Here, I've made a function, that takes a character array and a single element array as input.
The input of expression is like "56+78", and then someone suggested this approach of using ascii code and for loops to store the two "numeric" substrings as two numbers, and used the character and switch statement below. But, I don't understand the part of storing these substrings as numbers and the asciicode concept.
void calculate(char ch[], char op[]){
int i;
int num1 = 0, num2 = 0;
for(i=0; ch[i]!='\0';i++)
{
if((int)ch[i]>=48 && (int)ch[i]<=57){
num1 = num1*10+(((int)ch[i])-48);
}
else{
op[0]=ch[i];
break;
}}
i++;
for(; ch[i]!='\0';i++)
{
if((int)ch[i] >= 48 && (int)ch[i] <= 57){
num2 = num2*10+(((int)ch[i])-48);
}
}
cout<<"OUTPUT: ";
switch(op[0])
{
case '+':
cout<<num1 + num2<<endl;
break;
case '-':
cout<<num1 - num2<<endl;
break;
case '*':
cout<<num1 * num2<<endl;
break;
case '/':
cout<<num1 / num2<<endl;
break;
}
}
Related
in my code, i don't understand why zero doesn't print i did all possible solutions that I know but it doesn't print zero.
#include <iostream>
using namespace std;
int main(){
int digits;
int numberOne = 0;
int integer;
cout<<"Enter the number: ";
cin>>digits;
while (digits != 0) {
numberOne = (numberOne * 10) + (digits % 10);
digits /= 10;
}
for (integer = numberOne; integer > 0; integer = integer / 10){
switch (integer % 10) {
case 0:
cout<<"Zero\n";
break;
case 1:
cout<<"One\n";
break;
case 2:
cout<<"Two\n";
break;
case 3:
cout<<"Three\n";
break;
case 4:
cout<<"Four\n";
break;
case 5:
cout<<"Five\n";
break;
case 6:
cout<<"Six\n";
break;
case 7:
cout<<"Seven\n";
break;
case 8:
cout<<"Eight\n";
break;
case 9:
cout<<"Nine\n";
break;
}
}
return 0;
}
zero doesn't print how do I fix it?
Expected output is 900 (nine zero zero) but zero doesn't print in my case. help thanks.
Because in this line:
for (integer = numberOne; integer > 0; integer = integer / 10){
the loop continues only if integer>0. Therefore you never see "Zero".
Why doesn't it print zero? Look at your loop
for (integer = numberOne; integer > 0; integer = integer / 10){
If integer equals zero then the loop is never entered. You need to be a bit smarter in your loop. How about counting digits?
int num_digits = 0;
while (digits != 0) {
numberOne = (numberOne * 10) + (digits % 10);
digits /= 10;
++num_digits;
}
Now that you have the number of digits, you can use that in your second loop
for (integer = numberOne; num_digits > 0; integer = integer / 10, --num_digits) {
...
}
You still need to treat zero as a special case, because if the user enters 0 then num_digits will equal zero and you again won't print anything. I'll leave you to figure out how to fix that.
#include <iostream>
using namespace std;
int main(){
int digits;
int arrayLength = 10;
cout<<"Enter the number: ";
cin>>digits;
string reverse_number[arrayLength]; //array to store the string of numbers like "Zero"
int array_counter=0;
while (digits != 0) {
switch (digits % 10) {
case 0:
reverse_number[array_counter++] = "Zero";
break;
case 1:
reverse_number[array_counter++] = "One";
break;
case 2:
reverse_number[array_counter++] = "Two";
break;
case 3:
reverse_number[array_counter++] = "Three";
break;
case 4:
reverse_number[array_counter++] = "Four";
break;
case 5:
reverse_number[array_counter++] = "Five";
break;
case 6:
reverse_number[array_counter++] = "Six";
break;
case 7:
reverse_number[array_counter++] = "Seven";
break;
case 8:
reverse_number[array_counter++] = "Eight";
break;
case 9:
reverse_number[array_counter++] = "Nine";
break;
}
digits /=10;
}
int num_digits = array_counter;
for (int i = 0; i < num_digits; i++) {
cout << reverse_number[--array_counter] <<" ";
}
return 0;
}
The reason why your zeros are not showing while doing for 900 is because when you are reversing 900 to 009, there is no condition mentioned to handle the leading zeros in 009 which are just ignored. For this the best approach would be to store them somewhere else like in a string or an array. You can also just store the number words directly to an array while you are converting, that way you wont lose the leading zeros.
