Problem 1 Double r = 2.13; declares that r is a ………………………… Problem 2 cout << “x:

Question

Problem 1   Double r = 2.13; declares that r is a …………………………

Problem 2   cout << “x:   “ << r * 2.3; will write …………………………

Problem 3   char c = “4”; declares that c is a       …………………………

Problem 4   long t = “3”; declares that t is a         …………………………

Problem 5   int b;     b = 3.15 * 4.57;

cout << “b:   “ << “b”; will write    ………………

Problem 6   int c = 3; declares that c is a               …………………………

Problem 7 cout << “b:   “ << c+5; will write      ………………………… (refer to Problem 6)

Explanation / Answer

Problem 1 Double r = 2.13; declares that r is a
If we run Double r = 2.13; its throw error
right syntax is double r = 2.13;
Because c++ is case sensitive.

Problem 2 cout << “x: “ << r * 2.3; will write
If we run Double r = 2.13; then cout << “x: “ << r * 2.3;
This code throw error. Because c++ is case sensitive.
Right syntax is
double r = 2.13;
cout << “x: “ << r * 2.3;
This code Print x: 4.899

Problem 3 char c = “4”; declares that c is a

This code throw error because Character literals not accept double quotes, Character literals always single quotes
Right syntax char c = '4';

Problem 4 long t = “3”; declares that t is a
long t = “3”; not right syntax. SO its throw error.
A long int is a signed integral type that is at least 32 bits
Right syntax is long t = 3;

Problem 5 int b; b = 3.15 * 4.57;
cout << “b: “ << “b”; will write   
Its print or return b: b
Because we print b integer type variable in string so.
If we want right answer then syntax
b = 3.15 * 4.57;
int b;
cout << "b: " << b;
Now its print b: 14

Problem 6 int c = 3; declares that c is a   
C is a integer type variable and its right syntax.
if we print c variable then its print 3
Second thing we already define char c so if we try again define int c =3;
In this case code throw error.

Problem 7 cout << “b: “ << c+5; will write
If we comment char c = '4'; and run this line with
int c = 3;
cout << "b: " << c+5;
Now its print 8

IF we try with this code
char c = '4';
int c = 3;
cout << "b: " << c+5;
Now its throw error because we already define c variable.

Full ruining code.

#include <iostream>

using namespace std;

int main()
{

double r = 2.13;
//char c = '4';
long t = 3;
int b;   
b = 3.15 * 4.57;
int c = 3;
//cout << c;
cout << "b: " << c+5;
//cout << "b: " << b;
//cout << t;
//cout << c;
//cout << "x: " << r * 2.3;
return 0;
}

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