how to covert 'std::string' to 'char * in c++

Hi all


how to covert 'std::string' to 'char * in c++

You can "convert" a std::string to a const char* using the c_str() member function.

If you need a modifiable char* instead, then you have to copy over the characters.

i want it to be modifiable, can you help?

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i want it to be modifiable, can you help?

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Yes, but why not just use std::string?

As I said, you just need to copy over the characters.

Code:

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std::string contents("something");
std::vector<char> buff(contents.begin(), contents.end());

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&buff[0] could be used as modifiable char array

I have a standard function which nedds "char* " to be modifiable...

I didnt get what you mean by "copy over the characters."

DOes it mean i need to use "strcpy" function?

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DOes it mean i need to use "strcpy" function?

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Yes, or you could use what vart suggested.

ya i got it!!!!

Thanks alot ,laserlight & vast !!!!!

If the modified string will be larger than the original string and you're using a vector<char>, be sure to call vector's reserve() member function first to reserve enough space to store the new string...

>> I have a standard function which nedds "char* " to be modifiable
If it is a "standard" function, there's probably a better alternative that works on strings.

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Originally Posted by vart

Code:

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std::string contents("something");
std::vector<char> buff(contents.begin(), contents.end());

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&buff[0] could be used as modifiable char array

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Quote:

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Originally Posted by chintugavali

I have a standard function which nedds "char* " to be modifiable...

I didnt get what you mean by "copy over the characters."

DOes it mean i need to use "strcpy" function?

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Originally Posted by laserlight

Yes, or you could use what vart suggested.

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Not if you want to follow the standard. &buff[0] is a modifiable char array, but it is not guaranteed to be null terminated. On the other hand strcpy requires the string to be null terminated. So you should use &buff[0] if the string length is specified, and strcpy if the string must be null terminated.

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Not if you want to follow the standard. &buff[0] is a modifiable char array, but it is not guaranteed to be null terminated.

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Good point. In fact, it will not be null terminated since vart's example copied the actual string elements based on an iterator pair.

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On the other hand strcpy requires the string to be null terminated. So you should use &buff[0] if the string length is specified, and strcpy if the string must be null terminated.

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Or we could modify vart's example to:

Code:

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std::string contents("something");
std::vector<char> buff(contents.c_str(), contents.length() + 1);

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