The other answers have already addressed the main problem with your code.This answer focus on the post's title: how to print multiple numbers to words?
First, I would consider using an array of strings instead of a switch.
In order to retrieve a string for a given digit, you would just index that array.(see digits[c - '0'] below)
Also, if you are allowed to convert a number to a string, the code would reduce to walking that string and converting each character to a word.You could read the number directly as a string from the standard input, or use std::to_string on an integer.
Notice also that, when reading a character c from the a string of digits, c - '0' gives you an integer betweeen 0 and 9, which you can use as the array index.
Demo
#include <array>
#include <iostream>
#include <string> // to_string
std::array<std::string, 10> digits{
"Zero", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine"
};
void digits_to_words(int n) {
for (unsigned char c : std::to_string(n)) {
std::cout << digits[c - '0'] << "\n";
}
}
int main() {
for (int n : { 0, 5, 105, 900 }) {
std::cout << "Number: " << n << "\n";
digits_to_words(n);
std::cout << "\n";
}
}
poziomy= char;
pionowy= digit; ( no problems with this one)
So I need to convert char into a digit in function but obviusly I cannot do char=int, so I dont know how to pass on the converted char into digit properly.
I guees i can do two functions but maybe there is an easier way?
I thought of making a new variable poziomy_c but I dont know how to pass it to Ruch_gracza()
int Convert_digit (int cyfra)
{
switch (cyfra)
{
case 10: return 0;break;
case 9: return 1;break;
case 8: return 2;break;
case 7: return 3;break;
case 6: return 4;break;
case 5: return 5;break;
case 4: return 6;break;
case 3: return 7;break;
case 2: return 8;break;
case 1: return 9;break;
}
}
int Convert_letter (char literka)
{
switch (literka)
{
case 'A': return 0; break;
case 'B': return 1; break;
case 'C': return 2; break;
case 'D': return 3; break;
case 'E': return 4; break;
case 'F': return 5; break;
case 'G': return 6; break;
case 'H': return 7; break;
case 'I': return 8; break;
case 'J': return 9; break;
}
}
void Conwert(int &pionowy, char poziomy)
{
pionowy=Convert_digit(pionowy);
int poziomy_c;
poziomy_c=Convert_letter (poziomy);
}
void Ruch_gracza1 (int plansza[10][10])
{
int pionowy ;
char poziomy;
cout << "wprowadz wspolrzedne pola na ktorym lezy pion który chcesz ruszyc ( w pionie , potem w poziomie)" << endl;
cin >> pionowy >> poziomy;
Conwert (pionowy,poziomy);
cout << pionowy << endl;
cout << poziomy << endl;
}
You can use char arithmetic to make this a whole lot easier. Since 'A' to 'Z' will be contiguous in ASCII/Unicode, you can do literka - 'A' to get how far literka is from A (which is what your switch is doing):
int Convert_letter (char literka) {
if(!std::isalpha(literka)) { return literka; } // Not a letter
return std::toupper(literka) - 'A';
}
Or if you want a more robust solution to cover even less common character encodings:
int Convert_letter (char literka) {
if(!std::isalpha(literka)) { return literka; } // Not a letter
std::string alphabet = "abcdefghijklmnopqrstuvwxyz";
return std::distance(std::begin(alphabet), std::find(std::begin(alphabet), std::end(alphabet), literka));;
}
Convert_digit will look similar (except with std::isdigit instead of std::isalpha).
You can do as
char c = 'B';
int digit = c - 'A';
return digit;
You need some knowledge about the ASCII table and data type in C++.
