This question already has answers here:
How can I correctly assign a new string value?
(4 answers)
Closed 4 years ago.
Why does this not return a segmentation fault 11?
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char const *argv[])
{
char *test;
test = (char*) (malloc(sizeof(char)*3));
test = "foo";
printf("%s\n", test);
test = "foobar";
printf("%s\n", test);
return 0;
}
My outcome is
foo
foobar
I'm fairly new to C, but when I was compiling this using both gcc on mac and Windows Debugger on Windows 10, it doesn't crash like I expected.
My understanding is that by using (char*) (malloc(sizeof(char)*3)), I am creating a character array of length 3. Then, when I assign test to the string foobar I am writing to 6 array positions.
I'm left sitting here, staring at my apparently valid, runnable code scratching my head.
test = "foo";
Here you do not copy the string to the allocated memory, test no longer points to the allocated memory, instead it points to the string literal "foo". Same goes for "foobar". Also as pointed out in the comments the address of the allocated memory is lost and therefore it is a memory leak (since there is no way to retrieve the address of the memory).
If you want to copy a string to another destination you need to use strcpy or loop over every character.
If you write or read outside bounds of the allocated space you are invoking undefined behavior. That means that basicly everything can happen, also that it works.
Your program never writes to the location pointed to by the return from malloc(). All you've done with e.g. test = "foo"; is change what test points to, which by the way is a memory leak since you've then lost what malloc() returned.
To properly use the memory you allocated with malloc(), use strcpy(), snprintf(), etc.
Also, don't forget the null terminator in your C strings. To properly store e.g. "foobar" you need at least 7 bytes, not 6.
First thing is that you waste the memory allocated by malloc unnecessorily by storing the address of foo into that.
If you are going to point to string in code section then there is no need to allocate memory to the pointer.
When to allocate memory to pointer
e.g. when you intended to scan 'n' number of bytes from keyboard in pointer.
char *ptr,num_char;
scanf("%d",&num_char);
ptr = (char *)malloc(num_char*sizeof(char));
scanf("%s",ptr);
Related
#include <stdio.h>
int main ()
{
char str[40];
printf("Enter a string : \n");
gets(str);
printf("You entered: %s\n", str);
return 0;
};
in above code, if replace str to a pointer, char *str. Then NULL is out. Suppose gets defined by char *gets(char *str), it should use a pointer instead of array. All examples I saw are array not pointers. Thanks.
function gets() is depracted your libc/compiler might ignore it. try use fgets() instead.
#include <stdio.h>
int main ()
{
char str[40];
printf("Enter a string : \n");
if (fgets(str, sizeof(str), stdin) != NULL)
{
printf("You entered: %s\n", str);
}
return 0;
};
also if you want to don't use stack you need to give pointer that points allocated space. in code str also can be char *str = malloc(40); then change sizeof(str) to 40 since str is no longer stack.
Really interesting question, I have been asked this question a lot!
you should have a bit background of pointers and memory to understand what is happening.
first let's have a brief review about pointers and memory:
our computer have some memory and we can use it in programming, anything that we store (in runtime) for example an int, array of doubles, some complex struct and strings(that they are array of characters) should be somewhere in memory.
pointers contain address of somewhere in memory, some of them know about that memory (how to read/write value) some of them don't.
there is a special value for pointers (NULL) that means nowhere, if pointer is pointing to NULL, that pointer is pointing not nowhere (obviously nowhere is not a valid address in memory)
array is specific type of pointer, a const pointer that is pointing to already allocated memory in stack.
and about gets function: let's think we want to re-implement such function (namely my_gets) , how we suppose to do that? how to return a string (array of characters)?
these are options (as far as i know):
creating a local array in our function and fill it. then we should return it? no we cant! because that array is in stack of our function and after ending the function, our function data including this array will be popped automatically (handled by compiler).
although nobody forbid us from returning that array, but that would cause dangling pointer problem.
allocating some space rather than stack (heap) and fill that. this is perfectly fine and there is methods and do this! for example readline (not in ansi c, you can find it here) will do this. the drawback of this method is that you should take care of that memory and free it later, it also may be not to optimum way and you may should copy that string to your already allocated memory
the last way (and way that gets use) is getting a pointer that is already pointing to a valid memory and fill that memory. you already know that gets want a pointer as input, I add that, that pointer should point to a valid and accessible memory that gets can fill it. if pointer is pointing to NULL (or maybe uninitialized and pointing to some where random) gets will fail writing and cause undefined behavior (segmentation fault for example)
some final points:
array solution work because array name is pointer that pointing to valid memory (array in stack) so it's OK and easy to understand.
