CODEWITHSUNDEEP
c++arrays
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Reputation: 75

copy a c string into a char array

New to C, still getting a grasp on pointers. I'm trying to add a copy of a c string into a char array such that

char temp[40];
temp[0] = 'a';
temp[1] = 'b';
temp[2] = 'c';
temp[3] = NULL;
char *container[40] = {0};
strcpy(container[0], temp);
cout << container[0] << endl;

prints "abc". Trying to make a copy because temp is constantly being replaced with new characters and a pointer to temp obviously won't suffice as i want to do this after abc is printed.

char temp[40];
temp[0] = 'd';
temp[1] = 'e';
temp[2] = 'f';
temp[3] = NULL;
char *container[40] = {0};
strcpy(container[1], temp);
cout << container[1] << endl;

prints 'def'. P.S this is what i have so far, but it doesn't work. I'm not too sure if im using strcpy incorrectly, but any alternatives would be helpful.

Upvotes: 0

Views: 472

Answers (1)

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 726479

You are using strcpy correctly, but you are not giving it proper memory. That is why both your programs have undefined behavior - they write to memory pointed by uninitialized pointers.

To fix this, allocate memory to each element of container using malloc:

char temp[40];
temp[0] = 'a';
temp[1] = 'b';
temp[2] = 'c';
temp[3] = NULL;
char *container[40] = {0};
container[0] = (char*)malloc(strlen(temp)+1);
strcpy(container[0], temp);
cout << container[0] << endl;

You need to add 1 to the result of strlen in order to accommodate the null terminator of your C string. Note that the cast of malloc is necessary only because you are using C++; in C, the cast in front of malloc can be dropped.

Note: I assume that this is a learning exercise to learn ways to work with "raw" pointers, because C++ provides much better facilities for working with strings. In particular, your container could be replaced with a std::vector<std::string> which would grow dynamically as you add items to it, and manage its own memory, thus eliminating the need to use malloc and strcpy.

Upvotes: 3

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