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I have the following code in C now
int length = 50
char *target_str = (char*) malloc(length);
char *source_str = read_string_from_somewhere() // read a string from somewhere
// with length, say 20
memcpy(target_str, source_str, length);
The scenario is that target_str is initialized with 50 bytes. source_str is a string of length 20.
If I want to copy the source_str to target_str i use memcpy() as above with length 50, which is the size of target_str. The reason I use length in memcpy is that, the source_str can have a max value of length but is usually less than that (in the above example its 20).
Now, if I want to copy till length of source_str based on its terminating character ('\0'), even if memcpy length is more than the index of terminating character, is the above code a right way to do it? or is there an alternative suggestion.
Thanks for any help.
The scenario is that target_str is initialized with 50 bytes. source_str is a string of length 20.
If I want to copy the source_str to target_str i use memcpy() as above with length 50, which is the size of target_str.
currently you ask for memcpy to read 30 characters after the end of the source string because it does not care of a possible null terminator on the source, this is an undefined behavior
because you copy a string you can use strcpy rather than memcpy
but the problem of size can be reversed, I mean the target can be smaller than the source, and without protection you will have again a undefined behavior
so you can use strncpy giving the length of the target, just take care of the necessity to add a final null character in case the target is smaller than the source :
int length = 50
char *target_str = (char*) malloc(length);
char *source_str = read_string_from_somewhere(); // length unknown
strncpy(target_str, source_str, length - 1); // -1 to let place for \0
target_str[length - 1] = 0; // force the presence of a null character at end in case
If I want to copy the source_str to target_str i use memcpy() as above
with length 50, which is the size of target_str. The reason I use
length in memcpy is that, the source_str can have a max value of
length but is usually less than that (in the above example its 20).
It is crucially important to distinguish between
the size of the array to which source_str points, and
the length of the string, if any, to which source_str points (+/- the terminator).
If source_str is certain to point to an array of length 50 or more then the memcpy() approach you present is ok. If not, then it produces undefined behavior when source_str in fact points to a shorter array. Any result within the power of your C implementation may occur.
If source_str is certain to point to a (properly-terminated) C string of no more than length - 1 characters, and if it is its string value that you want to copy, then strcpy() is more natural than memcpy(). It will copy all the string contents, up to and including the terminator. This presents no problem when source_str points to an array shorter than length, so long as it contains a string terminator.
If neither of those cases is certain to hold, then it's not clear what you want to do. The strncpy() function may cover some of those cases, but it does not cover all of them.
Now, if I want to copy till length of source_str based on its terminating character ('\0'), even if memcpy length is more than the index of terminating character, is the above code a right way to do it?
No; you'd be copying the entire content of source_str, even past the null-terminator if it occurs before the end of the allocated space for the string it is pointing to.
If your concern is minimizing the auxiliary space used by your program, what you could do is use strlen to determine the length of source_str, and allocate target_str based on that. Also, strcpy is similar to memcpy but is specifically intended for null-terminated strings (observe that it has no "size" or "length" parameter):
char *target_str = NULL;
char *source_str = read_string_from_somewhere();
size_t len = strlen(source_str);
target_str = malloc(len + 1);
strcpy(target_str, source_str);
// ...
free(target_str);
target_str = NULL;
memcpy is used to copy fixed blocks of memory, so if you want to copy something shorter that is terminated by '\n' you don't want to use memcpy.
There is other functions like strncpy or strlcpy that do similar things.
Best to check what the implementations do. I removed the optimized versions from the original source code for the sake of readability.
