I'm currently attempting to learn the C side of C++.
I attempt to malloc a chunk of memory for a char array of 256 and then I assigned it a char* "Hello World!" but when I come to free the object I get an error.
Can anyone please explain to me the error.
#include <exception>
#include <stdexcept>
#include <iostream>
int main()
{
void* charVoidPointer = malloc( sizeof(char) * 256 ) ;
charVoidPointer = "Hello World";
std::cout << (char *)charVoidPointer;
free (charVoidPointer);
}
"Hello World" is statically allocated by the compiler. It is part of the program and exists at some place addressable by the program; call it address 12.
charVoidPointer initially points to some place allocated for you by malloc; call it address 98.
charVoidPointer = "Hello ..." causes charVoidPointer to point to the data in your program; address 12. You lose track of address 98 previously contained in charVoidPointer.
And you can't free memory not allocated by malloc.
To demonstrate more literally what I mean:
void* charVoidPointer = malloc(sizeof(char) * 256);
printf("the address of the memory allocated for us: %p\n", charVoidPointer);
charVoidPointer = "Hello World";
printf("no longer the address allocated for us; free will fail: %p\n",
charVoidPointer);
What you meant was:
strcpy(charVoidPointer, "Hello World");
Edit: Example of addressing memory for other types
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
int main()
{
// an array of 10 int
int *p = (int*)malloc(sizeof(int) * 10);
// setting element 0 using memcpy (works for everything)
int src = 2;
memcpy(p+0, &src, sizeof(int));
// setting element 1 using array subscripts. correctly adjusts for
// size of element BECAUSE p is an int*. We would have to consider
// the size of the underlying data if it were a void*.
p[1] = 3;
// again, the +1 math works because we've given the compiler
// information about the underlying type. void* wouldn't have
// the correct information and the p+1 wouldn't yield the result
// you expect.
printf("%d, %d\n", p[0], *(p+1));
free (p);
}
Experiment; Change the type from int to long, or double, or some complex type.
void* charVoidPointer = malloc( sizeof(char) * 256 ) ;
now charVoidPointer (weird name by the way - if you want chars, use char * and cast the pointer returned from malloc) points at a block of 256 characters. This block is uninitialized, so almost the only valid thing you can do is set them all to some value, or copy something in.
charVoidPointer = "Hello World";
now charVoidPointer points instead at a statically-allocated character array, and you've lost the address returned by malloc. There is no way to get it back, so this is a resource leak.
Your code should look something like:
char *charPointer = (char *)malloc(256);
strcpy(charPointer, "Hello World");
which copies the character array into your allocated block. Or, more concisely, just
char *charPointer = strdup("Hello World");
which will allocate a block just the right size and copy the string in. You still release the block with free.
Use strcpy(charVoidPointer, "Hello World"); because in your example you reassign your pointer.
You're assigning the pointer to the address of the string literal "Hello World", therefore the memory block you malloc'ed is leaked.
You should use
strcpy(charVoidPointer, "Hello World");
instead of the assignment operator.
Even better is to use strncpy(charVoidPointer, "Hello World", 255); to avoid overflowing the array you allocate.
Related
I have been struggling for a few hours with all sorts of C tutorials and books related to pointers but what I really want to know is if it's possible to change a char pointer once it's been created.
This is what I have tried:
char *a = "This is a string";
char *b = "new string";
a[2] = b[1]; // Causes a segment fault
*b[2] = b[1]; // This almost seems like it would work but the compiler throws an error.
So is there any way to change the values inside the strings rather than the pointer addresses?
When you write a "string" in your source code, it gets written directly into the executable because that value needs to be known at compile time (there are tools available to pull software apart and find all the plain text strings in them). When you write char *a = "This is a string", the location of "This is a string" is in the executable, and the location a points to, is in the executable. The data in the executable image is read-only.
