Pass char array by reference and return modified

Question

I have function definition lke below

void ConvertString(std::string &str)
{
size_t pos = 0;
    while ((pos = str.find("&", pos)) != std::string::npos) {
         str.replace(pos, 1, "and");
         pos += 3;
    }
}

Purpose of this function is to find & and replace it with and. function execution in fine. I written this for all generalised string at one instance I am calling this in following way

char mystr[80] = "ThisIsSample&String";
ConvertString((std::string)mystr);
printf(mystr);

In above call I am expecting console should be printed with new modified string with "and". But some of string modification is not working , any error in function?

Why not just use std::string? std::string mystr = "ThisIsSample&String"; — Mark Garcia
– Mark Garcia, Commented Aug 27, 2014 at 6:53
You're using Visual C++ right? Because this shouldn't compile. — Benjamin Lindley
– Benjamin Lindley, Commented Aug 27, 2014 at 6:53

Cheers and hth. - Alf · Accepted Answer · 2014-08-27 07:02:01Z

4

This code:

char mystr[80] = "ThisIsSample&String";
ConvertString((std::string)mystr);
printf(mystr);

… creates a temporary string object and passes that as argument.

Since the formal argument type is by reference to non-const, this should not compile, but Visual C++ supports it as a language extension (for class types only, IIRC).

Instead do like

string s = "Blah & blah";
ConvertString( s );
cout << s << endl;

By the way, C style casts are in general an invitation to bugs, because the basic nature of such a cast can change very silently from e.g. const_cast to reinterpret_cast when the code is maintained.

It's safe enough in the hands of an experienced programmer, like a power tool such as a chain saw can be safe in the hands of an experienced woodsman, but it's not a thing that a novice should use just to save a little work.

answered Aug 27, 2014 at 7:02

Cheers and hth. - Alf

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Comments

Some programmer dude · Accepted Answer · 2014-08-27 06:52:35Z

2

It's because you create a temporary std::string object (whose initial content is the content of the array mystr), and pass that temporary object by reference to the function. This temporary object is then destructed when the call id done.

answered Aug 27, 2014 at 6:52

Some programmer dude

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1 Comment

quantdev Over a year ago

It should not even compile

Basile Starynkevitch · Accepted Answer · 2014-08-27 06:53:55Z

1

Did you read some documentation of std::string and of printf?

You need

std::string mystr = "ThisIsSample&String";
ConvertString(mystr);
printf(mystr.c_str());

You obviously want to pass by reference a string variable (technically an l-value) to your ConvertString

answered Aug 27, 2014 at 6:53

Basile Starynkevitch

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Comments

Karolis Ryselis · Accepted Answer · 2014-08-27 07:01:45Z

1

I believe your problem is that you cast char array to string.

ConvertString((std::string)mystr);

this line creates a new variable of type std::string and passes it by reference. What you want is to convert it this way:

std::string convertedStr = (std::string)mystr;
ConvertString(convertedStr);
printf(convertedStr.c_str());

I am not very well aware of C++ pointer and reference syntax, but it's similar to this

edited Aug 27, 2014 at 7:01

answered Aug 27, 2014 at 6:55

Karolis Ryselis

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3 Comments

juanchopanza Over a year ago

It would look less weird to say std::string convertedStr = mystr;

Karolis Ryselis Over a year ago

@juanchopanza (std::string)mystr makes type conversion more explicit in my opinion

juanchopanza Over a year ago

If you want to be explicit you should use std::string convertedStr = std::string(mystr);. The C-style cast syntax looks pretty strange in C++.

lifeOfPI · Accepted Answer · 2014-08-27 07:18:27Z

0

what your are doing is not correct! you ~~cannot~~ should not convert a char* to a std::string with a cstyle-cast. what you should do is more like:

std::string mystr( "ThisIsSample&String" );
ConvertString(mystr);

edit: thx for -reputation... this code isn't even compiling... http://ideone.com/bCsmgf

edited Aug 27, 2014 at 7:18

answered Aug 27, 2014 at 6:55

lifeOfPI

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3 Comments

Benjamin Lindley Over a year ago

Yes, you can. It will result in a temporary std::string object, constructed from the char*.

lifeOfPI Over a year ago

@Benjamin Lindley: thx for your answer. But IMHO it is not a question if you can, it is a question about if it's correct.... my sentense "you cannot" is not about that your are not able to do so (you are able to do so many nasty things with c++), it should say: You cannot, because it will not work this way... thx anyway

juanchopanza Over a year ago

But the c-style cast is not the issue. The issue is passing an rvalue to a function taking a non-const lvalue reference. The C-style cast is just bad style.

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