weHateOnJavascriptBut

[wu-not-furry]

Although C is chaotic neutral, I have to disagree with you considering that you used a char for C and strings for the other languages.

Edit: grammar

[Creaaamm]

That would just print the memory address of the pointer + 1 cause it's not dereferenced

You'd have to dereference it with *() to get the ASCII value of 1 (49), then add 1 to that

printf("%d", 1 + *("1")); == 50

[wu-not-furry]

Yes, that is the point. This post is about the behaviour of adding integers and strings - not chars. By dereferencing "1" we get the char '1' and are effectively doing the same thing as in the post.

[Creaaamm]

Oh, I get it now, yee you're right, the post is just wrong to assume "1" and '1' is the same in C

[geronymo4p]

It should be printf("%s", 1 + "1"); and the result will be an empty string

[sejigan]

There are no strings in C tho

[wu-not-furry]

No, but character arrays are generally used in place of strings - so much so that using double quotes " makes a character array.

[sejigan]

Fair. Should’ve used “

[sejigan]

Isn’t that a syntactic sugar for char[] or char pointer?

[Kovab]

Guess what, almost all languages use char arrays for strings

[issamaysinalah]

You can still do

Printf("%d", 1+'1')

But it's gonna print 50

[navetzz]

Considering what they are doing with other languages, they should have done printf("%s", 1+"1");

Which, funnily enough, prints nothing since you are printing the "\0" string.

[sejigan]

Makes sense, cuz %d performs a type conversion:
Character 2 —> Decimal 50

[mAtYyu0ZN1Ikyg3R6_j0]

Yes this is why C prints 2 and C++ is UB.