Write multiple things in a switch statement in C - c++

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?

Related

how to print multiple numbers to words

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";
}
}

Can anyone explain this function definition above the switch statement?

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;
}
}

Different Output while using and not using 'break' in switch case in c++

I was solving the a question at HackerRank -
Question Link - https://www.hackerrank.com/challenges/maximum-element/problem
In one solution I used 'break' statement in switch-case, in another solution I didn't.
Solution was wrong when I didn't use break statement. What is the reason behind this ?
Input -
10
1 97
2
1 20
2
1 26
1 20
2
3
1 91
3
With break Statement -
#include <bits/stdc++.h>
using namespace std;
int main() {
int noOfTestCases;
cin>>noOfTestCases;
vector <int> st;
for(int x=0; x<noOfTestCases; x++){
int query;
cin>>query;
switch (query) {
case 1:
int number;
cin>>number;
if(st.empty()){
st.push_back(number);
}
else if(number > st[st.size()-1]){
st.push_back(number);
}
else{
st.push_back(st[st.size()-1]);
}
break;
case 2:
if(!st.empty()){
st.pop_back();
}
break;
case 3:
cout<<st[st.size()-1]<<endl;
}
}
}
//Output -
//26
//91
Without break statement -
#include <bits/stdc++.h>
using namespace std;
int main() {
int noOfTestCases;
cin>>noOfTestCases;
vector <int> st;
for(int x=0; x<noOfTestCases; x++){
int query;
cin>>query;
switch (query) {
case 1:
int number;
cin>>number;
if(st.empty()){
st.push_back(number);
}
else if(number > st[st.size()-1]){
st.push_back(number);
}
else{
st.push_back(st[st.size()-1]);
}
case 2:
if(!st.empty()){
st.pop_back();
}
case 3:
cout<<st[st.size()-1]<<" "<<query<<endl;
}
}
}
//Output -
//0
//0
//0
//0
//0
//0
//0
//0
//0
//0
Consider the following code
switch (x) {
case 1:
std::cout << "one\n";
case 2:
std::cout << "two\n";
break;
case 3:
std::cout << "three\n";
}
If x is 1 it will print both one and two. It will then exit the switch block due to the break statement. Note that x will not be compared with 2 after printing "one", it will directly fall-through to printing "two".
So I had a misconception about switch case statement -
What I thought -
I thought that in switch case if the expression is equal to any case D, then further cases - E, F, G, will not be executed, irrespective of the break statement.
Reality - If an expression is equal to any case D, then further cases - E, F, G will also be executed if we don't use break statement.
Thanks Jeffrey and Paul Sanders to clarify this problem.

How to use a while loop in a switch case statement?(c++)

So when I put the limits of the integer that the switch case uses I keep getting a infinite number of the output of default for example:
int (num);
cout<<"Choose a number between 1-5"<<endl;
cin>>num;
while (num<1 || num>5)
{
switch (num)
{
case 1:
cout<<" Good"<<endl;
break;
case 2 :
cout<<" Okay"<<endl;
break;
case 3:
cout<<" Decent"<<endl;
break;
case 4:
cout<<" Nice Try"<<endl;
break;
case 5:
cout<<"Failed"<<endl;
break;
default
cout<<" Not Valid"<<endl;
break;
{
{
So for this example how would I make the user choose from 1-5 and if not repeat the loop in order for them to try again.
This will loop until valid number is given then test it in the switch case.
int num;
cout<<"Choose a number between 1-5"<<endl;
while(cin>>num && (num < 1 || num > 5));
switch (num)
{
... //for brevity
}
This will loop through the switch cases until a valid input is given.
int num;
bool keepLooping = true;
cout<<"Choose a number between 1-5"<<endl;
while(cin>>num && keepLooping)
{
keepLooping = false;
switch(num)
{
... //for brevity
default:
keepLooping = true;
}
}

Having problems with switch statements

I've been trying to get this code working, but somehow I can't do it..
#include <iostream.h>
#include <stdio.h>
int main() {
int a,b,c;
int y=3;
int i=2;
int g[] = {20};
int m,k;
int Z;
printf("Enter a number for a");
scanf("%d", &a);
printf("Enter a number for b");
scanf("%d", &b);
printf("Enter a number for c");
scanf("%d", &c);
m=y;
do
{
Z = a+b-c;
switch(Z)
{
case '0':
case '1':
k=17;
m+=b;
break;
case '2':
m+=b;
m=a;
break;
case '3':
m=a-c;
m+=b;
m=a;
break;
case '7':
m+=b;
break;
default:
m=a;
break;
}
g[m] = m%i;
m--;
}while(m>b);
}
This is the scheme that I had to turn into coding. http://ff.tu-sofia.bg/PIK/Izpiti/MidTest07.html
y and i are 3 and 2 by default, the array g should contain 20 integers, and the users have to type values for a, b and c.
Z is an integer and your cases are looking for strings. get rid of the quotes around the numbers in the cases.
The array g does not contain 20 integers it contains one element 20 in this case. I think what you meant was g[20] = {}
Also How is Z calculated ? Is it (a+b)-c or a+(b-c)? You need parentheses to make your intent clearer
Z = a+b-c;
switch(Z)
{
case 0:
case 1:
k=17;
m+=b;
break;
case 2:
m+=b;
m=a;
break;