Initializing char arrays on the free store

OK. I have some code, but there has to be an easier way. I want to declare a char *-style string on the free store and initialize it at the same time. What I'd like to do, in other words, is something like this:

Code:

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char * theString = "TEST STRING";

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But with dynamic memory, like:

Code:

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char * theString = new char("TEST STRING");

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But that obviously does not work. Does anybody know what I'm talking about and can help me with it? Thanks!

char *test = new char[strlen(someotherchar) + 1];
strcpy(test, someotherchar);
Remember to include string.h

Thanks for the response....
Is there a similar solution for other types of arrays, e.g. int's?

Quote:

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Originally posted by roktsyntst
Thanks for the response....
Is there a similar solution for other types of arrays, e.g. int's?

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yeah, almost the same way

for example

int main(){

int *array;

int size;
cout << "How Big Do You want your array?";
cin >> size;

array = new int[size];

}

Thanks, but I'm talking about initializing...

i.e. on the stack you can say:

Code:

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int array[5] = { 1, 5, 9, 12, 13 }; // Initialize it with a set of numbers right away

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But on the heap you have to say:

Code:

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array = new int[5];

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Is there some place to put the initializer list or do you have to go one by one and copy it from some other array? Thanks again!

You have to copy them in. You may be able to do more than one at a time, depending on where they are coming from.

In C:

Code:

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#include <stdio.h>
#include <string.h>

int main(void)
{
  char *msg = strdup("Hello world in C\n");
  if(msg) printf(msg);
  free(msg);
  return 0;
}

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In C++:

Code:

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#include <iostream>
#include <string>

int main(void)
{
  string msg = "Hello world in C++\n";
  std::cout << msg;
}

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