Why is my string not being printed?

Question

I have some code that, in its smallest complete form that exhibits the problem (being a good citizen when it comes to asking questions), basically boils down to the following:

#include <string>
#include <iostream>
int main (void) {
    int x = 11;
    std::string s = "Value was: " + x;
    std::cout << "[" << s << "]" << std::endl;
    return 0;
}

and I'm expecting it to output

[Value was: 11]

Instead, instead of that, I'm getting just:

[]

Why is that? Why can I not output my string? Is the string blank? Is cout somehow broken? Have I gone mad?

You need to include `` for this to "work" in a compliant manner, by the way. — GManNickG, Feb 03 '11 at 06:15
`gcc (Ubuntu 4.4.3-4ubuntu5) 4.4.3` compiles that without errors - maybe it's just forgiving. In any case, adding string doesn't help so I'll add it to the question so we're more kosher. However, Xeo beat me to it so thanks :-) — paxdiablo, Feb 03 '11 at 06:34

Oswald · Accepted Answer · 2011-02-03T07:30:25.327

8

"Value was: " is of type const char[12]. When you add an integer to it, you are effectively referencing an element of that array. To see the effect, change x to 3.

You will have to construct an std::string explicitly. Then again, you cannot concatenate an std::string and an integer. To get around this you can write into an std::ostringstream:

#include <sstream>

std::ostringstream oss;
oss << "Value was: " << x;
std::string result = oss.str();

edited Feb 03 '11 at 07:30

answered Feb 03 '11 at 06:15

Oswald

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In this case, since the number is `11` and the length of the string is 11, the resulting pointer points right at the null terminator. – caf Feb 03 '11 at 06:36
2

It's of the type `const char[12]`, actually. – GManNickG Feb 03 '11 at 06:36
`char[11]` actually (I think). – paxdiablo Feb 03 '11 at 06:50
@paxdiablo: `const char[12]`. It's zero-terminated implicitly. – Oswald Feb 03 '11 at 06:55
Sorry 'bout that, you're actually right about the 12, I miscounted. However, you're not quite so right about the `const`. While it's undefined to try and change them, string literals are `char`, not `const char`. Except the wide ones which aren't `char`, but I'm still pretty sure those aren't `const` either, even though I've never used them. Not that it affects my upvote for this answer :-) – paxdiablo Feb 03 '11 at 06:58
or alternatively the statement can be modified as std::string s = string("Value was: ") + x; – mukeshkumar Feb 03 '11 at 07:06
It used to be `char[]`, but according to http://www.glenmccl.com/ansi_014.htm it was decided to change it to `const char[]`. I don't know if this change already applies or if it's for the upcoming C++ standard. – Oswald Feb 03 '11 at 07:10
1

@hype: No, it cannot. This gives a _no match for ‘operator+’ in std::basic_string_ error. – Oswald Feb 03 '11 at 07:15
2

@pax: §2.13.4/1, C++03: "An ordinary string literal has type “array of n const char” and static storage duration (3.7), where n is the size of the string" – GManNickG Feb 03 '11 at 07:35

score 4 · Answer 2 · edited Feb 03 '11 at 07:02

4

You can't add a character pointer and an integer like that (you can, but it won't do what you expect).

You'll need to convert the x to a string first. You can either do it out-of-band the C way by using the itoa function to convert the integer to a string:

char buf[5];
itoa(x, buf, 10);

s += buf;

Or the STD way with an sstream:

#include <sstream>

std::ostringstream oss;
oss << s << x;
std::cout << oss.str();

Or directly in the cout line:

std::cout << text << x;

edited Feb 03 '11 at 07:02

paxdiablo

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answered Feb 03 '11 at 06:13

Mahmoud Al-Qudsi

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Wouldn't he still run into problems because x is of type int, not string or char*? – helloworld922 Feb 03 '11 at 06:15
That gives me `[Value was ]` (had to pipe through `od -xcb` to see what was going on there. – paxdiablo Feb 03 '11 at 06:16
When you do `S += x;` it is converting integer 11 to char 11 and tries to get the corresponding ASCII character. – Naveen Feb 03 '11 at 06:17

score 4 · Answer 3 · answered Feb 03 '11 at 07:49

Amusing :) That's what we pay for C-compatibility and the lack of a built-in string.

Anyway, I think the most readable way to do it would be:

std::string s = "Value was: " + boost::lexical_cast<std::string>(x);

Because the lexical_cast return type is std::string here, the right overload of + will be selected.

score 2 · Answer 4 · answered Feb 03 '11 at 06:13

2

C++ doesn't concatenate strings using the the + operator. There's also no auto-promote from data types to string.

answered Feb 03 '11 at 06:13

helloworld922

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score 2 · Answer 5 · edited Feb 03 '11 at 06:40

2

In C/C++, you cannot append an integer to a character array using the + operator because a char array decays to a pointer. To append an int to a string, use ostringstream:

#include <iostream>
#include <sstream>

int main (void) {  
  int x = 11;
  std::ostringstream out;
  out << "Value was: " << x;
  std::string s = out.str();
  std::cout << "[" << s << "]" << std::endl;
  return 0;
}

edited Feb 03 '11 at 06:40

paxdiablo

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answered Feb 03 '11 at 06:16

Vijay Mathew

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Why is my string not being printed?

5 Answers5