assigning char value to an int

[theju112]

#include<iostream.h>

int main()
{
	int a;
	cin>>a;
	cout<<a;
	
}

hey guys,
i had the idea that when a character is assigned to an int variable,
what would be stored in the int variable would be the ascii value of the character.

so in the above code,if i input 'a' when the program runs, i thought 97
will be the output but its not so

please explain or show me somewhere i can read about htis

[grumpy]

i had the idea that when a character is assigned to an int variable,
what would be stored in the int variable would be the ascii value of the character.

That is completely wrong.

When reading an int using formatted I/O, the digits 0-9 and sign characters (+/-) are interpreted in order to get the value. So "+152" will result in an int getting the value 152. If other characters are encountered, reading stops, and that character is left in the stream.

If doing a formatted read of an integer in other bases, the same idea applies, except the range of characters interprets as digits changes (octal 0-8, hex 0-9 and A-E or a-e).

If you want input 'a' to get the value 97, then make a of type char. Then the character is read directly from the stream, not interpreted in terms of formatted digits. (This assuming your compiler implementation works with ASCII characters, which is common but not actually guaranteed).

[programiz]

You cannot assign character to an integer variable.
For more information visit: Input and Output in C

[laserlight]

You cannot assign character to an integer variable.

That is not true.

For more information visit: Input and Output in C

It looks like you are linking to your own resource. That can be fine, except that the page that you linked to is concerned with C, yet this is C++. Now, C I/O can be used in C++, but theju112's post is not about that, so your link is irrelevant.