I am practicing allocation memory using malloc() with pointers, but 1 observation about pointers is that, why can strcpy() accept str variable without *:
char *str;
str = (char *) malloc(15);
strcpy(str, "Hello");
printf("String = %s, Address = %u\n", str, str);
But with integers, we need * to give str a value.
int *str;
str = (int *) malloc(15);
*str = 10;
printf("Int = %d, Address = %u\n", *str, str);
it really confuses me why strcpy() accepts str, because in my own understanding, "Hello" will be passed to the memory location of str that will cause some errors.
In C, a string is (by definition) an array of characters. However (whether we realize it all the time or not) we almost always end up accessing arrays using pointers. So, although C does not have a true "string" type, for most practical purposes, the type pointer-to-char (i.e. char *) serves this purpose. Almost any function that accepts or returns a string will actually use a char *. That's why strlen() and strcpy() accept char *. That's why printf %s expects a char *. In all of these cases, what these functions need is a pointer to the first character of the string. (They then read the rest of the string sequentially, stopping when they find the terminating '\0' character.)
In these cases, you don't use an explicit * character. * would extract just the character pointed to (that is, the first character of the string), but you don't want to extract the first character, you want to hand the whole string (that is, a pointer to the whole string) to strcpy so it can do its job.
In your second example, you weren't working with a string at all. (The fact that you used a variable named str confused me for a moment.) You have a pointer to some ints, and you're working with the first int pointed to. Since you're directly accessing one of the things pointed to, that's why you do need the explicit * character.
The * is called indirection or dereference operator.
In your second code,
*str = 10;
assigns the value 10 to the memory address pointed by str. This is one value (i.e., a single variable).
OTOTH, strcpy() copies the whole string all at a time. It accepts two char * parameters, so you don't need the * to dereference to get the value while passing arguments.
You can use the dereference operator, without strcpy(), copying element by element, like
char *str;
str = (char *) malloc(15); //success check TODO
int len = strlen("Hello"); //need string.h header
for (i = 0; i < len; i ++)
*(str+i)= "Hello"[i]; // the * form. as you wanted
str[i] = 0; //null termination
Many string manipulation functions, including strcpy, by convention and design, accept the pointer to the first character of the array, not the pointer to the whole array, even though their values are the same.
This is because their types are different; e.g. a pointer to char[10] has a different type from that of a pointer to char[15], and passing around the pointer to the whole array would be impossible or very clumsy because of this, unless you cast them everywhere or make different functions for different lengths.
For this reason, they have established a convention of passing around a string with the pointer to its first character, not to the whole array, possibly with its length when necessary. Many functions that operate on an array, such as memset, work the same way.
Well, here's what happens in the first snippet :
You are first dynamically allocating 15 bytes of memory, storing this address to the char pointer, which is pointer to a 1-byte sequence of data (a string).
Then you call strcpy(), which iterates over the string and copy characters, byte per byte, into the newly allocated memory space. Each character is a number based on the ASCII table, eg. character a = 97 (take a look at man ascii).
Then you pass this address to printf() which reads from the string, byte per byte, then flush it to your terminal.
In the second snippet, the process is the same, you are still allocating 15 bytes, storing the address in an int * pointer. An int is a 4 byte data type.
When you do *str = 10, you are dereferencing the pointer to store the value 10 at the address pointed by str. Remind what I wrote ahead, you could have done *str = 'a', and this index 0 integer would had the value 97, even if you try to read it as an int. you can event print it if you would.
So why strcpy() can take a int * as parameter? Because it's a memory space where it can write, byte per byte. You can store "Hell" in an int, then "o!" in the next one.
It's just all about usage easiness.
See there is a difference between = operator and the function strcpy.
* is deference operator. When you say *str, it means value at the memory location pointed by str.
Also as a good practice, use this
str = (char *) malloc( sizeof(char)*15 )
It is because the size of a data type might be different on different platforms. Hence use sizeof function to determine its actual size at the run time.
Related
I am not able to understand the difference between strcpy function and the method of equating the addresses of the strings using a pointer.The code given below would make my issue more clear. Any help would be appreciated.
