Re: Concatenating multiple std::string's

From:

red floyd <redfloyd@gmail.com>

Newsgroups:

comp.lang.c++

Date:

Tue, 9 Nov 2010 16:05:29 -0800 (PST)

Message-ID:

<6f2de667-c27c-4069-87d7-307b8bd9bcb3@p7g2000prb.googlegroups.com>

On Nov 9, 3:56 pm, A B <ab47...@gmail.com> wrote:

Hi,

I recently came across this strange behavior (from my point of view)
with C++ strings. I thought it was supposed to be intuitively easy to
use, unlike C-style char arrays. In particular, operator+ was
supposed to concatenate strings on the fly and let the compiler figure
out the low-level details. Here is a short code sample that I tested:

#include <string>
#include <iostream>

using namespace std;

int main() {

    const string name("sample");
    const string ext("txt");

    const char* const srcName = (string("file_") + name + string(".=

+ ext).c_str();

    cout << "srcName, when created: " << srcName << endl;

    const string trgNameStr = name + string("_1.") + ext;

    cout << "srcName, after trgNameStr has been created: " << srcName
<< endl;

    const char* const trgName = trgNameStr.c_str();

    cout << "trgName, when created: " << trgName << endl;

    return 0;

}

Under GCC 4.4.3, this produces

srcName, when created: file_sample.txt
srcName, after trgNameStr has been created: sample_1.txt
trgName, when created: sample_1.txt

So, the mere fact of creating a "target file name" string modifies the
"source file name" string, and makes it equal to the target??? Despite
the fact that everything is constant? This is not the behavior I
expected from my limited C++ knowledge. Could one of the gurus please
clarify?

You've got UB. I believe that once the temporaries involved in
creating
srcName go away, you've got a pointer to deallocated memory.

Sort of like:

char *s = new char[100];
strcpy(s,"Hello");
char *t = s;
delete[] s;