What you need to do (as the other answers have pointed out) is create that memory in a location that is not read only--on the heap, or in the stack frame. If you declare a local array, then space is made on the stack for each element of that array, and the string literal (which is stored in the executable) is copied to that space in the stack.
char a[] = "This is a string";
you can also copy that data manually by allocating some memory on the heap, and then using strcpy() to copy a string literal into that space.
char *a = malloc(256);
strcpy(a, "This is a string");
Whenever you allocate space using malloc() remember to call free() when you are finished with it (read: memory leak).
Basically, you have to keep track of where your data is. Whenever you write a string in your source, that string is read only (otherwise you would be potentially changing the behavior of the executable--imagine if you wrote char *a = "hello"; and then changed a[0] to 'c'. Then somewhere else wrote printf("hello");. If you were allowed to change the first character of "hello", and your compiler only stored it once (it should), then printf("hello"); would output cello!)
No, you cannot modify it, as the string can be stored in read-only memory. If you want to modify it, you can use an array instead e.g.
char a[] = "This is a string";
Or alternately, you could allocate memory using malloc e.g.
char *a = malloc(100);
strcpy(a, "This is a string");
free(a); // deallocate memory once you've done
A lot of folks get confused about the difference between char* and char[] in conjunction with string literals in C. When you write:
char *foo = "hello world";
...you are actually pointing foo to a constant block of memory (in fact, what the compiler does with "hello world" in this instance is implementation-dependent.)
Using char[] instead tells the compiler that you want to create an array and fill it with the contents, "hello world". foo is the a pointer to the first index of the char array. They both are char pointers, but only char[] will point to a locally allocated and mutable block of memory.
The memory for a & b is not allocated by you. The compiler is free to choose a read-only memory location to store the characters. So if you try to change it may result in seg fault. So I suggest you to create a character array yourself. Something like: char a[10]; strcpy(a, "Hello");
It seems like your question has been answered but now you might wonder why char *a = "String" is stored in read-only memory. Well, it is actually left undefined by the c99 standard but most compilers choose to it this way for instances like:
printf("Hello, World\n");
c99 standard(pdf) [page 130, section 6.7.8]:
The declaration:
char s[] = "abc", t[3] = "abc";
defines "plain" char array objects s and t whose elements are initialized with character string literals.
This declaration is identical to char
s[] = { 'a', 'b', 'c', '\0' }, t[] = { 'a', 'b', 'c' };
The contents of the arrays are modifiable. On the other hand, the declaration
char *p = "abc";
defines p with type "pointer to char" and initializes it to point to an object with type "array of char" with length 4 whose elements are initialized with a character string literal. If an attempt is made to use p to modify the contents of the array, the behavior is undefined.
You could also use strdup:
The strdup() function returns a pointer to a new string which is a duplicate of the string s.
Memory for the new string is obtained with malloc(3), and can be freed with free(3).
For you example:
char *a = strdup("stack overflow");
All are good answers explaining why you cannot modify string literals because they are placed in read-only memory. However, when push comes to shove, there is a way to do this. Check out this example:
#include <sys/mman.h>
#include <unistd.h>
#include <stddef.h>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
int take_me_back_to_DOS_times(const void *ptr, size_t len);
int main()
{
const *data = "Bender is always sober.";
printf("Before: %s\n", data);
if (take_me_back_to_DOS_times(data, sizeof(data)) != 0)
perror("Time machine appears to be broken!");
memcpy((char *)data + 17, "drunk!", 6);
printf("After: %s\n", data);
return 0;
}
int take_me_back_to_DOS_times(const void *ptr, size_t len)
{
int pagesize;
unsigned long long pg_off;
void *page;
pagesize = sysconf(_SC_PAGE_SIZE);
if (pagesize < 0)
return -1;
pg_off = (unsigned long long)ptr % (unsigned long long)pagesize;
page = ((char *)ptr - pg_off);
if (mprotect(page, len + pg_off, PROT_READ | PROT_WRITE | PROT_EXEC) == -1)
return -1;
return 0;
}
I have written this as part of my somewhat deeper thoughts on const-correctness, which you might find interesting (I hope :)).
Hope it helps. Good Luck!
You need to copy the string into another, not read-only memory buffer and modify it there. Use strncpy() for copying the string, strlen() for detecting string length, malloc() and free() for dynamically allocating a buffer for the new string.
