can we declare size to a pointer
#include<iostream>
#include<cstring>
using namespace std;
int main()
{
char (*ptr)=new char[3];
strcpy(ptr,"ert");
cout<<ptr<<endl;
return 0;
}
what is the meaning of this line char *ptr=new char[3] if it allocates size to ptr.since i have given the size as 3 and the string as "ert"it has to show error since the string length is too long but it doesn't .can we allocate size to pointers if so how?
You need 4 characters:
char *ptr=new char[4];
strcpy(ptr,"ert");
One extra space for the nul terminator:
|e|r|t|\0|
It's not the size of the pointer that you've declared, but the size of the character array that the pointer points to.
strcpy() does not know the length of the array that the pointer points to - it just knows it's got a pointer to the first byte it can copy into, and trusts that you know there's enough room for the copy to be made. Thus it's very fast, but it's also rather dangerous and should be used only when you're sure the destination is large enough.
strncpy() is worth looking into for some extra safety, but you still have to know that the target pointer points to something large enough for the size you specify (it protects more against the size of the source than the size of the target).
The lesson to learn here is that C and C++ won't give you any help - the compiler trusts you to get your buffer sizes right, and won't do any checking on your behalf either at compile time or runtime. This allows programs to run extremely fast (no runtime checking) but also requires the programmer to be a lot more careful. If you're writing in C++ which your tags suggest, for normal string handling you should definitely be using the std::string class unless you have a specific reason to need C-style string handling. You may well have such a reason from time to time, but don't do it unless you have to.
This statement
char (*ptr)=new char[3];
at first allocates in the heap unnamed character array with 3 elements and then the address of the first element of the array is assigned to pointer ptr.
The size of the pointer will not be changed whether you initialize it as in the statement above or the following way
char (*ptr)=new char;
that is sizeof( ptr ) will be the same and equal usually either to 4 or 8 bytes depending on the environment where the program will be compiled.
C++ does not check bounds of arrays. So in this statement
strcpy(ptr,"ert");
you have undefined behaviour of the program because string literal "ert" has four elements including the terminating zero.
Related
I am looking for something which give me size which taken by str character pointer.
int main()
{
char * str = (char *) malloc(sizeof(char) * 100);
int size = 0;
size = /* library function or anything use to find size */
printf("Total size of str array - %d\n", size);
}
I want prove that give memory is 100 bytes.
Is any one have any idea about this ?
A raw pointer only knows it points to a single element of it's type. If that thing it points to happens to be part of an array, the pointer doesn't know and there's no way to get that information from it.
You want to instead use types that do know their size, like for example; std::string, std::array or std::vector.
The C and C++ standards do not provide a way to get, from an address, the amount of memory that was requested in the call to malloc that returned that address.
Some C or C++ implementations provide a way to get the amount of memory that was provided at the given address, such as malloc_size. The amount provided may be greater than the amount that was requested.
If the memory contains a string, which is an array of characters terminated by a null character, then you can determine the length of the string by counting characters up to the null character. This function is provided by the standard strlen function. This length is different from the space allocated unless, of course, the string happens to fill the space.
There is no (good, standard, portable) way to tell from a pointer value alone whether it's the first element of an array or not, nor how many elements follow it. That information has to be tracked separately.
If you're writing in C++, don't do your own memory management if you can help it. Use a standard container type like std::vector or std::map (or std::string for text). If you must do your own memory management, use the new and delete operators instead of the *alloc and free library functions, and wrap a class around those operations that also keeps track of how many elements have been allocated (which, like std::vector and std::map, is returned via a read-only size() method).
This is my code
char *t = type1[j];
int ln = strlen(t);
char *s = new char[ln];
cout<<"ln "<<ln<<" s "<<strlen(s)<<endl;
When I run it:
Why are the lengths different?
When you dynamically allocate an array with new[], you get a pointer to an uninitialized block of memory. You can't call strlen on it, as that function expects a pointer to a character array that contains a null-terminated string. It works by reading each byte in the array until it finds a null byte and returning the count of how many bytes it read before it found a null byte. And since the array is uninitialized, calling strlen does make sense. In fact, it invokes undefined behavior.
