Critical Declaration question

bcompt143 · #1 11-Nov-2008, 01:47

CPP / C++ / C Code:

#include<iostream.h>
#Include<conio.h>

void main ( ) 
{
  int x;
  char y;
  cout<<"enter any number";
  cin>>x;
  cout<<"enter any character;
  cin>>y;
  cout<<x;
  cout<<y;
 getch();
}

In the above program if we enter integer in variable x it display, the number itself and if we enter character in y it display itself. but if we enter character in the place of x, it display the any number and if we enter integer in the place of y it display the number. my question is, why character accept and display the integer number. tell me the brief reason please.

ocicat · #2 11-Nov-2008, 02:29

Quote:

Originally Posted by bcompt143

In the above program...

...which I assume you simply entered by hand without actually compiling. Errors include:

CPP / C++ / C Code:
#Include<conio.h>
...is not a valid preprocessor command.
CPP / C++ / C Code:
cout<<x; cout<<y;
...outputs the integer value of x followed by the character value of y on the same line with no separation.
...& it is not nice to suppress the return value of main().

Your question is not entirely clear, but you should review how many bits comprise an int & char. On 32-bit systems, the following example:

CPP / C++ / C Code:

#include <iostream>

using namespace std;

int main() {
    char c = 'a';
    int i = 'a';

    cout << c << "\t" << sizeof(c) << endl;
    cout << i << "\t" << sizeof(i) << endl;

    return 0;
}

...yields the following output:

Code:

a       1
97      4

...which points out that the char value of 'a' comprises one byte on a 32-bit system, & the int value (or ASCII value...) of 'a' is 97 consumes 4 bytes on a 32-bit system.

What happens when a 1-byte char value is assigned to a 4-byte int? Since ASCII values typically are 7-bits, the most significant bit is set to zero. When assigned to an int, the three most significant bytes of the int are also set to zero.

bcompt143 · #3 11-Nov-2008, 02:50

My question is not that, I want to ask, why varialbe declared as character accept integer and display the integer, and why not the variable declared as integer is does not display the character.

Peter_APIIT · #4 11-Nov-2008, 03:47

Quote:

character accept integer and display the integer,

As long as it is a character, it will display the character.

Integer is not allow to store character but you can display it ANSII value.

cpit · #5 11-Nov-2008, 04:04

A variable of type character allows you to assign chars and numbers (like integer, float, etc...). However, you can´t use a variable of type char (even if it was assigned a number) to make calculations.

I hope this help you.

ocicat · #6 11-Nov-2008, 06:22

Quote:

Originally Posted by bcompt143

I want to ask, why varialbe declared as character accept integer and display the integer, and why not the variable declared as integer is does not display the character.

Present a case where this occurs. The code posted doesn't compile, nor have you provided a specific example which supports your question.

Actually, Peter_APIIT is right. Everything is stored as a numerical value. In the case of chars, the value is interpreted as an ASCII character, & if printable, the corresponding character glyph will be displayed. If the numeric value represents an unprintable ASCII character, nothing should be displayed.

n00pster · #7 11-Nov-2008, 09:45

you can assign a number to char because numbers also have an ascii values and they can be considered as just plain characters which does not have any other functionalities of numbers.

But you obviously cannot masquerade a charater as a number because a number is as a subset of chars with many special features which other chars dont have.

so this is expected to throw an error.

n00pster · #8 11-Nov-2008, 09:54

Quote:

Originally Posted by ocicat

Present a case where this occurs. The code posted doesn't compile, nor have you provided a specific example which supports your question.
.

eg: enter any alphabet when it ask for a number and it will break the normal execution. (Atleast if you are using vc++ 2008 )

run this again and enter a number when it asks for a number and enter a number again when it asks for a char. There is no problem in this case.

CPP / C++ / C Code:

#include <iostream>
//#include <exception>
using namespace std;

int main ( ) 
{
	int x;
	char y;
	cout<<"enter any number\n";
	//try{
	cin>>x;
	//}
	//catch (exception& e) {
	//	cout << e.what();
	//}
	cout<<"enter any character\n";
	cin>>y;
	cout<<"number = "<<x;
	cout<<"  char = "<<y;
	cout << endl;
	return 0;
}

n00pster · #9 11-Nov-2008, 11:18

you might want to use this code if you want to handle this issue.

CPP / C++ / C Code:

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

int main ( ) 
{
	int x;
	char y;
	while ((cout << "Enter any number ") && !(cin >> x)) {
		cout << "That's not a number; ";
		cin.clear();
		cin.ignore(numeric_limits<streamsize>::max(), '\n');
	}
	cout<<"enter any character\n";
	cin>>y;
	cout<<"number = "<<x;
	cout<<"  char = "<<y;
	cout << endl;
	return 0;
}

(check this out: http://www.parashift.com/c++-faq-lit....html#faq-15.3)

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