For example (C++ like pseudocode):
int stringLength = strlen( sourceString );
char* newBuffer = malloc( stringLength + 1 );
// you should check if newBuffer is 0 here to test for memory allocaton failure - omitted
strncpy( newBuffer, sourceString, stringLength );
newBuffer[stringLength] = 0;
// you can now modify the contents of newBuffer freely
free( newBuffer );
newBuffer = 0;
char *a = "stack overflow";
char *b = "new string, it's real";
int d = strlen(a);
b = malloc(d * sizeof(char));
b = strcpy(b,a);
printf("%s %s\n", a, b);
I'm trying to figure out how to use malloc correctly in C, and have run into an error that I'm having trouble with.
My code:
#include <stdio.h>
#include <stdlib.h>
int main() {
char * str;
str = (char *)malloc(10);
str = "Hello World";
str[0] = 'R';
return EXIT_SUCCESS;
}
My Valgrind output:
==23136== Process terminating with default action of signal 10 (SIGBUS)
==23136== Non-existent physical address at address 0x100000F92
==23136== at 0x100000F66: main (test.c:12)
I know that the issue is due to me trying to allocate the letter 'R' to str, but I was under the impression that the advantage of using malloc in this situation (as opposed to char str[10] = "Hello World"), was the ability to edit the contents of my string.
Thanks!
str = "Hello World"; makes str point to a constant char string "Hello World", and the memory you have malloced will become memory leak.
You copy a string with strcpy from <string.h>, not by re-assigning a pointer.
But take care that the target buffer will actually hold strlen(source) + 1 characters (0-terminator). "Hello World" is 11+1.
Also, trying to modify that improperly assigned string literal is UB.
Anyway, Don't cast the result of malloc (and friends).
Finally, return EXIT_SUCCESS is superfluous (since C99 main has an implicit return 0; at the end).
You are throwing away the return value of malloc. Instead you are setting it to a value in read only memory
Try replacing
str = "Hello World";
with
strcpy(str, "Hello World");
You will need to include the appropriate header file
I am puzzled by this response.Can anyone help me on this and point out where I am making a mistake? The output at codepad is "memory clobbered before allocated block"
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
char *s = (char *)malloc(10 * sizeof(char));
s = "heel";
printf("%s\n",s);
printf("%c\n",s[2]);
printf("%p\n",s);
printf("%d\n",s);
free(s);
return 0;
}
You're trying to free constant memory with:
free(s); // cannot free constant "heel"
What you're doing is allocating a piece of memory and storing its location (char *s). You are then overwriting that reference with one to a string constant "heel" (memory leak), which cannot be freed. To make this behave as desired, you should be copying the constant string to the memory you allocated:
strcpy(s, "heel");
Here is an example for getting user input:
char *input = malloc(sizeof(char) * 16); // enough space for 15 characters + '\0'
fgets(input, 16, stdin);
// do something with input
free(input);
To expand on #TimCooper's answer:
first you do: char *s = (char *)malloc(10 * sizeof(char));
then: s = "heel";
The first line allocates memory and assigns the location of that memory to s. But the second line reassigns s to the memory location of constant string heel on the stack!
Which means you try and free() memory on the stack, which is illegal. AND you leak memory, since what you first allocated to s is now inaccessible.
If you want to write a string into the memory pointed by s, you should use something like strcpy() (or, better, strncpy()).
You cannot free(s) - it's constant memory.
Try to change s = "heel"; with strcpy(s,"heel");
char *s = (char *)malloc(10 * sizeof(char));
s = "heel";
Doesn't do what you think, or what you would expect with more modern languages
The first line allocates some memory for 10chars and returns the address of it.
The second line changes that address to point to a constant block of memory allocated at compile time, containing "heel" losing the link to the allocated memory - leaking it
I'm specifically focused on when to use malloc on char pointers
char *ptr;
ptr = "something";
...code...
...code...
ptr = "something else";
Would a malloc be in order for something as trivial as this? If yes, why? If not, then when is it necessary for char pointers?
