Initialise struct pointers in a function in C - c

I'm fairly new to C and I'm a little confused as to the correct way to initialise struct variables which are pointers, within a function. Is this style sufficient, or do I need to allocate memory before I assign s->str?
Thank you kindly for your replies, apologies if the question is unclear as I am very new to this language.
typedef struct Mystruct{
const char* str1;
const char* str2;
}mystruct;
mystruct* mystruct_new(const char* str1, const char* str2){
mystruct *s = (mystruct*)(malloc(sizeof(mystruct)));
s->str1 = str1;
s->str2 = str2;
return s;
}

your function is legal and doesn't do anything bad. Nevertheless, you should document it to mention that the strings are not copied, only the pointers are.
So if the passed data has a shorter life than the structure itself, you may meet undefined behaviour. Example:
mystruct*func()
{
char a[]="foo";
char b[]="bar";
return mystruct_new(a,b);
}
mystruct*func2()
{
char *a="foo";
char *b="bar";
return mystruct_new(a,b);
}
int main()
{
mystruct *s = func();
printf(s->a); // wrong, memory could be trashed
mystruct *s2 = func2();
printf(s2->a); // correct
mystruct *s3 = mystruct_new("foo","bar");
printf(s3->a); // also correct, string literals have global scope
}
the above code is undefined behaviour for the first print because s->a points to some memory that is no longer allocated (local to func).
The second print is OK because s2->a points to a string literal which has infinite life span.
So maybe your function is more useful like this:
mystruct* mystruct_new(const char* str1, const char* str2){
mystruct *s = malloc(sizeof(mystruct));
s->str1 = strdup(str1);
s->str2 = strdup(str2);
return s;
}
now the memory is allocated for the strings. Don't forget to free it when discarding the structure, better done in another utility function.

If the strings being passed in to str and str2 will always be string constants than yes you can get away with doing it this way. My guess however is that this is not the case. So you would be better off making a copy of each string with strdup and assigning those to the struct members:
mystruct* mystruct_new(const char* str1, const char* str2){
mystruct *s = malloc(sizeof(mystruct));
s->str1 = strdup(str1);
s->str2 = strdup(str2);
return s;
}
Just make sure to free each of those fields before freeing the struct.

Think of it this way: when you allocate memory for the struct, you get the pointer member variables for free. So in essence, when you do this:
mystruct *s = malloc(sizeof(mystruct)); //don't cast result of malloc.
Then you can treat s->str1 in the exact same way as you would any regular char* variable, say
char *str1 = NULL;
If you want it to point to something, then you have to allocate memory for the pointers. Consider this:
mystruct* mystruct_new(const char* str1, const char* str2){
mystruct *s = malloc(sizeof(mystruct);
char* someString = getMyString(); //gets some arbitrary string
char* str1 = NULL;//just for demonstration
int length = strlen(someString) + 1;
//for struct members
s->str1 = malloc(sizeof(char) * length);
strcpy(s->str1, someString);
//For regular pointers
str1 = malloc(sizeof(char) * length);
strcpy(str1, someString);
return s;
}
Also note that if you just assign to a pointer by using the = operator instead of allocating memory, then it will only copy the address to the original value. This may or may not be what you want depending on the context. Generally, if you know the memory location will stay in scope and you don't need (or don't mind) to change the original string, then it is preferred to simply assign it. Otherwise, it is advisable to make a copy.
//Makes a copy of the string
s->str1 = malloc(sizeof(char) * length);
strcpy(s->str1, someString);
//copies the address of the original value only!
s->str1 = someString;

Use strncpy() instead of strcpy(). The latter is subject to buffer overruns.
For example in this code snippet given by another user, use the strncpy() in place of strcpy()
mystruct* mystruct_new(const char* str1, const char* str2){
mystruct *s = malloc(sizeof(mystruct);
char* someString = getMyString(); //gets some arbitrary string
char* str1 = NULL;//just for demonstration
int length = strlen(someString) + 1;
//for struct members
s->str1 = malloc(sizeof(char) * length);
strcpy(s->str1, someString);
//For regular pointers
str1 = malloc(sizeof(char) * length);
strcpy(str1, someString); // replace with strncpy(str1, someString, bufsize); where bufsize is the maximum number of characters in your string + 1 for the terminator '\0'.
return s;
}