This is an example memcpy implementation: https://git.musl-libc.org/cgit/musl/tree/src/string/memcpy.c
void *memcpy(void *restrict dest, const void *restrict src, size_t n)
{
unsigned char *d = dest;
const unsigned char *s = src;
for (; n; n--) *d++ = *s++;
return dest;
}
It's clear that here, both pieces of memory are visited for n times. regardless of the size of source or destination string, which causes copying of memory past your string if it was shorter. Which is bad and can cause various unwanted behavior.
this is strlcpy from: https://git.musl-libc.org/cgit/musl/tree/src/string/strlcpy.c
size_t strlcpy(char *d, const char *s, size_t n)
{
char *d0 = d;
size_t *wd;
if (!n--) goto finish;
for (; n && (*d=*s); n--, s++, d++);
*d = 0;
finish:
return d-d0 + strlen(s);
}
The trick here is that n && (*d = 0) evaluates to false and will break the looping condition and exit early.
Hence this gives you the wanted behaviour.
Use strlen to determine the exact size of source_string and allocate accordingly, remembering to add an extra byte for the null terminator. Here's a full example:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void) {
char *source_str = "string_read_from_somewhere";
int len = strlen(source_str);
char *target_str = malloc(len + 1);
if (!target_str) {
fprintf(stderr, "%s:%d: malloc failed", __FILE__, __LINE__);
return 1;
}
memcpy(target_str, source_str, len + 1);
puts(target_str);
free(target_str);
return 0;
}
Also, there's no need to cast the result of malloc. Don't forget to free the allocated memory.
As mentioned in the comments, you probably want to restrict the size of the malloced string to a sensible amount.
Have a
typedef struct person {
char name[20]
char surname[20]
} person_t;
I need to create a string like XXXXXX:YYYYYY with the function like
char* personToString(person_t *p). I tried to make it:
char* personToString(person_t* p) {
int n1,n2;
n1=strlen(p->name);
n2=strlen(p->surname);
char *p = (char*) malloc((n1+n2+2)*sizeof(char));
strcat(p,puser->name);
strcat(p,":");
strcat(p,puser->surname);
return p;
}
This give me a reasonable output but I have some errors testing with valgrind! I also think that there is a way more classy to write the function!
When you malloc memory for p the memory will hold garbage values. Strcat will append a string after the null character, but in an uninitialized string will hold random values.
Replace the first strcat with strcpy.
You need to
strcpy(p,puser->name);
not
strcat(p,puser->name);
malloc does not initialize the buffer to zero, so strcat is searching for a null byte in p first and probably not finding one, reading past the end of the buffer and thus crashing.
Instead of one strcpy plus two strcat you can also write one call to sprintf:
sprintf(p, "%s:%s", puser->name, puser->surname);
First you should call string copy, then strcat:
strcat(p,puser->name);
should be:
strcpy(p,puser->name);
because memory allocated with malloc function keeps values garbage, by doing strcat for first you are concatenating after garbage -- it also brings Undefined behaviour in your code.
You can use void* calloc (size_t num, size_t size); instead of malloc(), calloc function initialized allocated memory with 0 (then strcat() no problem).
Also dynamically allocated memory you should deallocate memory block using void free (void* ptr);) explicitly.
This looks good to me,
char* personToString( struct person_t *p )
{
int len = strlen(p->name) + strlen(p->surname) + 2; // holds ':' + NULL
char *str = malloc( len ); // Never cast malloc's return value in C
// Check str for NULL
if( str == NULL )
{
// we are out of memory
// handle errors
return NULL;
}
snprintf( str, len, "%s:%s", p->name, p->surname);
return str;
}
NOTE:
Never cast malloc's return value in C.
Use snprintf when multiple strcat is needed, its elegant.
free the return value str here in caller.
Fixed struct and char variables.
I have a string:
char * someString;
If I want the first five letters of this string and want to set it to otherString, how would I do it?
#include <string.h>
...
char otherString[6]; // note 6, not 5, there's one there for the null terminator
...
strncpy(otherString, someString, 5);
otherString[5] = '\0'; // place the null terminator
Generalized:
char* subString (const char* input, int offset, int len, char* dest)
{
int input_len = strlen (input);
if (offset + len > input_len)
{
return NULL;
}
strncpy (dest, input + offset, len);
return dest;
}
char dest[80];
const char* source = "hello world";
if (subString (source, 0, 5, dest))
{
printf ("%s\n", dest);
}
char* someString = "abcdedgh";
char* otherString = 0;
otherString = (char*)malloc(5+1);
memcpy(otherString,someString,5);
otherString[5] = 0;
UPDATE:
Tip: A good way to understand definitions is called the right-left rule (some links at the end):
Start reading from identifier and say aloud => "someString is..."
