Question

ILS thE Sum of 3- m divided by the quantity m 4. (8 Points) Consider the following fragment and determine its output: #include <iostream> #include <string> using namespace std: Assume the function prototype is given int main() int í. 4, j 6, k = 9; string st "abcde fghijklmnopgrstuvwxyz" cout << st.find ("abcdef") <<"* func (i, i, k) 13-3 1: -5 cout << st[?] << endl; // OMG..what does this do? ???? return 0 ) I end of main program void func(int &a, int b, int &c) a c + 2*b; s run. Choose the correct output after the program above i A) 0*22* 6-8 e B) 0*21 6*+9*f C) 0*20*6*-9*g D) 0*21 6*-9*f E) 0*21 6*-8 g

Answer 1

it is needed to oberve that the given program is in c++ 98 syntax which is a little different from trubo c++.so there's a slight chane in statements and function usage that you need to observe and this explanation is done by ignoring some program ststements in comment section (don't know why they're ommited..may be purposefully to deviate or something i don't know !)

and here's the program  

#include<iostream>
#include<string>
using namespace std;

void func(int,int,int); //prototype of the user defined function

int main()
{
int i=4,j=6,k=9;
string st[]="abcdefghjklmnopqrstuvwxyz";
cout<<st.find("abcdef")<<"*"; // find a sub string
func(i,j,k);
cout<<i<<"*"<<j<<"*"<<k<<"*";
cout<<st[--j]<<endl;
return 0;
}

void func(int &a,int b,int &c)
{
a=c+2*b;
c =-5*b+a;
}

the resulting output of this program is:

0*21*6*-9*f

Explanation

find() function:

this function will find the substring equal to a given character sequence from an existing string

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the statement

cout<<st.find("abcdef")<<"*";

gives output as

0*

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now into the statement

func(i,j,k);

will take the execution to its declaration part

void func(int &a,int b,int &c)
{
a=c+2*b;
c =-5*b+a;
}

this function on execution yield the values of a as 21 and c as -9 which are function parameters for i and k repectively. j remains intact during this execution

here "a" and "c" are call by reference parameters and "b" is a call by value parameter

call by value parameters will have access by address location whereas

call by reference parameters will have access by direct values(by variable names)

so the statement

cout<<i<<"*"<<j<<"*"<<k<<"*";

yields the output

21*6*-9*

and finally the following statement

cout<<st[--j]<<endl; // confused?

here " - - j" is very much similar to j= j -1 so it yields the value of j as 5

so replacing the same in the program statement will lok like

cout<<st[5]<<endl; //this looks cool !

now taking this into our mind ,keep an eye on the string st defined as alphabet sequence a through z refering the string as st[0]=a,st[1]=b,st[2]=c,st[3]=d,st[4]=e,.......and so on;

st[5] will yield the ouput "f"

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the final answer is:

D) 0*21*6*-9*f