copy a c string into a char array

Question

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.

container[0] is just a pointer to a char. You must point it to enough memory before you call strcpy. — Fiddling Bits
– Fiddling Bits, Commented Jun 2, 2015 at 20:22
C or C?? Which is it? These are C techniques, but the C++ language. And you use both terms in your question.... — Lightness Races in Orbit
– Lightness Races in Orbit, Commented Jun 2, 2015 at 20:46

Sergey Kalinichenko · Accepted Answer · 2015-06-02 21:00:55Z

3

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.

edited Jun 2, 2015 at 21:00

answered Jun 2, 2015 at 20:22

Sergey Kalinichenko

729k85 gold badges1.2k silver badges1.6k bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Lightness Races in Orbit Over a year ago

Might be worthwhile warning against doing any of this in the first place? Also you misspelt "container".

zzzzz Over a year ago

question. would this be the equiv. to char *container[] = {"abc" ,"def"}. behind the scenes are the parameters being stored as cstrings and the memory is allocated for it? note: i realise they're static values.

Sergey Kalinichenko Over a year ago

@zzzzz Yes, this would be equivalent in terms of a memory layout, except for two things: the container would have two elements instead of 40, and you would need to declare it const char *container[40] with a const, because C++ requires C strings to be declared constant.

Collectives™ on Stack Overflow

copy a c string into a char array

1 Answer 1

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related