Simply, a char is an integer from -128 ... 127. If you declare a char variable name ch like this:
char ch = 'B';
C++ will understand that ch = 66 (look at ASCII table). So that we can do arithmetic operator with ch like an integer variable.
ch - 'A'; // result 1, because 'A' = 65
ch - 65; // same result with ch - 'A'
Finally, you can write your function like this:
int functionChar2Int(char x){
return x - 'A';
}
I was playing around with VS 2012 and came across a very tricky problem (for me at least). I have a simple program that does mathematical operations on inputs from the command line. However, if you input arguments with decimal places somewhere in the program it seems to forget the numbers after the decimal place.
double add (char const *inp, char const *inp2) { //addition function
double val = std::strtol(inp, NULL, 0);
double val2 = std::strtol(inp2, NULL, 0);
return val + val2;
}
and then calling the function :
double result = add(argv[1], argv[2]);
printf("%f \n", result);
and call from command line
test 52.2 44.6
returns 96 instead of 96.8.
You want strtod, not strtol. strtol only parses integers.
strtod only takes two arguments, because floating point numbers are always in base 10 (as far as strtod is concerned, anyway.)
I think you are turning your strings into longs (I'm guessing that's what strtol() is...). I was building a calculator and I had that problem - I wrote a function to handel it:
double string_to_double(string str)
{
int length = (int)str.length();
double output = 0; //This holds value of output. Will be added for each 10's digit
int decimalPos = length - 1;
for (int i = 0; i < length; i++)
{
if(str.at(i) == '.') // Find if there is a decimal point
{
decimalPos = i - 1; //Sets Decimal Position
str.erase(str.begin()+i);
length--;
}
}
for (int i = 0; i < length; i++)
{
switch (str.at(i))
{
case '1':
output += 1*pow(10, decimalPos - i);
break;
case '2':
output += 2*pow(10, decimalPos - i);
break;
case '3':
output += 3*pow(10, decimalPos - i);
break;
case '4':
output += 4*pow(10, decimalPos - i);
break;
case '5':
output += 5*pow(10, decimalPos - i);
break;
case '6':
output += 6*pow(10, decimalPos - i);
break;
case '7':
output += 7*pow(10, decimalPos - i);
break;
case '8':
output += 8*pow(10, decimalPos - i);
break;
case '9':
output += 9*pow(10, decimalPos - i);
break;
case '0':
break;
}
}
return output;
}
...but there is probably a better way
So i need to have my switch statement go through and write out five two if the user says 52 but i cannot get it pass my 0-9. if they type 0-9 it works perfect but if i try to do any number past that it makes a blank. help!
#include <stdio.h>
int main (void)
{
int x;
printf("Please enter an integer: ");
scanf("%d", &x);
printf("\nYou have entered:\n\n");
for(x;x<0;x++);
switch (x)
{
case 0:
printf("zero");
break;
case 1:
printf("one");
break;
case 2:
printf("two");
break;
case 3:
printf("three");
break;
case 4:
printf("four");
break;
case 5:
printf("five");
break;
case 6:
printf("six");
break;
case 7:
printf("seven");
break;
case 8:
printf("eight");
break;
case 9:
printf("nine");
break;
}
printf("\n\n");
return 0;
}
do {
switch (x%10) {
...
}
x = x / 10;
} while (x>0) ;
or to get it in the right order use recursion
void f(int x) {
if (x==0) return;
f(x/10);
switch(x%10) { ... }
}
This question has been asked-and-answered before.
The for-loop you wrote is empty, because the body of the loop ends with the semi-colon:
for(x;x<0;x++) /* Empty Body!!*/ ;
The way a typical for loop works is:
for( /*Initialize*/; /*Test*/; /*Change*/)
{
/* Body */
}
In your case, I think you want:
for(int i=0; i < x; ++i)
{
switch(x)
{
[...]
}
}
This will:
Initialize i to 0
Test if i is LESS THAN x (the number you entered)
Keep increasing i by 1, until it gets up to x.
I'm not going to do your homework for you but consider the following pseudo-code:
print_text_digits(x)
{
if (x >= 10) print_text_digits(x / 10);
switch (x) {
print "zero" through "nine" as appropriate
}
}
main()
{
scan number into x;
print_text_digits(x);
}
This relies on a recursive routine so that you get your digits processed one at a time, with the might significant digit printed first.
You could solve this with recursion.
void printDigit(int x) {
int digit = x%10;
if(digit!=x)
printDigit(x/10);
switch(digit) {
...
}
}
This will print the most significant figure first, unlike the while loops most people are mentioning.