Related

Confused between passing strings to a function (C)

Why this works:
#include <stdio.h>
void slice(char *st, int m, int n)
{
int i = 0;
while ((i + m) < n)
{
st[i] = st[i + m];
i++;
}
st[i-1] = '\0';
}
int main()
{
char st[] = "Hello";
slice(st, 1, 6);
printf("The value of string is %s\n", st);
return 0;
}
And this doesn't:
#include <stdio.h>
void slice(char *st, int m, int n)
{
int i = 0;
while ((i + m) < n)
{
st[i] = st[i + m];
i++;
}
st[i-1] = '\0';
}
int main()
{
char*st = "Hello";
slice(st, 1, 6);
printf("The value of string is %s\n", st);
return 0;
}
In first I initialized my string using:
char st[]="Hello"; (using array)
And in latter I used:
char*st="Hello"; (using pointer)
I'm kind of getting confused between these 2 initialization types, what's the key difference between declaring a string by using char st[]="Hello"; and by using char*st = "Hello";.
With char st[] = "Hello";, st[] is a modifiable array of characters. The call slice(st, 1, 6); takes the array st and converts to a pointer to the first element of the array. slice() then receives that pointer, a pointer to modifiable characters.
With char *st = "Hello";, st is a pointer that points to a string literal "Hello". With the call slice(st, 1, 6);, the function receives a copy of the pointer - a pointer to the string literal. Inside slice(), code st[i] = ... is attempting to modify a string literal, that is undefined behavior (UB). It might work, it might fail, it might work today and fail tomorrow - it is not defined.
Do not attempt to modify a string literal.
... passing strings to a function ...
In both cases, code does not pass a string to slice(), but a pointer to a string. Knowing that subtle distinction helps in understanding what is truly happening.
This is an artifact of old syntax in C:
char * s = "Hello world!";
is a non-const character pointer to const memory. It is still permitted by syntax, but the string is still not a mutable object. To be pedantic it should really be written as:
const char * s = "Hello world!";
In contrast:
char s[] = "Hello world!";
allocates a local (on the stack), mutable array and copies the string data to it (from wherever the non-mutable copy is stored in memory). Your function can then do as it likes to your local copy of the string.
The type char [] is different from the type char* (char* is a variable - int. but char[] is an array which is not a variable). However, an array name can be used as a pointer to the array.
So we can say that st[] is technically similar to *str .
the problem in the 2nd version of your code
If you have read-only strings then you can use const char* st = "hello"; or simply char* st = "hello"; . So the string is most probably be stored in a read-only memory location and you'll not be able to modify it.
However, if you want to be able to modify it, use the malloc function:
char *st= (char*) malloc(n*sizeof(char)); /* n-The initial size that you need */
// ...
free(st);
**So to allocate memory for st, count the characters ("hello"-strlen(st)=5) and add 1 for this terminating null character , and functions like scanf and strcpy will add the null character;
so the code becomes :
#include <stdio.h>
void slice(char *st, int m, int n)
{
int i = 0;
while ((i + m) < n)
{
st[i] = st[i + m];
i++;
}
st[i-1] = '\0';
}
int main()
{
char *st =malloc(6*sizeof(char)) ;
const char *cpy="hello";
strcpy(st, cpy); /* copies the string pointed by cpy (including the null character) to the st. */
slice(st, 1, 6);
printf("The value of string is %s\n", st);
return 0;
}
you can fill your string also by a for loop or by scanf() .
in the case of a large allocation you must end your code with free(st);