If we don't want to use array, we can point our pointer to a valid memory, we need to use malloc/calloc to allocate a block of memory. see this:
#include <stdio.h>
#include <stdlib.h>
int main()
{
int size = 40 * sizeof(char);
char* p = malloc(size);
printf("Enter a string : \n");
if (fgets(p, size, stdin) != NULL) {
printf("You entered: %s\n", p);
}
free(p);
return 0;
}
gets is not secure because it doesn't care how much memory we have, it writes until and string ends and it may cause buffer overflow, better option (as people said) is fgets because it care memory size and will not exceed that. but my answer doesn't care it's fgets or gets.
This question already has answers here:
Expanding an array with malloc
(4 answers)
In C, are arrays pointers or used as pointers?
(6 answers)
Closed 2 years ago.
I'm trying to reallocate memory using the realloc function, I saw that you need to use malloc before but I don't understand if you MUST use it because let's say I'm creating the following string:
char string[] = "fun";
would the realloc function work if I try to add more space?
that brings me to my question, I'm trying to simply add one letter at the end of the string, lets say 'p', but for some reason the program crushes on the realloc line every time I run it.
Here is my full code:
int main()
{
char string[] = "fun" ;
str_func(string);
printf("%s", string);
return 0;
}
void str_func(char* str)
{
str = (char*)realloc(str, strlen(str) + 2);
strcat(str, "p");
}
I also tried making a pointer to 'string' and sending the pointer, which results the same thing.
would the realloc function work if I try to add more space?
No, because that array is no allocated on the heap - in your case it is very likely allocated on the stack and can't get resized. Simply put: realloc doesn't recognize the pointer and doesn't know what to do with it, but tries to do something anyway, hence the crash.
You can only call realloc on a pointer that was previously passed to malloc, or on a null pointer. That's just how these functions work.
For details, see What gets allocated on the stack and the heap?.
I saw that you need to use malloc before but I don't understand if you MUST use it
If you need to use malloc before you can realloc something, then by definition you must only realloc things originally allocated with malloc.
You're trying to find some space between "need" and "must" that doesn't exist.
... for some reason the program crushes on the realloc
You already said you know you need to use malloc. Then you didn't use malloc, and you're asking why this is a problem. You could at least try doing the thing you "know" you need to do, to see if that solves the problem.
The program should probably look like
int main()
{
/* array is an automatic local variable. It wasn't dynamically allocated
in the first place, so can't be dynamically re-allocated either.
You cannot (and don't need to) free it either, it just goes out of scope
like any other automatic variable.
*/
char array[] = "fun";
/* you need to use malloc (or one of the other dynamic allocation functions)
before you can realloc, as you said yourself */
char *dynamic = malloc(1+strlen(array));
memcpy(dynamic, array, 1+strlen(array));
/* realloc can move your data, so you must use the returned address */
dynamic = str_func(dynamic);
printf("old:'%s', new:'%s'\n", array, dynamic);
/* not really essential since the program is about to exit anyway */
free(dynamic);
}
char* str_func(char* str)
{
char* newstr = realloc(str, strlen(str) + 2);
if (newstr) {
strcat(newstr, "p");
return newstr;
} else {
/* we failed to make str larger, but it is still there and should be freed */
return str;
}
}
Your original condition isn't quite correct: actually the pointer passed to realloc
... must be previously allocated by malloc(), calloc() or realloc() and not yet freed with a call to free or realloc
[OR] If ptr is NULL, the behavior is the same as calling malloc(new_size).
The realloc function only works with things that were originally created with a small group of allocation functions (such as malloc, calloc, or realloc itself), or the null pointer. Since string is none of those things, your code is not well-defined.
I have already written a couple of C programs and consider this awkward to ask. But why do I receive a segmentation fault for the following code being supposed to replace "test" by "aaaa"?
#include <stdio.h>
int main(int argc, char* argv[])
{
char* s = "test\0";
printf("old: %s \n", s);
int x = 0;
while(s[x] != 0)
{
s[x++] = 'a'; // segmentation fault here
}
printf("new: %s \n", s); // expecting aaaa
return 0;
}
This assignment is writing to a string literal, which is stored in a read-only section of your executable when it is loaded in memory.
