C++ Concatenating const char * with string, only const char * prints

Question

I am trying to cout a const char *

This is how I convert an int to a string and concatenate it with the const char*

char tempTextResult[100];
const char * tempScore = std::to_string(6).c_str();
const char * tempText = "Score: ";
strcpy(tempTextResult, tempText);
strcat(tempTextResult, tempScore);
std::cout << tempTextResult;

The result when printing is: Score:

Does anyone know why the 6 is not printing?

Thanks in advance.

Any reasons you don't just write `std::string textResult = std::string("Score: ") + std::to_string(6); std::cout << textResult << std::endl`? — Simon Kraemer, Feb 08 '16 at 17:22
@SimonKraemer Well, it wasn't about printing the value. I needed it to render the text in my game made with SDL. — Hugius, Feb 08 '16 at 17:24

score 6 · Answer 1 · answered Feb 08 '16 at 17:13

6

As the docs for c_str say, "The pointer returned may be invalidated by further calls to other member functions that modify the object." This includes the destructor.

const char * tempScore = std::to_string(6).c_str();

This makes tempScore point to a temporary string that no longer exists. You should do this:

std::string tempScore = std::to_string(6);
...
strcat(tempTextResult, tempScore.c_str());

Here, you're calling c_str on a string that continues to exist.

answered Feb 08 '16 at 17:13

David Schwartz

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And of course, accumulating the whole thing in a `std::string` would be even better. – Sebastian Redl Feb 08 '16 at 17:15
@SebastianRedl Well, that's kind of impossible because SDL_ttf needs a const char * as text to render... – Hugius Feb 08 '16 at 17:16
3

@OpenGLManiac You just managed to use `strcat`, which also wants a `const char*` I think you can handle `SDL_ttf`. – Sebastian Redl Feb 08 '16 at 17:17

2785528 · Answer 2 · 2016-02-08T17:45:38.307

0

You have marked this post as C++.

One possible C++ approach: (not compiled, not tested)

std::string result;  // empty string
{
   std::stringstream ss;
   ss << "Score: "  // tempText literal
      << 6;         // tempScore literal
   // at this point, the values placed into tempTextResult 
   //    are contained in ss
   result = ss.str();    // because ss goes out of scope
}
// ss contents are gone

// ...   many more lines of code

// ... now let us use that const char* captured via ss
std::cout << result.c_str() << std::endl;
//                  ^^^^^^^ - returns const char*

edited Feb 08 '16 at 17:45

answered Feb 08 '16 at 17:21

2785528

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Well, I needed it to be a const char *, and this looks like a string(I don't know what stringstream is so I am not sure) – Hugius Feb 08 '16 at 17:25
@OpenGLManiac Ok ... I've appended the c_str() so that the result out of ss is const char*. – 2785528 Feb 08 '16 at 17:28
Added reminder that the limited scope of ss (via braces) requires that I create and capture the ss contents to a result string .. for later use. – 2785528 Feb 08 '16 at 17:48
@OpenGLManiac -- std::stringstream can be considered a ram based stream ... simpler to use, works very much like fstream (but without open and close) and ram is somewhat faster than (most) alternatives. – 2785528 Feb 08 '16 at 17:52

C++ Concatenating const char * with string, only const char * prints

2 Answers2