There's no way to look at a dynamically allocated buffer and find out how big that buffer is. You need to keep track of the size of the buffer separately. Since you used new char[ln] to create the buffer, you know that you have a buffer large enough to hold ln variables of type char.
As I understood the correct programming style tells that if you want to get string (char []) from another function is best to create char * by caller and pass it to string formating function together with created string length. In my case string formating function is "getss".
void getss(char *ss, int& l)
{
sprintf (ss,"aaaaaaaaaa%d",1);
l=11;
}
int _tmain(int argc, _TCHAR* argv[])
{
char *f = new char [1];
int l =0;
getss(f,l);
cout<<f;
char d[50] ;
cin>> d;
return 0;
}
"getss" formats string and returns it to ss*. I thought that getss is not allowed to got outside string length that was created by caller. By my understanding callers tells length by variable "l" and "getcc" returns back length in case buffer is not filled comleatly but it is not allowed go outside array range defined by caller.
But reality told me that really it is not so important what size of buffer was created by caller. It is ok, if you create size of 1, and getss fills with 11 characters long. In output I will get all characters that "getss" has filled.
So what is reason to pass length variable - you will always get string that is zero terminated and you will find the end according that.
What is the reason to create buffer with specified length if getss can expand it?
How it is done in real world - to get string from another function?
Actually, the caller is the one that has allocated the buffer and knows the maximum size of the string that can fit inside. It passes that size to the function, and the function has to use it to avoid overflowing the passed buffer.
In your example, it means calling snprintf() rather than sprintf():
void getss(char *ss, int& l)
{
l = snprintf(ss, l, "aaaaaaaaaa%d", 1);
}
In C++, of course, you only have to return an instance of std::string, so that's mostly a C paradigm. Since C does not support references, the function usually returns the length of the string:
int getss(char *buffer, size_t bufsize)
{
return snprintf(buffer, bufsize, "aaaaaaaaaa%d", 1);
}
You were only lucky. Sprintf() can't expand the (statically allocated) storage, and unless you pass in a char array of at least length + 1 elements, expect your program to crash.
In this case you are simply lucky that there is no "important" other data after the "char*" in memory.
The C runtime does not always detect these kinds of violations reliably.
Nonetheless, your are messing up the memory here and your program is prone to crash any time.
Apart from that, using raw "char*" pointers is really a thing you should not do any more in "modern" C++ code.
Use STL classes (std::string, std::wstring) instead. That way you do not have to bother about memory issues like this.
In real world in C++ is better to use std::string objects and std::stringstream
char *f = new char [1];
sprintf (ss,"aaaaaaaaaa%d",1);
Hello, buffer overflow! Use snprintf instead of sprintf in C and use C++ features in C++.
By my understanding callers tells length by variable "l" and "getcc" returns back length in case buffer is not filled comleatly but it is not allowed go outside array range defined by caller.
This is spot on!
But reality told me that really it is not so important what size of buffer was created by caller. It is ok, if you create size of 1, and getss fills with 11 characters long. In output I will get all characters that "getss" has filled.
This is absolutely wrong: you invoked undefined behavior, and did not get a crash. A memory checker such as valgrind would report this behavior as an error.
So what is reason to pass length variable.
The length is there to avoid this kind of undefined behavior. I understand that this is rather frustrating when you do not know the length of the string being returned, but this is the only safe way of doing it that does not create questions of string ownership.
One alternative is to allocate the return value dynamically. This lets you return strings of arbitrary length, but the caller is now responsible for freeing the returned value. This is not very intuitive to the reader, because malloc and free happen in different places.
The answer in C++ is quite different, and it is a lot better: you use std::string, a class from the standard library that represents strings of arbitrary length. Objects of this class manage the memory allocated for the string, eliminating the need of calling free manually.
For cpp consider smart pointers in your case propably a shared_ptr, this will take care of freeing the memory, currently your program is leaking memory since, you never free the memory you allocate with new. Space allocate by new must be dealocated with delete or it will be allocated till your programm exits, this is bad, imagine your browser not freeing the memory it uses for tabs when you close them.