I believe you need this:
#include <stdio.h>
int main (void)
{
int count = 0;
int x, count2;
printf("Please enter an integer: ");
scanf("%d", &x);
printf("\nYou have entered:\n\n");
int aux = x;
while(aux>0) {
aux=aux/10;
count++;
}
count2 = count;
while(count) {
aux = x;
for(int i=count-1;i>0;i--)
aux=aux/10;
for(int i=count2-count;i>0;i--)
aux%=10;
switch (aux) {
case 0:
printf("zero");
break;
case 1:
printf("one");
break;
case 2:
printf("two");
break;
case 3:
printf("three");
break;
case 4:
printf("four");
break;
case 5:
printf("five");
break;
case 6:
printf("six");
break;
case 7:
printf("seven");
break;
case 8:
printf("eight");
break;
case 9:
printf("nine");
break;
}
count--;
if(count) printf(" ");
}
printf("\n\n");
return 0;
}
Now, with an input 52 it will propelly return five two.
#include <stdio.h>
static const char * const num[] = {
"zero ", "one ", "two " , "three ", "four ",
"five ", "six ", "seven ", "eight ", "nine "
};
void printNum(int x)
{
if (x < 10) {
printf(num[x]);
return;
}
printNum(x / 10);
printNum(x % 10);
}
int main (void)
{
int x;
printf("Please enter an integer: ");
scanf("%d", &x);
printf("\nYou have entered:\n\n");
printNum(x);
return 0;
}
Here's what you need to do. Similar to what #simonc said, it's a good idea to convert the user input to a string, then loop through the characters. For each character, convert it into an int and then take that into the switch statement.
EDIT---------------------------
Here's a way with strictly using integers.
First find how many digits your integer has. You can do this by a method mentioned here.
Then do integer division starting from the largest power of 10 that divides the integer you have and divide the divisor by 10 until you reach 1. For example:
If user input is 213, it has 3 digits. We divide this by 100 first.
213/100 = 2
We take the 2 and put it into the switch statement, outputting 2. Now we're done with the hundreds digit, so now we take the 13 from 213 and divide it by 10.
13/10 = 1 So now we output one.
Keep doing this process until you get to the ones digit.
Does this make sense?
My program reduce expression to one value.
I have a problem with changing "char sign" as a character of action. Could you show me some simple solution or idea how to do it?
I tried:
(tab[i]-'0') 'sign' (tab[i+1]-'0');
This is the full code:
#include <cstdlib>
#include <iostream>
using namespace std;
int main(int argc, char *argv[])
{
char* tab = "12+";
int b = sizeof (tab);
char* tmp = new char[b] ;
tmp [b-1] = '\0';
int k = b/3;
for(int i=0; i<k; i++){
if(isdigit(tab[i]) && isdigit(tab[i+1]) ){
if(tab[i+2]=='+' || tab[i+2]=='-' || tab[i+2]=='*'){
char sign = tab[i+2];
int n = (tab[i]-'0') + (tab[i+1]-'0'); //here is a problem, i want to replice + as a char sign which will be recognized
tmp[i] = n+'0';
}
else goto LAB;
}
else if (isdigit(tab[i]) && isdigit(tab[i+2])){
}
else if (isdigit(tab[i+1]) && isdigit(tab[i+2])){
}
else
LAB:
tmp[i]= tab[i];
}
cout<<"Import "<<tmp[0]-'0'<<endl;
system("PAUSE");
return EXIT_SUCCESS;
}
You cannot replace an operator, or function name with a symbol which is given by user, because compiler has to know which function should be called and there is no mechanism to convert a '+' character to operator+(int, int) call embedded in a language. You have to write it by yourself.
The simplest solution is to write explicitly every operator you want to support:
int n;
switch(tab[i+2]){
case '+':
n = (tab[i]-'0') + (tab[i+1]-'0');
break;
case '-':
n = (tab[i]-'0') - (tab[i+1]-'0');
break;
// etc...
}
int n;
switch(sign){
case '-':
n = (tab[i]-'0') - (tab[i+1]-'0');
break;
case '+':
n = (tab[i]-'0') + (tab[i+1]-'0');
break;
case '*':
n = (tab[i]-'0') * (tab[i+1]-'0');
break;
default:
// handle error here;
break;
}