Assume that an int variable age has been declared and already given a value. Ass
ID: 3529083 • Letter: A
Question
Assume that an int variable age has been declared and already given a value. Assume further that the user has just been presented with the following menu: S: hangar steak, red potatoes, asparagus T: whole trout, long rice, brussel sprouts B: cheddar cheeseburger, steak fries, cole slaw (Yes, this menu really IS a menu!) Write some code that reads the String (S or T or B) that the user types in into a String variable choice that has already been declared and prints out a recommended accompanying drink as follows: if the value of age is 21 or lower, the recommendation is "vegetable juice" for steak, "cranberry juice" for trout, and "soda" for the burger. Otherwise, the recommendations are "cabernet", "chardonnay", and "IPA" for steak, trout, and burger respectively. Regardless of the value of age , your code should print "invalid menu selection" if the character read into choice was not S or T or B. ASSUME the availability of a variable, stdin , that references a Scanner object associated with standard input.Explanation / Answer
#include<iostream>
#include<string>
using namespace std;
int main()
{
int age;
age=18; // initialize some value
std::string input;
std::cout<<"Enter your manu : ";
std::cin>>input;
if(input.compare("S")==0)
{
if(age<=21) std::cout<<"accompanying drink is vegetable juice"<<std::endl;
else std::cout<<"accompanying drink is cabernet"<<std::endl;
}
else if(input.compare("T")==0)
{
if(age<=21) std::cout<<"accompanying drink is cranberry juice"<<std::endl;
else std::cout<<"accompanying drink is chardonnay"<<std::endl;
}
else if(input.compare("B")==0)
{
if(age<=21) std::cout<<"accompanying drink is soda"<<std::endl;
else std::cout<<"accompanying drink is IPA"<<std::endl;
}
else std::cout<<"Invalid menu selection"<<std::endl;
return 0;
}
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