In the special case of strings I would recommend what OP's said, go with a String. With Cpp11 this will be moved (not copied) and you don't need to use new and have no worries with delete.
std::string myFunc() {
std::string str
//work with str
return str
}
In C++ you don't have to build a string. Just output the parts separately
std::cout << "aaaaaaaaaa" << 1;
Or, if you want to save it as a string
std::string f = "aaaaaaaaaa" + std::to_string(1);
(Event though calling to_string is a bit silly for a constant value).
I'm new to pointers in C++. I'm not sure why I need pointers like char * something[20] as oppose to just char something[20][100]. I realize that the second method would mean that 100 block of memory will be allocated for each element in the array, but wouldn't the first method introduce memory leak issues.
If someone could explain to me how char * something[20] locates memory, that would be great.
Edit:
My C++ Primer Plus book is doing:
const char * cities[5] = {
"City 1",
"City 2",
"City 3",
"City 4",
"City 5"
}
Isn't this the opposite of what people just said?
You allocate 20 pointers in the memory, then you will need to go through each and every one of them to allocate memory dynamically:
something[0] = new char[100];
something[1] = new char[20]; // they can differ in size
And delete them all separately:
delete [] something[0];
delete [] something[1];
EDIT:
const char* text[] = {"These", "are", "string", "literals"};
Strings specified directly in the source code ("string literals", which are always const char *) are quite different to char *, mainly because you don't have to worry about alloc/dealloc of them. They are also generally handled very different in memory, but this depends on the implementation of your compiler.
You're right.
You'd need to go through each element of that array and allocate a character buffer for each one.
Then, later, you'd need to go through each element of that array and free the memory again.
Why you would want to faff about with this in C++ is anyone's guess.
What's wrong with std::vector<std::string> myStrings(20)?
It will allocate space for twenty char-pointers.
They will not be initialized, so typical usage looks like
char * something[20];
for (int i=0; i<20; i++)
something[i] = strdup("something of a content");
and later
for (int i=0; i<20; i++)
if (something[i])
free(something[i]);
You're right - the first method may introduce memory leak issues and the overhead of doing dynamic allocations, plus more reads. I think the second method is usually preferable, unless it wastes too much RAM or you may need the strings to grow longer than 99 chars.
How the first method works:
char* something[20]; // Stores 20 pointers.
something[0] = malloc(100); // Make something[0] point to a new buffer of 100 bytes.
sprintf(something[0], "hai"); // Make the new buffer contain "hai", going through the pointer in something[0]
free(something[0]); // Release the buffer.
char* smth[20] does not allocate any memeory on heap. It allocates just enough space on the stack to store 20 pointers. The value of those pointers is undefined, so before using them, you have to initialize them, like this:
char* smth[20];
smth[0] = new char[100]; // allocate memory for 100 chars, store the address of the first one in smth[0]
//..some code..
delete[] smth[0];
First of all, this almost inapplicable in C++. The normal equivalent in C++ would be something like: std::vector<std::string> something;
In C, the primary difference is that you can allocate each string separately from the others. With char something[M][N], you always allocate exactly the same number of strings, and the same space for each string. This will frequently waste space (when the strings are shorter than you've made space for), and won't allow you to deal with any more strings or longer of strings than you've made space for initially.
char *something[20] let's you deal with longer/shorter strings more efficiently, but still only makes space for 20 strings.
The next step (if you're feeling adventurous) is to use something like:
char **something;
and allocate the strings individually, and allocate space for the pointers dynamically as well, so if you get more than 20 strings you can deal with that as well.
I'll repeat, however, that for most practical purposes, this is restricted to C. In C++, the standard library already has data structures for situations like these.
C++ has pointers because C has pointers.
Why do we use pointers?
To track dynamically-allocated memory. The memory allocation functions in C (malloc, calloc, realloc) and the new operator in C++ all return pointer values.