Now go to right of someString (statement has ended with a semicolon, nothing to say).
Now go left of identifier (* is encountered) => so say "...a pointer to...".
Now go to left of "*" (the keyword char is found) => say "..char".
Done!
So char* someString; => "someString is a pointer to char".
Since a pointer simply points to a certain memory address, it can also be used as the "starting point" for an "array" of characters.
That works with anything .. give it a go:
char* s[2]; //=> s is an array of two pointers to char
char** someThing; //=> someThing is a pointer to a pointer to char.
//Note: We look in the brackets first, and then move outward
char (* s)[2]; //=> s is a pointer to an array of two char
Some links:
How to interpret complex C/C++ declarations and
How To Read C Declarations
You'll need to allocate memory for the new string otherString. In general for a substring of length n, something like this may work for you (don't forget to do bounds checking...)
char *subString(char *someString, int n)
{
char *new = malloc(sizeof(char)*n+1);
strncpy(new, someString, n);
new[n] = '\0';
return new;
}
This will return a substring of the first n characters of someString. Make sure you free the memory when you are done with it using free().
You can use snprintf to get a substring of a char array with precision:
#include <stdio.h>
int main()
{
const char source[] = "This is a string array";
char dest[17];
// get first 16 characters using precision
snprintf(dest, sizeof(dest), "%.16s", source);
// print substring
puts(dest);
} // end main
Output:
This is a string
Note:
For further information see printf man page.
You can treat C strings like pointers. So when you declare:
char str[10];
str can be used as a pointer. So if you want to copy just a portion of the string you can use:
char str1[24] = "This is a simple string.";
char str2[6];
strncpy(str1 + 10, str2,6);
This will copy 6 characters from the str1 array into str2 starting at the 11th element.
I had not seen this post until now, the present collection of answers form an orgy of bad advise and compiler errors, only a few recommending memcpy are correct. Basically the answer to the question is:
someString = allocated_memory; // statically or dynamically
memcpy(someString, otherString, 5);
someString[5] = '\0';
This assuming that we know that otherString is at least 5 characters long, then this is the correct answer, period. memcpy is faster and safer than strncpy and there is no confusion about whether memcpy null terminates the string or not - it doesn't, so we definitely have to append the null termination manually.
The main problem here is that strncpy is a very dangerous function that should not be used for any purpose. The function was never intended to be used for null terminated strings and it's presence in the C standard is a mistake. See Is strcpy dangerous and what should be used instead?, I will quote some relevant parts from that post for convenience:
Somewhere at the time when Microsoft flagged strcpy as obsolete and dangerous, some other misguided rumour started. This nasty rumour said that strncpy should be used as a safer version of strcpy. Since it takes the size as parameter and it's already part of the C standard lib, so it's portable. This seemed very convenient - spread the word, forget about non-standard strcpy_s, lets use strncpy! No, this is not a good idea...
Looking at the history of strncpy, it goes back to the very earliest days of Unix, where several string formats co-existed. Something called "fixed width strings" existed - they were not null terminated but came with a fixed size stored together with the string. One of the things Dennis Ritchie (the inventor of the C language) wished to avoid when creating C, was to store the size together with arrays [The Development of the C Language, Dennis M. Ritchie]. Likely in the same spirit as this, the "fixed width strings" were getting phased out over time, in favour for null terminated ones.
The function used to copy these old fixed width strings was named strncpy. This is the sole purpose that it was created for. It has no relation to strcpy. In particular it was never intended to be some more secure version - computer program security wasn't even invented when these functions were made.
