Concatenating string and int in c++

Question

I am doing some exercises in C++ when I came upon something not so clear for me:

cout << "String" + 1 << endl;

outputs : tring

I suggest it is something with pointer arithmetic, but does that mean that everytime I print something in quotes that is not part of previous defined array,I actually create a char array ?

The type of `"String"` is `const char*` so yes it's pointer arithmetic, but you're not creating an array, that one is baked into the data section of the exectuble (AFAIK). — Borgleader, Feb 10 '17 at 16:08
Do you want an answer on why this happened or how to get what you want? — NathanOliver, Feb 10 '17 at 16:08
`"String"` is a string literal of static lifetime, and you're correct pointer arithmetic is happening. Careful not to add more than the length of the string as the program will start reading garbage and invoke ub. — George, Feb 10 '17 at 16:08
[Related question](http://stackoverflow.com/questions/28957950/why-in-the-code-4561-output-is-56/). — juanchopanza, Feb 10 '17 at 16:09
@PeteBecker Right, but in practice it "always" decays. You are definitely correct though. SO keeping me on my toes as usual. — Borgleader, Feb 10 '17 at 16:10
@Borgleader -- no, it doesn't "always" decay. It's exactly this kind of sloppiness that gets beginners (and sometimes more advanced programmers) in trouble. — Pete Becker, Feb 10 '17 at 16:11
You could as well write `&"String"[1]` and, since addition is commutative, `&1["String"]`. — Peter - Reinstate Monica, Feb 10 '17 at 16:14
And in any case, welcome to the brittle charm of raw data (in case you come from, say, C# or Java). Does it feel so 70s? It is. Peek, poke, you name it. — Peter - Reinstate Monica, Feb 10 '17 at 16:18
@PeteBecker if "String" is const char[7], shouldn't I get for sizeof(string) = 8, instead of my result which is 28 — stoychos, Feb 10 '17 at 16:20
@stoychos [`sizeof("String")`](http://ideone.com/PV0kT6) is 7 — PeterT, Feb 10 '17 at 16:22
@stoychos Generally `sizeof(std::string)` will be between 16 and 32 (assuming 64 bit pointers). — NathanOliver, Feb 10 '17 at 16:26
@stoychos string literal is not equal to `std::string` unless you add `S` at the end — Slava, Feb 10 '17 at 16:26
@stoychos -- `"String"` has type `const char[7]`. It's type is **not** `std::string`. — Pete Becker, Feb 10 '17 at 16:27

score 7 · Accepted Answer · answered Feb 10 '17 at 16:08

7

A quoted string (formally a string literal) is an array of const char, regardless of whether your printing it or doing anything else with it.

answered Feb 10 '17 at 16:08

Pete Becker

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score 5 · Answer 2 · answered Feb 10 '17 at 16:16

5

Code:

cout << "String" + 1 << endl;

has the same effect as this:

const char *ptr = "String";
cout << ptr + 1 << endl;

so no you do not create a new array, you just change pointer and pass it to std::cout

answered Feb 10 '17 at 16:16

Slava

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The compiler will optimize ptr + 1 to a single number. – Yashas Feb 10 '17 at 16:23
@YashasSamaga and? – Slava Feb 10 '17 at 16:25
Nothing else but I wonder if the compiler throws away the first character for doing cout< – Yashas Feb 10 '17 at 16:26
@YashasSamaga it may it may not, that does not really matter in this situation – Slava Feb 10 '17 at 16:28

Concatenating string and int in c++

2 Answers2