To mimic pass-by-reference semantics (C only). In C, all function arguments are passed by value; the formal parameter and the actual parameter are distinct objects, and modifying a formal parameter doesn't affect the actual parameter. We get around this by passing pointers to the function. C++ introduced reference types, which serve the same purpose, but are a bit cleaner and safer than using pointers.
To build dynamic, self-referential data structures. A struct cannot contain an instance of itself, but it can contain a pointer to an instance. For example, the following code
struct node
{
data_t data;
struct node *next;
};
creates a data type for a simple linked-list node; the next member explicitly points to the next element in the list. Note that in C++, the STL containers for stacks and queues and vectors all use pointers under the hood, isolating you from the bookkeeping.
There are literally dozens of other places where pointers come up, but those are the main reasons you use them.
Your array of pointers could be used to store strings of varying length by allocating just enough memory for each, rather than relying on some maximum size (which will eventually be exceeded, leading to a buffer overflow error, and in any case will lead to internal memory fragmentation). Naturally, in C++ you'd use the string data type (which hides all the pointer and memory management behind the class API) instead of pointers to char, but someone has decided to confuse you by starting with low-level details instead of the big picture.
I'm not
sure why I need pointers like char *
something[20] as oppose to just char
something[20][100]. I realize that the
second method would mean that 100
block of memory will be allocated for
each element in the array, but
wouldn't the first method introduce
memory leak issues.
The second method will suffice if you're only referencing your buffer(s) locally.
The problem comes when you pass the array name to another function. When you pass char something[10] to another function, you're actually passing char* something because the array length doesn't go along for the ride.
For multidimensional arrays, you can declare a function that takes in an array of determinate length in all but one direction, e.g. foo(char* something[10]).
So why use the first form rather than the second? I can think of a few reasons:
You don't want to have the restriction that the entire buffer must reside in continuous memory.
You don't know at compile-time that you'll need each buffer, or that the length of each buffer will need to be the same size, and you want the flexibility to determine that at run-time.
This is a function declaration.
char * something[20]
Assuming this is 32Bit, this allocates 80 bytes of data on the stack.
4 Bytes for each pointer address, 20 pointers total = 4 x 20 = 80 bytes.
The pointers are all uninitialized, so you need to write additional code to allocate/free
the buffers for doing this.
It roughly looks like:
[0] [4 Bytes of Uninitialized data to hold a pointer/memory address...]
[1] [4 Bytes of ... ]
...
[19]
char something[20][100]
Allocates 2000 bytes on the stack.
100 Bytes for each something, 20 somethings total = 100 x 20 = 2000 bytes.
[0] [100 bytes to hold characters]
[1] [100 bytes to hold characters]
...
[19]
The char *, has a smaller memory overhead, but you have to manage the memory.
The char[][] approach, has bigger memory overhead, but you don't have additional memory management.
With either approach, you have to be careful when writing to the buffer allocated not to exceed/overwrite the memory alloc'd for it.
Basic question.
char new_str[]="";
char * newstr;
If I have to concatenate some data into it or use string functions like strcat/substr/strcpy, what's the difference between the two?
I understand I have to allocate memory to the char * approach (Line #2). I'm not really sure how though.
And const char * and string literals are the same?
I need to know more on this. Can someone point to some nice exhaustive content/material?
The excellent source to clear up the confusion is Peter Van der Linden, Expert C Programming, Deep C secrets - that arrays and pointers are not the same is how they are addressed in memory.
With an array, char new_str[]; the compiler has given the new_str a memory address that is known at both compilation and runtime, e.g. 0x1234, hence the indexing of the new_str is simple by using []. For example new_str[4], at runtime, the code picks the address of where new_str resides in, e.g. 0x1234 (that is the address in physical memory). by adding the index specifier [4] to it, 0x1234 + 0x4, the value can then be retrieved.
Whereas, with a pointer, the compiler gives the symbol char *newstr an address e.g. 0x9876, but at runtime, that address used, is an indirect addressing scheme. Supposing that newstr was malloc'd newstr = malloc(10);, what is happening is that, everytime a reference in the code is made to use newstr, since the address of newstr is known by the compiler i.e. 0x9876, but what is newstr pointing to is variable. At runtime, the code fetches data from physical memory 0x9876 (i.e. newstr), but at that address is, another memory address (since we malloc'd it), e.g 0x8765 it is here, the code fetches the data from that memory address that malloc assigned to newstr, i.e. 0x8765.
