Here, at same address two different datatypes are stored .

1. don't obscure your code. 2. Use a pasting website to share code. 3. you are doing a narrowing conversion, of course the value printed will be different 0x0401 will be seen as as 0x01 0x00000401 (big endian notation) is what it is stored as in memory. when you take a int pointer, it points to 0x00000401 when you take a char pointer, it points to 0x01, when you do (p0+1) it points to the next byte which is 0x04

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11.10.2021

Peace Автор вопроса

Manav | avoid unnecessary messaging me
1. don't obscure your code. 2. Use a pasting websi...

https://pastebin.com/QYeQcjzy compiler is returning smile emoji as output of last two print statements . why ? it is the same code in C++.

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11.10.2021

Peace Автор вопроса

Peace
https://pastebin.com/QYeQcjzy ...

@usernamestring

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11.10.2021

Manav | avoid unnecessary messaging me

Peace
@usernamestring

be patient, there is a reason why i haven't replied to you yet. I need to see the output myself.

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11.10.2021

Peace Автор вопроса

Manav | avoid unnecessary messaging me
be patient, there is a reason why i haven't replie...

😁😁ohhh..

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11.10.2021

@^@ ^

Peace
@usernamestring

Lol, what emoji do you see? a smiley face?

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11.10.2021

Peace Автор вопроса

@^@ ^
Lol, what emoji do you see? a smiley face?

yes.

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11.10.2021

@^@ ^

Peace
yes.

This-> 😄, or this-> 🙂 ?

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11.10.2021

Nader Jafari · Accepted Answer

Nader Jafari

C doesnt keep track of data types in memory you used the dirty trick of casting the pointer so it just remains as 0x401 in memory where the extra bits gets removed when you de ref. the new char ptr

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11.10.2021

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Here, at same address two different datatypes are stored .

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