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I am learning about implicit conversions in C++. And i read the following example:

char a;
std::cin>>a; //I can enter an integer like 56 here
std::cout<<a<<std::endl; //for the input 56 it will display 5 because of its ASCII value

I understood the above example by reading about it in different books and posts on SO. For example, if i provide the input J, then the program successfully prints J on the console. Similarly if i provide the input say 56 then the output will be 5 because of its ASCII value.

But then i tried the opposite as shown below:

int a;
std::cin>>a;//if i provide the input as the character J then why is the output 0 instead of the corresponding code point of `J`
std::cout<<a<<std::endl;

For the above snippet, if i provide the input 56 then the output is correctly printed as 56. But if i provide the input as J then the output is 0.

So my question is in the above 2nd snippet why the code point corresponding to the character J is not printed and instead we get 0 printed on the console. I mean, a is an integer variable so it is able to store the code point corresponding to the character J and then when we do cout<<a; we should be getting that code point as the output instead of 0. What is happening here. Is this related to implicit conversion like a char can be promoted to an int or something else.

Anoop Rana
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    Your comments in the first passage are incorrect. The output is `5` because `'5'` is the first character you entered and `'5'` got stored in `a`, not because the ASCII value of `'5'` is `56`. – Nathan Pierson Apr 18 '22 at 05:34
  • When you read characters, you will read actual characters. If you give the input `56` then the character `'5'` will be stored in the variable `a`, not the integer value `56` (which is the ASCII encoded value for the character `'8'`). – Some programmer dude Apr 18 '22 at 05:35
  • @Someprogrammerdude Ok, can you explain in an answer what exactly happens in the 2nd snippet. I get your point about the 1st snippet. – Anoop Rana Apr 18 '22 at 05:38
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    @273K Since C++11 it does initialize the variables to zero. See e.g. [this old answer of mine](https://stackoverflow.com/questions/13378989/why-does-stringstream-change-value-of-target-on-failure/13379073#13379073). – Some programmer dude Apr 18 '22 at 05:38
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    @Richard In the second example the extraction fails since you don't give an integer as input, which will set the variable to zero and set the `failbit` state in the stream. – Some programmer dude Apr 18 '22 at 05:40
  • `std::cin>>a;` will stop parsing as soon as it finds something that cannot be turned into a digit in an `int`. If the parsing stops immediately, nothing is extracted from the stream and the stream is placed into fail state. Depending on the C++ Standard you are compiling to you will either get a 0, newer Standards, or unchanged value, older Standards (this could be undefined behaviour. Can't remember, unfortunately). In the case of inputting j, instant failure, probably a zero value stored in `a`, and a failed stream. – user4581301 Apr 18 '22 at 05:40

2 Answers2

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When istream operator<< expects something but gets something else, its failbit is set and in this state, the stream will not do anything.
So if there is a chain, it is stopped when a wrong input is met.
About handling this here.

But something else is also happening.
Since C++11, if extraction fails, zero is written to its value.

int a; // default-initializing fundamental type (no init / garbage value)
std::cin >> a;

If a char is encountered, a will become 0.

And also for integers, if the entered value is more or less than the type can hold, failbit is set and it will become std::numeric_limits<T>::max/min() or max() if T is unsigned.

MisaghM
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The << operator works different depending on what is on the right hand side. In the first example it reads the char ‘5’ with a ASCII value of 53 and throws the ‘6’ away. In the second case it reads the int fifty-six. When you enter “J” that’s not an int, so trying to read an Int gets a result of 0.

And the >> operator does the same thing: When you output ‘5’ it’s written, it is written as a char, and the int 56 is written as an Int.

gnasher729
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  • You mixed up the two operators. And the '6' is not thrown away but remains for the next input operation. – interjay Apr 18 '22 at 07:27
  • I don't think that the char `'5'` is read with an ascii value `53` as explained by Someprogrammerdude in the comments of my question in the 1st snippet. – Anoop Rana Apr 18 '22 at 08:16