The char new_str[] and char *newstr are used interchangeably, since an zeroth element index of the array decays into a pointer and that explains why you could newstr[5] or *(newstr + 5) Notice how the pointer expression is used even though we have declared char *newstr, hence *(new_str + 1) = *newstr; OR *(new_str + 1) = newstr[1];
In summary, the real difference between the two is how they are accessed in memory.
Get the book and read it and live it and breathe it. Its a brilliant book! :)
Please go through this article below:
Also see in case of array of char like in your case, char new_str[] then the new_str will always point to the base of the array. The pointer in itself can't be incremented. Yes you can use subscripts to access the next char in array eg: new_str[3];
But in case of pointer to char, the pointer can be incremented new_str++ to fetch you the next character in the array.
Also I would suggest this article for more clarity.
This is a character array:
char buf [1000];
So, for example, this makes no sense:
buf = &some_other_buf;
This is because buf, though it has characteristics of type pointer, it is already pointing to the only place that makes sense for it.
char *ptr;
On the other hand, ptr is only a pointer, and may point somewhere. Most often, it's something like this:
ptr = buf; // #1: point to the beginning of buf, same as &buf[0]
or maybe this:
ptr = malloc (1000); // #2: allocate heap and point to it
or:
ptr = "abcdefghijklmn"; // #3: string constant
For all of these, *ptr can be written to—except the third case where some compiling environment define string constants to be unwritable.
*ptr++ = 'h'; // writes into #1: buf[0], #2: first byte of heap, or
// #3 overwrites "a"
strcpy (ptr, "ello"); // finishes writing hello and adds a NUL
The difference is that one is a pointer, the other is an array. You can, for instance, sizeof() array. You may be interested in peeking here
If you're using C++ as your tags indicate, you really should be using the C++ strings, not the C char arrays.
The string type makes manipulating strings a lot easier.
If you're stuck with char arrays for some reason, the line:
char new_str[] = "";
allocates 1 byte of space and puts a null terminator character into it. It's subtly different from:
char *new_str = "";
since that may give you a reference to non-writable memory. The statement:
char *new_str;
on its own gives you a pointer but nothing that it points to. It can also have a random value if it's local to a function.
What people tend to do (in C rather than C++) is to do something like:
char *new_str = malloc (100); // (remember that this has to be freed) or
char new_str[100];
to get enough space.
If you use the str... functions, you're basically responsible for ensuring that you have enough space in the char array, lest you get all sorts of weird and wonderful practice at debugging code. If you use real C++ strings, a lot of the grunt work is done for you.
The type of the first is char[1], the second is char *. Different types.
Allocate memory for the latter with malloc in C, or new in C++.
char foo[] = "Bar"; // Allocates 4 bytes and fills them with
// 'B', 'a', 'r', '\0'.
The size here is implied from the initializer string.
The contents of foo are mutable. You can change foo[i] for example where i = 0..3.
OTOH if you do:
char *foo = "Bar";
The compiler now allocates a static string "Bar" in readonly memory and cannot be modified.
foo[i] = 'X'; // is now undefined.
char new_str[]="abcd";
This specifies an array of characters (a string) of size 5 bytes (one byte for each character plus one for the null terminator). So it stores the string 'abcd' in memory and we can access this string using the variable new_str.
char *new_str="abcd";
This specifies a string 'abcd' is stored somewhere in the memory and the pointer new_str points to the first character of that string.
To differentiate them in the memory allocation side:
// With char array, "hello" is allocated on stack
char s[] = "hello";
// With char pointer, "hello" is stored in the read-only data segment in C++'s memory layout.
char *s = "hello";
// To allocate a string on heap, malloc 6 bytes, due to a NUL byte in the end
char *s = malloc(6);
s = "hello";
If you're in c++ why not use std::string for all your string needs? Especially anything dealing with concatenation. This will save you from a lot of problems.