1')
{
cout << a << " is not a string representing a binary number>" << endl;
repeat = 0;
break;
}
else
repeat = 1;
}
}
return 0;
}
I don't think that pow suits for integer calculation. In this case, you can use shift operator.
a[i] = '1' sets the value of a[i] to '1' and return '1', which is always true.
You shouldn't access a[length], which should be meaningless.
fixed code:
int main()
{
input();
int decimal, x= 0, length, total = 0;
length = a.length();
// atempting to make it put the digits through a formula backwords.
for (int i = length - 1; i >= 0; i--)
{
// Trying to make it only add the 2^x if the number is 1
if (a[i] == '1')
{
//should make total equal to the old total plus 2^x if a[i] = 1
total = total + (1 << x);
}
//trying to let the power start at 0 and go up each run of the loop
x++;
}
cout << endl << total;
int stop;
cin >> stop;
return 0;
}
I would use this approach...
#include <iostream>
using namespace std;
int main()
{
string str{ "10110011" }; // max length can be sizeof(int) X 8
int dec = 0, mask = 1;
for (int i = str.length() - 1; i >= 0; i--) {
if (str[i] == '1') {
dec |= mask;
}
mask <<= 1;
}
cout << "Decimal number is: " << dec;
// system("pause");
return 0;
}
Works for binary strings up to 32 bits. Swap out integer for long to get 64 bits.
#include <iostream>
#include <stdio.h>
#include <string>
using namespace std;
string getBinaryString(int value, unsigned int length, bool reverse) {
string output = string(length, '0');
if (!reverse) {
for (unsigned int i = 0; i < length; i++) {
if ((value & (1 << i)) != 0) {
output[i] = '1';
}
}
}
else {
for (unsigned int i = 0; i < length; i++) {
if ((value & (1 << (length - i - 1))) != 0) {
output[i] = '1';
}
}
}
return output;
}
unsigned long getInteger(const string& input, size_t lsbindex, size_t msbindex) {
unsigned long val = 0;
unsigned int offset = 0;
if (lsbindex > msbindex) {
size_t length = lsbindex - msbindex;
for (size_t i = msbindex; i <= lsbindex; i++, offset++) {
if (input[i] == '1') {
val |= (1 << (length - offset));
}
}
}
else { //lsbindex < msbindex
for (size_t i = lsbindex; i <= msbindex; i++, offset++) {
if (input[i] == '1') {
val |= (1 << offset);
}
}
}
return val;
}
int main() {
int value = 23;
cout << value << ": " << getBinaryString(value, 5, false) << endl;
string str = "01011";
cout << str << ": " << getInteger(str, 1, 3) << endl;
}
I see multiple misstages in your code.
Your for-loop should start at i = length - 1 instead of i = length.
a[i] = '1' sets a[i] to '1' and does not compare it.
pow(x,2) means and not . pow is also not designed for integer operations. Use 2*2*... or 1<<e instead.
Also there are shorter ways to achieve it. Here is a example how I would do it:
std::size_t fromBinaryString(const std::string &str)
{
std::size_t result = 0;
for (std::size_t i = 0; i < str.size(); ++i)
{
// '0' - '0' == 0 and '1' - '0' == 1.
// If you don't want to assume that, you can use if or switch
result = (result << 1) + str[i] - '0';
}
return result;
}