Allocating memory to a struct member pointer in C

I have a structure with a member that I need to pass to a function by reference. In that function, I'd like to allocate memory & assign a value. I'm having issues somewhere along the line - it seems that after the code returns from allocateMemory, the memory that I had allocated & the values that I assigned go out of scope (this may not be exactly what is happening, but it appears to be the case).
#include <stdio.h>
#include <stdlib.h>
typedef struct myStruct_t
{
char *myString;
} myStruct;
void allocateMemory(void *str);
int main(void) {
myStruct tmp = {
.myString = NULL
};
myStruct *p = &tmp;
allocateMemory(p->myString);
//allocateMemory(&(p->myString)); //also tried this
printf("%s", p->myString);
return 0;
}
void allocateMemory(void *str)
{
str = malloc(8);
((char *)str)[0] = 'a';
((char *)str)[1] = 0;
}
If I print the value of str inside of allocateMemory, the 'a' is successfully printed, but if I attempt to print p->myString in main, my string is empty.
Can anyone tell me what I'm doing wrong?
You need to pass address of the structure member and then you can change (aka allocate memory) to it. In your version of the function, you are not taking a pointer not reference of a pointer, so you can change the content of memory referenced by the pointer but not the pointer itself.
So change your function to
void allocateMemory(char **ret_str)
{
char *str = malloc(8);
str[0] = 'a';
str[1] = 0;
*ret_str = str;
}
And then call it as
allocateMemory(&p->myString)
An alternative way of writing the same function Rohan did, eliminating the need to define any new variables:
void allocateMemory(char **str, size_t size) {
*str = malloc(size);
(*str)[0] = 'a';
(*str)[1] = '\0';
}
Note that I pass a size parameter to justify using malloc() in the first place.

Append char to string with no value

How would I append a char to string that has no initial value in the following method I've started:
void append(char a) {
const char *str;
char *ret = malloc (strlen(str)+2);
strncpy(str,strlen(str),ret);
ret[strlen(str)-2] = a;
ret[strlen(str)-1] = 0;
printf("%s", str);
}
I've tried a few different answers to other questions but none have worked, please help.
Since the pointer str is not initialized, you can't add characters to what it points at. What makes sense depends on what you're going to do next. Since you don't return a value from the function, you can't access the string to which a is appended unless append calls some other function.
char *str; is not a string. It is a mnemonic that says that *str will give you a value which is supposedly a character.
str is a pointer that points at random. Make it point to an allocated bunch of memory first.
Do something like- str = malloc(100); and then do a *(str+1) = a and then a *(str+2) = '\0' to NULL terminate the string.
how about something like:
char * append(char a, const char * str)
{
char *ret = malloc (strlen(str)+2);
strncpy(str,strlen(str),ret);
ret[strlen(str)-2] = a;
ret[strlen(str)-1] = 0;
return ret;
}

strcpy using pointers

I'm trying to write strcpy on my own using pointers and I get an error during runtime.
void str_cpy(char **destination, const char *source) {
// char *s1 = *destination;
while (*source != '\0') {
**destination++ = *source++; //Get an error here
}
**destination = '\0';
}
I call the function as follows:
char *str = NULL;
str_cpy(&str, "String");
Is it not OK?
Thanks!
No, it's not okay. Why? Because str is a NULL pointer. It's pointing to nothing. When you try to write values into it, where will they go? It's not pointing to any allocated memory!
You first have to allocate memory for str. You can do:
char *str = malloc(strlen("String") + 1); // + 1 for the '\0' character at the end of C-style strings
Or you can do:
char str[256]; // make str large enough to hold 256 chars. Note that this is not as safe as the above version!
Also, destination should be a single pointer, not a double pointer. Well, it's not technically wrong to use a double pointer, it's just unnecessary.
Optionally, you can allocate the memory in the str_cpy function, like so:
void str_cpy(char **destination, const char *source) {
*destination = malloc(strlen(source) + 1);
// ... continue as normal
For simplicity's sake, this can be done in one line in a function.
void mystrcpy(char *dest, const char *src) {
while (*dest++ = *src++);
}
This being said, you do need to allocate memory for dest beforehand using malloc or just simply by having a character array like char dest[256].
I don't see any need to pass a pointer-to-pointer:
void str_cpy(char *dst, const char *src) {
while (*src != '\0') {
*dst++ = *src++;
}
*dst = '\0';
}
And you need to allocate memory for dst before passing:
const char *src = "String";
char *str = malloc(strlen(src)+1); //plus one for null byte
str_cpy(dst, src);
You should likely allocate some memory for that pointer before passing it off to a function that fills what it points to (which in this case, is NULL).
Example:
char *str = malloc(128);
if (str)
{
str_cpy(&str, "String");
free(str);
str = NULL;
}
I advise not doing this without also providing target-buffer size information (i.e. if you're writing your own, then boundary-check the target buffer, otherwise your version has the same security flaws as strcpy() which are bad enough as it is).
Note: Unless you're planning on changing the address held by the pointer passed as the target, you need not use a double pointer either. The double pointer usage you have prevents the traditional strcpy() usage pattern of:
char str[128];
str_cpy(&str, "Hello"); // error.
An array address cannot be passed as a pointer-to-pointer, so your code cannot fill a static array without an intermediate pointer:
char str[128];
char *p = str;
str_cpy(&p, "Hello"); //ok. passing address of pointer.
If this is not intentional (and I don't see why it could be unless you have ideas of internally emulating strdup() on a NULL pointer passage) You should address this.
Here is a complete implementation.
Good article from here. Describes timing and performance. I did not measure myself though.
http://www.howstuffworks.com/c35.htm
char* mystrcpy(char *dst, const char *src) {
char *ptr = dst;
while ((*dst++ = *src++) ) ;
return ptr;
}
int main(int argc, char *argv[]) {
const char *src = "This is C.\0";
char *dst = malloc(sizeof(char)*(strlen(src)+1)); //+1 for the null character
dst = mystrcpy(dst, src);
printf("%s",dst);
return 1;
}
Recently I faced same problem of above one using double pointer strcpy implementation
It might helpful to others below code
void strcpy_i( char **dst, const char *src )
{
*dst=(char *)malloc((strlen(src)+1)*sizeof(char));
char *tmp=*dst;
if(tmp == NULL || src == NULL)
return ;
while((*tmp++=*src++)!='\0');
}
int main()
{
char v[]="Vinay Hunachyal";
char *d=NULL;
strcpy_i(&d,v);
printf("%s",d);
return 0;
}
#include<stdio.h>
void main()
{
void mystrcpy(char *,char *);
char s1[100],s2[100];
char *p1;
char *p2;
p1=s1;
p2=s2;
printf("Enter the string to copy to s2...?\n");
scanf("%s",p1);
mystrcpy(p2,p1);
printf("S2 after copying = %s",p2);
}
void mystrcpy(char *p2,char *p1)
{
while(*p1!='\0')
{
*p2=*p1;
p2++;
p1++;
}
*p2='\0';
}
Its my solution..Simple to understand..

