How to robustly copy text to char* without any errors

Question

I have 2 questions..

is it necessary to add a termination character when executing the following commands against a char *string ?

strcpy();

strncpy();
Is it necessary to allocate memory before before doing any operation with the above to function against the char *string ?

for example..

char *str;
str = malloc(strlen(texttocopy));
strcpy(texttocopy, str); // see the below edit

Please explain.

EDIT : in the above code I inverted the argument. it is just typo i made while asking the question here. The correct way should be

strcpy(str, texttocopy); // :)

noted.. I was hurry to read the below answer that was edited each second. :) — Muneer, Mar 12 '14 at 07:50

score 3 · Accepted Answer · answered Mar 12 '14 at 07:43

3

The strcpy function always adds the terminator, but strncpy may not do it in some cases.

And for the second question, yes you need to make sure there is enough memory allocated for the destination. In your example you have not allocated enough memory, and will have a buffer overflow. Remember that strlen returns the length of the string without counting the terminator. You also have inverted the arguments to strcpy, the destination is the first argument.

answered Mar 12 '14 at 07:43

Some programmer dude

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I hae put `malloc(strlen(texttocopy))` . this will not allocate enough memory? So according to you,, do i need to change it as `malloc(strlen(texttocopy) + 1)` ? – Muneer Mar 12 '14 at 07:45
2

@BlueBird You need `strlen(texttocopy) + 1` – Some programmer dude Mar 12 '14 at 07:46
@BlueBird And change the arguments for the `strcpy` call. – Some programmer dude Mar 12 '14 at 07:47

score 1 · Answer 2 · answered Mar 12 '14 at 08:02

'strcpy' function copies data from source to destination address including with '\0' termination character . 'strncpy' function copies data as the same way but if there is no termination character '\0' exists in the first n bytes to be copied, termination character will not be copied then and you will need to add it by yourself to terminate the string.
You will always have to statically or dynamically allocate a memory space to play with. Therefore, you should declare a character array or dynamically allocate a chunk of memory first then you can play nice with your strings

Let's say you need 10 bytes for instance: Static allocation: char buf[10]; Dynamic allocation: char* buf = (char*) malloc(10); — Unavailable, Mar 12 '14 at 08:13

How to robustly copy text to char* without any errors

2 Answers2