Somehow strncpy still made it into the first C standard in 1989. A whole lot of highly questionable functions did - the reason was always backwards compatibility. We can also read the story about strncpy in the C99 rationale 7.21.2.4:
The strncpy function
strncpy was initially introduced into the C library to deal with fixed-length name fields in
structures such as directory entries. Such fields are not used in the same way as strings: the
trailing null is unnecessary for a maximum-length field, and setting trailing bytes for shorter
5 names to null assures efficient field-wise comparisons. strncpy is not by origin a “bounded
strcpy,” and the Committee preferred to recognize existing practice rather than alter the function
to better suit it to such use.
The Codidact link also contains some examples showing how strncpy will fail to terminate a copied string.
I think it's easy way... but I don't know how I can pass the result variable directly then I create a local char array as temp and return it.
char* substr(char *buff, uint8_t start,uint8_t len, char* substr)
{
strncpy(substr, buff+start, len);
substr[len] = 0;
return substr;
}
strncpy(otherString, someString, 5);
Don't forget to allocate memory for otherString.
#include <stdio.h>
#include <string.h>
int main ()
{
char someString[]="abcdedgh";
char otherString[]="00000";
memcpy (otherString, someString, 5);
printf ("someString: %s\notherString: %s\n", someString, otherString);
return 0;
}
You will not need stdio.h if you don't use the printf statement and putting constants in all but the smallest programs is bad form and should be avoided.
Doing it all in two fell swoops:
char *otherString = strncpy((char*)malloc(6), someString);
otherString[5] = 0;
char largeSrt[] = "123456789-123"; // original string
char * substr;
substr = strchr(largeSrt, '-'); // we save the new string "-123"
int substringLength = strlen(largeSrt) - strlen(substr); // 13-4=9 (bigger string size) - (new string size)
char *newStr = malloc(sizeof(char) * substringLength + 1);// keep memory free to new string
strncpy(newStr, largeSrt, substringLength); // copy only 9 characters
newStr[substringLength] = '\0'; // close the new string with final character
printf("newStr=%s\n", newStr);
free(newStr); // you free the memory
Try this code:
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
char* substr(const char *src, unsigned int start, unsigned int end);
int main(void)
{
char *text = "The test string is here";
char *subtext = substr(text,9,14);
printf("The original string is: %s\n",text);
printf("Substring is: %s",subtext);
return 0;
}
char* substr(const char *src, unsigned int start, unsigned int end)
{
unsigned int subtext_len = end-start+2;
char *subtext = malloc(sizeof(char)*subtext_len);
strncpy(subtext,&src[start],subtext_len-1);
subtext[subtext_len-1] = '\0';
return subtext;
}
Ok, so I have a char stringA and char stringB, and I want to be able to insert stringB into stringA at point x.
char *stringA = "abcdef";
char *stringB = "123";
with a product of "ab123cdef"
does anyone know how to do this? Thanks in advance
char * strA = "Blahblahblah", * strB = "123", strC[50];
int x = 4;
strncpy(strC, strA, x);
strC[x] = '\0';
strcat(strC, strB);
strcat(strC, strA + x);
printf("%s\n", strC);
Explanation:
You declare the two strings you will be joining, plus a third string into which you will put them.
You declare an integer to tell the program after how many characters you wish to insert the second string into the first.
The strncpy function copies the first x characters into strC. You have to add the null ('\0') character at the end, otherwise you'll probably get rubbish.
Strcat to copy the second string.
Another strcat to copy the remaining part of the first string (strA + x).
Hope that helps.
Remark: remember to make strC long enough to contain both strA and strC, otherwise you'll produce a segmentation fault. You may do this by declaring the string as an array, like in my example.
stringA and stringB are both pointers - they contain the starting address for a blob of memory. The memory they are pointing to contain continuous strings of characters: "abcdef" and "123" respectively. Since strings are contiguous blocks memory (meaning that the memory location of a given character is one byte after the previous) you can't insert more characters into the middle of a string without first moving some characters. In your case you can't even really do this, since the amount of memory allocated for each string is exactly large enough to hold JUST that string (ignoring padding).