Is this a correct method using free() in C?

If I have this function foo() and I'm calling it from another function foo2(); must I free the memory in the calling function like this?
char *foo(char *str1){
char *str2;
str2 = malloc((sizeof(char) * strlen(str1)) + 1);
memcpy(str2, str1, strlen(str1) + 1)
return str2;
}
void foo2(void){
char *str1 = "Hello world!"
char *str2;
str2 = foo(str1);
...some stuff
free(str2); //Should I do this here?
}
Yes, except you don't test that your malloc is successful
You can free the allocated memory anywhere given the fact that you did the allocation, that the allocation was successful, and that the memory was not already freed.
Yeah that would work, however you have a bug:
char *foo(const char *str1)
{
char *str2 = (char *)malloc((sizeof(char) * strlen(str1)) + 1);
if (str2 != NULL)
memcpy(str2, str1, strlen(str1) + 1);
return str2;
}
void foo2
{
char *str1 = "Hello world!"
char *str2;
str2 = foo(str1);
...some stuff
free(str2);
}
Conventionally you would document the fact that the caller is responsible for freeing the returned pointer using free().
Also your foo() function is nothing more than strdup().
Yes, it is right. foo() allocates some memory and it must be freed by the caller. It's not a very good design but it works. It could be better if foo() accepts two parameters: output buffer and its size.
void foo(char* input, char* output, int* bufferSize);
If output is NULL the required size is written in bufferSize by foo().
Yes. The memory needs to be freed at some point. As you can see, you now have a tight coupling between the foo function and any of its callers. One common alternative is to pass in a pointer to a char array and the size of it. The function then returns whether it filled out the array properly. This moves the responsibility of both allocating and freeing to the caller.
bool * foo(char * str, size_t size)
{
if(size < FOO_REQUIRED_SIZE) {
return FALSE;
} else {
...
return TRUE;
}
}
void foo2(void)
{
char str[FOO_REQUIRED_SIZE];
foo(str, ARRAY_SIZE(str));
}
Instead of simply returning FALSE if the function failed, you can furthermore document the required size of the array. Or alternatively provide a function which calculates the required size and returns it.
size_t getRequiredBufSizeForFoo()
{
// Calculate how many bytes required
return ...;
}
void foo2(void)
{
size_t len = getRequiredBufSizeForFoo();
char str[len];
foo(str, len);
}

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