Also, note that the \0 in the literal is redundant.
One way to fix this (as suggested in comments) without copying the string: declare your variable as an array:
char s[] = "test";
This will cause the function to allocate at least 5 bytes of space for the string on the stack, which is normally writeable memory.
Also, you should generally declare a pointer to a string literal as const char*. This will cause the compiler to complain if you try to write to it, which is good, since the system loader will often mark the memory it points to as read-only.
Answering the OP's question in the comment posted to #antron.
What you need is to allocate a character array, then use strcpy() to initialize it with your text, then overwrite with a-s.
Allocation can be done statically (i.e., char s[10];) but make sure the space is enough to store the length of your init string (including the terminating \0).
Alternatively, you can dynamically allocate memory using malloc() and free it using free(). This enables you to allocate exactly enough space to hold your init string (figure it out in run-time using strlen()).
I'm told that when initializing a string like so
char str[] = "Hello world!";
The compiler will allocate an area in constants memory(read only for the program) and then copy the string to the array which resides in the stack. My question is, can I read or point to the original string after modifying the copy I'm given, and how? And if not, why does the string even exist outside of the stack in the first place?
It's done this way for space efficiency. When you write:
char str[] = "Hello world!";
it's compiled effectively as if you'd written:
static char str_init[] = "Hello world!";
char str[13];
strncpy(str, str_init, 13);
An alternative way to implement this might be equivalent to:
char str[13];
str[0] = 'H';
str[1] = 'e';
...
str[11] = '!';
str[12] = 0;
But for long strings, this is very inefficient. Instead of 1 byte of static data for each character of the string, it will use a full word of instruction (probably 4 bytes, but maybe more on some architectures) for each character. This will quadruple the size of the initialization data unnecessarily.
Because the program has to remember the string somewhere, i.e., your so-called "constant memory". Otherwise how can it know what values to assign when allocating the variable? Think about a variable with a given initial value. The variable is not allocated until declared. But the initial value must be stored somewhere else.
When this statement is compiled
char str[] = "Hello world!";
the compiler does not keep the string literal in the program. it is used only to initialize the array.
If you want to keep the string literal then you have to write the following way
char *s = "Hello world!";
char str[13];
strcpy( str, s );
When the program runs, "Hello world" will be stored in the constant part of the memory as a string literal, after that, the program will reserve enough space in the stack and copy character by character from the constant part of the memory. Unfortunately, you don't have access to the constant part that stores the string literal because you are telling the program that you want the values to be modifiable (string stored in stack), so it gives what you asked.
Most of your question has been addressed in the other answers, However, I did not see anyone address this one specifically:
Regarding your question: ...can I read or point to the original string after modifying the copy I'm given, and how?
The following sequence demonstrates how you can read the original after modifying a copy:
char str[] = "hello world"; //creates original (stack memory)
char *str2 = 0;//create a pointer (pointer created, no memory allocated)
str2 = StrDup(str); populate pointer with original (memory allocated on heap)
str2[5]=0; //edit copy: results in "hello" (i.e. modified) (modifying a location on the heap)
str; //still contains "hello world" (viewing value on the stack)
EDIT (answering comment question)
The answer above only addressed the specific question about accessing an original string after a copy has been modified. I just showed one possible set of steps to address that. You can edit the original string too:
char str[] = "Hello world!"; //creates location in stack memory called "str",
//and assigns space enough for literal string:
//"Hello world!", 13 spaces in all (including the \0)
strcpy(str, "new string"); //replaces original contents with "new string"
//old contents are no longer available.
So, using these steps, the original values in the variable str are changed, and are no longer available.
The method I outline in my original answer, (at top) shows a way whereby you can make an editable copy, while maintaining the original variable.
In your comment question, you are referring to things such as system memory and constant memory. Normally, system memory refers to RAM implementations on a system (i.e. how much physical memory). By constant memory, my guess is that you are referring to memory used by variables created on the stack. (read on)
First In a development, or run-time environment, there is stack memory. This is usually defaulted to some maximum value, such as 250,000 bytes perhaps. It is a pre-build settable value in most development environments, and is available for use by any variable you create on the stack. Example:
int x[10]; //creates a variable on the stack
//using enough memory space for 10 integers.
int y = 1; //same here, except uses memory for only 1 integer value
Second There is also what is referred to a heap memory. The amount of heap memory is system dependent, the more physical memory your system has available, the more heap memory you can use for variable memory space in your application. Heap memory is used when you dynamically allocate memory, for example using malloc(), calloc(), realloc().
int *x=0; //creates a pointer, no memory allocation yet...
x = malloc(10); //allocates enough memory for 10 integers, but the
//memory allocated is from the _heap_
//and must be freed for use by the system
//when you are done with it.
free(x);
I have marked the original post (above) with indications showing what type of memory each variable is using. I hope this helps.