What you are going to have to do is copy the strings to another block of memory, one that you have allocated for that purpose, and copy them so that the second string starts x characters into the first string.
Several other SO users have posted code-solutions but I think you should try and find the exact solution on your own (and hopefully my high-level explanation of what's going on will help).
Here a more general solution.
Note that destination must have enough space in memory to add seed (e.g. an heap-allocated array of size grater than strlen(seed)+strlen(destination) ). So, concerning the question, you have to create a larger array.
/*
destination: a NULL terminated string
pos: where insert seed
seed: a NULL terminated string
*/
void insertString(char* destination, int pos, char* seed)
{
char * strC;
strC = (char*)malloc(strlen(destination)+strlen(seed)+1);
strncpy(strC,destination,pos);
strC[pos] = '\0';
strcat(strC,seed);
strcat(strC,destination+pos);
strcpy(destination,strC);
free(strC);
}
int insert_pos = 5;
int product_length = strlen(stringA) + strlen(stringB) + 1;
char* product = (char*)malloc(product_length);
strncpy(product, stringA, insert_pos);
product[insert_pos] = '\0';
strcat(product, stringB);
strcat(product, stringA + insert_pos);
...
free(product);
char *strInsert(char *str1, const char *str2, int pos) {
size_t l1 = strlen(str1);
size_t l2 = strlen(str2);
if (pos < 0) pos = 0;
if (pos > l1) pos = l1;
char *p = str1 + pos;
memmove(p + l2, p, l1 - pos);
memcpy (p, str2, l2);
return str1;
}
Can u Give solution for this code of typecasting, LPCTSTR(here lpsubkey) to Char*
for below code snippet ,
char* s="HKEY_CURRENT_USER\\";
strcat(s,(char*)lpSubKey);
printf("%S",s);
here it makes error of access violation ,so what will be the solution for that?.
...thanks in advance
There are several issues with your code that might well lead to the access violation. I don't think any have anything to do with the cast you mentioned.
You are assigning a pointer to the first element of a fixed size char array to a char * and then attempt to append to this using strcat. This is wrong as there is no additional space left in the implicitly allocated string array. You will need to allocate a buffer big enough to hold the resulting string and then copy the string constant in there before calling strcat. For example, like so:
char *s = (char*)malloc(1024 * sizeof(char));
strcpy(s, "HKEY_CURRENT_USER\\");
strcat(s, T2A(lpSubKey));
printf("%s", s);
free(s);
Please note that the fixed size array I'm allocating above is bad practise. In production code you should always determine the correct size of the array on the go to prevent buffer overflows or use functions like strncat and strncpy to ensure that you are not copying more data into the buffer than the buffer can hold.
These are not the same thing. What are you trying to do?
The problem is you are trying to append to a string that you have not reserved memory for.
Try:
char s[1024] = "HKEY_CURRENT_USER";
strcat(s,(char*)lpSubKey );
printf("%S",s);
Do be careful with the arbitrary size of 1024. If you expect your keys to be much longer your program will crash.
Also, look at strcat_s.
ATL and MFC has set of macros to such conversion, where used next letters:
W - wide unicode string
T - generic character string
A - ANSI character string
OLE - BSTR string,
so in your case you need T2A macros
strcat does not attempt to make room for the combination. You are overwriting memory that isn't part of the string. Off the top of my head:
char *strcat_with_alloc(char *s1, char *s2)
{
if (!s1 || !s2) return NULL;
size_t len1 = strlen(s1);
size_t len2 = strlen(s2);
char *dest = (char *)malloc(len1 + len2 + 1);
if (!dest) return NULL;
strcpy(dest, s1);
strcat(dest, s2);
return dest;
}
now try:
char* s="HKEY_CURRENT_USER\\";
char *fullKey = strcat_with_alloc(s,(char*)lpSubKey);
if (!fullKey)
printf("error no memory");
else {
printf("%S",fullKey);
free(fullKey);
}