Trying to add a string object to an integer

Question

This is probably a very basic question for which I have been searching on google for the last 20 mins. I am not sure if i am phrasing it correctly, but I am not getting an explanation that I understand.

Basically, I have a string object and when I add an integer value x, it shortens the string by x characters.

Here is the code:

#include <iostream>
#include <string>

void Print::print(std::string str)
{
   std::cout << str << std::endl;    
}


print("formatString:" + 5);

The output is: tString:

Now i realise that the above is incorrect and during my search I have found ways correct the behaviour, but I haven’t found what is actually happening internally for me to get the above result.

Thanks

String literals are const char[N], not std::string.

Qaz
– Qaz

2014-04-08 23:02:48 +00:00
Commented Apr 8, 2014 at 23:02 — Qaz
– Qaz, Commented Apr 8, 2014 at 23:02

Deduplicator · Accepted Answer · 2014-04-08 23:22:00Z

2

The answer is simple: Pointer arithmetic.

Your string literal (array of const char including implicit 0-terminator), decays to a const char* on use, which you increment and pass to your print()-function, thus invoking the std::string-constructor for string literals.

So, yes, you start with a string object (0-terminated array of const char), but not a std::string object.

edited Apr 8, 2014 at 23:22

answered Apr 8, 2014 at 23:02

Deduplicator

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Comments

Marius Bancila · Accepted Answer · 2014-04-08 23:10:46Z

1

Basically, I have a string object

No, you do not have a string object. "formatString:" is not a std::string, but a "string" literal. It is in fact a const char*. A const char* has a operator + defined that takes an integer and advances the value of the pointer with a number of positions. In your case it's 5.

To get a compiler error you'd have to wrap the literal in a std::string.

print(std::string("formatString:") + 5);

answered Apr 8, 2014 at 23:10

Marius Bancila

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2 Comments

Danvil Over a year ago

"To get a compiler error you'd have to wrap the literal in a std::string." - LOL

cHao Over a year ago

A string literal is actually a const char[] (C++11, §2.14.5/8). The array is just decaying to a pointer here.</pedantry>

Vlad from Moscow · Accepted Answer · 2014-04-08 23:20:55Z

0

"formatString:" is a string literal that has type const char[14] That is it is an array of const char with size equal to 14 (the array includes the terminating zero). In expressions like this

"formatString:" + 5

the array is implicitly converted to a pointer to its first element. So if for example const char *p denotes this pointer then the expression looks as

p + 5

The result of the expression is a pointer that points to the element of the array with index 5. That is there is used the pointer arithmetic.

P + 5 points to the first symbol of string "tString" And this expression is used by the constructor of class std::string.

answered Apr 8, 2014 at 23:20

Vlad from Moscow

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Comments

Community · Accepted Answer · 2020-06-20 09:12:55Z

0

Examine the following,

#include <iostream>

void print(std::string str)
{
    std::cout << str << std::endl;
}

int main(int argc, char* argv[])
{
    //following two lines created implicitly by the compiler
    const char* pstr = "formatString";
    std::string tmp(pstr + 5);  //string c-tor: string (const char* s);
    
    // now tmp: --> "tString"
    
    print(tmp);
    
    return 0;
}

pstr is a pointer and you are doing pointer arithmetic when you use + operation.

Note:Compiler may create different internal structure, but it is a instructive way to think the above two lines.

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Apr 8, 2014 at 23:48

Fredrick Gauss

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