I'm having difficulty learning C language's malloc and pointer:
What I learned so far:
Pointer is memory address pointer.
malloc() allocate memory locations and returns the memory address.
I'm trying to create a program to test malloc and pointer, here's what I have:
#include<stdio.h>
main()
{
char *x;
x = malloc(sizeof(char) * 5);
strcpy(*x, "123456");
printf("%s",*x); //Prints 123456
}
I'm expecting an error since the size I provided to malloc is 5, where I put 6 characters (123456) to the memory location my pointer points to. What is happening here? Please help me.
Update
Where to learn malloc and pointer? I'm confused by the asterisk thing, like when to use asterisk etc. I will not rest till I learn this thing! Thanks!
You are invoking undefined behaviour because you are writing (or trying to write) beyond the bounds of allocated memory.
Other nitpicks:
Because you are using strcpy(), you are copying 7 bytes, not 6 as you claim in the question.
Your call to strcpy() is flawed - you are passing a char instead of a pointer to char as the first argument.
If your compiler is not complaining, you are not using enough warning options. If you're using GCC, you need at least -Wall in your compiler command line.
You need to include both <stdlib.h> for malloc() and <string.h> for strcpy().
You should also explicitly specify int main() (or, better, int main(void)).
Personally, I'm old school enough that I prefer to see an explicit return(0); at the end of main(), even though C99 follows C++98 and allows you to omit it.
You may be unlucky and get away with invoking undefined behaviour for a while, but a tool like valgrind should point out the error of your ways. In practice, many implementations of malloc() allocate a multiple of 8 bytes (and some a multiple of 16 bytes), and given that you delicately do not step over the 8 byte allocation, you may actually get away with it. But a good debugging malloc() or valgrind will point out that you are doing it wrong.
Note that since you don't free() your allocated space before you return from main(), you (relatively harmlessly in this context) leak it. Note too that if your copied string was longer (say as long as the alphabet), and especially if you tried to free() your allocated memory, or tried to allocate other memory chunks after scribbling beyond the end of the first one, then you are more likely to see your code crash.
Undefined behaviour is unconditionally bad. Anything could happen. No system is required to diagnose it. Avoid it!
If you call malloc you get and adress of a memory region on heap.
If it returns e.g. 1000 you memory would look like:
Adr Value
----------
1000 1
1001 2
1002 3
1003 4
1004 5
1005 6
1006 0
after the call to strcpy(). you wrote 7 chars (2 more than allocated).
x == 1000 (pointer address)
*x == 1 (dereferenced the value x points to)
There are no warnings or error messages from the compiler, since C doesn't have any range-checking.
My three cents:
Use x, as (*x) is the value that is stored at x (which is unknown in your case) - you are writing to unknown memory location. It should be:
strcpy(x, "123456");
Secondly - "123456" is not 6 bytes, it's 7. You forgot about trailing zero-terminator.
Your program with it's current code might work, but not guaranteed.
What I would do:
#include<stdio.h>
main()
{
char str[] = "123456";
char *x;
x = malloc(sizeof(str));
strcpy(x, str);
printf("%s",x); //Prints 123456
free(x);
}
Firstly, there is one problem with your code:
x is a pointer to a memory area where you allocated space for 5 characters.
*x it's the value of the first character.
You should use strcpy(x, "123456");
Secondly, the memory after your 5 bytes allocated, can be valid so you will not receive an error.
#include<stdio.h>
main()
{
char *x;
x = malloc(sizeof(char) * 5);
strcpy(x, "123456");
printf("%s",x); //Prints 123456
}
Use this...it will work
See difference in your & mine program
Now here you are allocating 5 bytes & writing 6 byte so 6th byte will be stored in next consecutive address. This extra byte can be allocated to some one else by memory management so any time that extra byte can be changed by other program because 6th byte is not yours because you haven't malloc'd that.. that's why this is called undefined behaviour.