As was indicated by others, you don't need to use malloc just to do:
const char *foo = "bar";
The reason for that is exactly that *foo is a pointer — when you initialize foo you're not creating a copy of the string, just a pointer to where "bar" lives in the data section of your executable. You can copy that pointer as often as you'd like, but remember, they're always pointing back to the same single instance of that string.
So when should you use malloc? Normally you use strdup() to copy a string, which handles the malloc in the background. e.g.
const char *foo = "bar";
char *bar = strdup(foo); /* now contains a new copy of "bar" */
printf("%s\n", bar); /* prints "bar" */
free(bar); /* frees memory created by strdup */
Now, we finally get around to a case where you may want to malloc if you're using sprintf() or, more safely snprintf() which creates / formats a new string.
char *foo = malloc(sizeof(char) * 1024); /* buffer for 1024 chars */
snprintf(foo, 1024, "%s - %s\n", "foo", "bar"); /* puts "foo - bar\n" in foo */
printf(foo); /* prints "foo - bar" */
free(foo); /* frees mem from malloc */
malloc is for allocating memory on the free-store. If you have a string literal that you do not want to modify the following is ok:
char *literal = "foo";
However, if you want to be able to modify it, use it as a buffer to hold a line of input and so on, use malloc:
char *buf = (char*) malloc(BUFSIZE); /* define BUFSIZE before */
// ...
free(buf);
Use malloc() when you don't know the amount of memory needed during compile time. In case if you have read-only strings then you can use const char* str = "something"; . Note that the string is most probably be stored in a read-only memory location and you'll not be able to modify it. On the other hand if you know the string during compiler time then you can do something like: char str[10]; strcpy(str, "Something"); Here the memory is allocated from stack and you will be able to modify the str. Third case is allocating using malloc. Lets say you don'r know the length of the string during compile time. Then you can do char* str = malloc(requiredMem); strcpy(str, "Something"); free(str);
malloc for single chars or integers and calloc for dynamic arrays. ie pointer = ((int *)malloc(sizeof(int)) == NULL), you can do arithmetic within the brackets of malloc but you shouldnt because you should use calloc which has the definition of void calloc(count, size)which means how many items you want to store ie count and size of data ie int , char etc.
Everytime the size of the string is undetermined at compile time you have to allocate memory with malloc (or some equiviallent method). In your case you know the size of your strings at compile time (sizeof("something") and sizeof("something else")).
I have been struggling for a few hours with all sorts of C tutorials and books related to pointers but what I really want to know is if it's possible to change a char pointer once it's been created.
This is what I have tried:
char *a = "This is a string";
char *b = "new string";
a[2] = b[1]; // Causes a segment fault
*b[2] = b[1]; // This almost seems like it would work but the compiler throws an error.
So is there any way to change the values inside the strings rather than the pointer addresses?
When you write a "string" in your source code, it gets written directly into the executable because that value needs to be known at compile time (there are tools available to pull software apart and find all the plain text strings in them). When you write char *a = "This is a string", the location of "This is a string" is in the executable, and the location a points to, is in the executable. The data in the executable image is read-only.
What you need to do (as the other answers have pointed out) is create that memory in a location that is not read only--on the heap, or in the stack frame. If you declare a local array, then space is made on the stack for each element of that array, and the string literal (which is stored in the executable) is copied to that space in the stack.
char a[] = "This is a string";
you can also copy that data manually by allocating some memory on the heap, and then using strcpy() to copy a string literal into that space.
char *a = malloc(256);
strcpy(a, "This is a string");
Whenever you allocate space using malloc() remember to call free() when you are finished with it (read: memory leak).
Basically, you have to keep track of where your data is. Whenever you write a string in your source, that string is read only (otherwise you would be potentially changing the behavior of the executable--imagine if you wrote char *a = "hello"; and then changed a[0] to 'c'. Then somewhere else wrote printf("hello");. If you were allowed to change the first character of "hello", and your compiler only stored it once (it should), then printf("hello"); would output cello!)
No, you cannot modify it, as the string can be stored in read-only memory. If you want to modify it, you can use an array instead e.g.
char a[] = "This is a string";
Or alternately, you could allocate memory using malloc e.g.
char *a = malloc(100);
strcpy(a, "This is a string");
free(a); // deallocate memory once you've done
A lot of folks get confused about the difference between char* and char[] in conjunction with string literals in C. When you write:
char *foo = "hello world";
...you are actually pointing foo to a constant block of memory (in fact, what the compiler does with "hello world" in this instance is implementation-dependent.)
Using char[] instead tells the compiler that you want to create an array and fill it with the contents, "hello world". foo is the a pointer to the first index of the char array. They both are char pointers, but only char[] will point to a locally allocated and mutable block of memory.
The memory for a & b is not allocated by you. The compiler is free to choose a read-only memory location to store the characters. So if you try to change it may result in seg fault. So I suggest you to create a character array yourself. Something like: char a[10]; strcpy(a, "Hello");
It seems like your question has been answered but now you might wonder why char *a = "String" is stored in read-only memory. Well, it is actually left undefined by the c99 standard but most compilers choose to it this way for instances like:
printf("Hello, World\n");
c99 standard(pdf) [page 130, section 6.7.8]:
The declaration:
char s[] = "abc", t[3] = "abc";
defines "plain" char array objects s and t whose elements are initialized with character string literals.
This declaration is identical to char
s[] = { 'a', 'b', 'c', '\0' }, t[] = { 'a', 'b', 'c' };
The contents of the arrays are modifiable. On the other hand, the declaration
char *p = "abc";
defines p with type "pointer to char" and initializes it to point to an object with type "array of char" with length 4 whose elements are initialized with a character string literal. If an attempt is made to use p to modify the contents of the array, the behavior is undefined.
You could also use strdup:
The strdup() function returns a pointer to a new string which is a duplicate of the string s.
Memory for the new string is obtained with malloc(3), and can be freed with free(3).
For you example:
char *a = strdup("stack overflow");
All are good answers explaining why you cannot modify string literals because they are placed in read-only memory. However, when push comes to shove, there is a way to do this. Check out this example:
#include <sys/mman.h>
#include <unistd.h>
#include <stddef.h>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
int take_me_back_to_DOS_times(const void *ptr, size_t len);
int main()
{
const *data = "Bender is always sober.";
printf("Before: %s\n", data);
if (take_me_back_to_DOS_times(data, sizeof(data)) != 0)
perror("Time machine appears to be broken!");
memcpy((char *)data + 17, "drunk!", 6);
printf("After: %s\n", data);
return 0;
}
int take_me_back_to_DOS_times(const void *ptr, size_t len)
{
int pagesize;
unsigned long long pg_off;
void *page;
pagesize = sysconf(_SC_PAGE_SIZE);
if (pagesize < 0)
return -1;
pg_off = (unsigned long long)ptr % (unsigned long long)pagesize;
page = ((char *)ptr - pg_off);
if (mprotect(page, len + pg_off, PROT_READ | PROT_WRITE | PROT_EXEC) == -1)
return -1;
return 0;
}
I have written this as part of my somewhat deeper thoughts on const-correctness, which you might find interesting (I hope :)).
Hope it helps. Good Luck!
You need to copy the string into another, not read-only memory buffer and modify it there. Use strncpy() for copying the string, strlen() for detecting string length, malloc() and free() for dynamically allocating a buffer for the new string.
For example (C++ like pseudocode):
int stringLength = strlen( sourceString );
char* newBuffer = malloc( stringLength + 1 );
// you should check if newBuffer is 0 here to test for memory allocaton failure - omitted
strncpy( newBuffer, sourceString, stringLength );
newBuffer[stringLength] = 0;
// you can now modify the contents of newBuffer freely
free( newBuffer );
newBuffer = 0;
char *a = "stack overflow";
char *b = "new string, it's real";
int d = strlen(a);
b = malloc(d * sizeof(char));
b = strcpy(b,a);
printf("%s %s\n", a, b);