//code to take input of strings in an array of pointers
#include <stdio.h>
#include <strings.h>
int main()
{
//suppose the array of pointers is of 10 elements
char *strings[10],string[50],*p;
int length;
//proper method to take inputs:
for(i=0;i<10;i++)
{
scanf(" %49[^\n]",string);
length = strlen(string);
p = (char *)malloc(length+1);
strcpy(p,string);//why use strcpy here instead of p = string
strings[i] = p; //why use this long way instead of writing directly strcpy(strings[i],string) by first defining malloc for strings[i]
}
return 0;
}
A short introduction into the magic of pointers:
char *strings[10],string[50],*p;
These are three variables with distinct types:
char *strings[10]; // an array of 10 pointers to char
char string[50]; // an array of 50 char
char *p; // a pointer to char
Then the followin is done (10 times):
scanf(" %49[^\n]",string);
Read C string from input and store it into string considering that a 0 terminator must fit in also.
length = strlen(string);
Count non-0 characters until 0 terminator is found and store in length.
p = (char *)malloc(length+1);
Allocate memory on heap with length + 1 (for 0 terminator) and store address of that memory in p. (malloc() might fail. A check if (p != NULL) wouldn't hurt.)
strcpy(p,string);//why use strcpy here instead of p = string
Copy C string in string to memory pointed in p. strcpy() copies until (inclusive) 0 terminator is found in source.
strings[i] = p;
Assign p (the pointer to memory) to strings[i]. (After assignment strings[i] points to the same memory than p. The assignment is a pointer assignment but not the assignment of the value to which is pointed.)
Why strcpy(p,string); instead of p = string:
The latter would assign address of string (the local variable, probably stored on stack) to p.
The address of allocated memory (with malloc()) would have been lost. (This introduces a memory leak - memory in heap which cannot be addressed by any pointer in code.)
p would now point to the local variable in string (for every iteration in for loop). Hence afterwards, all entries of strings[10] would point to string finally.
char *strings[10]---- --------->1.
strcpy(strings[i],string) ----->2.
strings[i] = string ----------->3.
p = (char *)malloc(length+1); -|
strcpy(p,string); |-> 4.
strings[i] = p;----------------|
strings is an array of pointers, each pointer must point to valid memory.
Will lead undefined behavior since strings[i] is not pointing to valid memory.
Works but every pointer of strings will point to same location thus each will have same contents.
Thus create the new memory first, copy the contents to it and assign that memory to strings[i]
strcpy copies a particular string into allocated memory. Assigning pointers doesn't actually copy the string, just sets the second pointer variable to the same value as the first.
strcpy(char *destination, char *source);
copies from source to destination until the function finds '\0'. This function is not secure and should not be used - try strncpy or strlcpy instead. You can find useful information about these two functions at https://linux.die.net/man/3/strncpy - check where your code is going to run in order to help you choose the best option.
In your code block you have this declaration
char *strings[10],string[50],*p;
This declares three pointers, but they are quite different. *p is an ordinary pointer, and must have space allocated for it (via malloc) before you can use it. string[50] is also a pointer, but of length 50 (characters, usually 1 byte) - and it's allocated on the function stack directly so you can use it right away (though the very first use of it should be to zero out the memory unless you've used a zeroing allocator like Solaris' calloc. Finally, *strings[10] is a double pointer - you have allocated an array of 10 pointers, each element of which (strings[1], strings[9] etc) must be allocated for before use.
The only one of those which you can assign to immediately is string, because the space is already allocated. Each of those pointers can be addressed via subscripts - but in each case you must ensure that you do not walk off the end otherwise you'll incur a SIGSEGV "segmentation violation" and your program will crash. Or at least, it should, but you might instead get merely weird results.
Finally, pointers allocated to must be freed manually otherwise you'll have memory leaks. Items allocated on the stack (string) do not need to be freed because the compiler handles that for you when the function ends.
Guys i have few queries in pointers. Kindly help to resolve them
char a[]="this is an array of characters"; // declaration type 1
char *b="this is an array of characters";// declaration type 2
question.1 : what is the difference between these 2 types of declaration ?
printf("%s",*b); // gives a segmentation fault
printf("%s",b); // displays the string
question.2 : i didn't get how is it working
char *d=malloc(sizeof(char)); // 1)
scanf("%s",d); // 2)
printf("%s",d);// 3)
question.3 how many bytes are being allocated to the pointer c?
when i try to input a string, it takes just a word and not the whole string. why so ?
char c=malloc(sizeof(char)); // 4)
scanf("%c",c); // 5)
printf("%c",c);// 6)
question.4 when i try to input a charcter why does it throw a segmentation fault?
Thanks in advance.. Waiting for your reply guys..
printf("%s",*b); // gives a segmentation fault
printf("%s",b); // displays the string
the %s expects a pointer to array of chars.
char *c=malloc(sizeof(char)); // you are allocating only 1 byte aka char, not array of char!
scanf("%s",c); // you need pass a pointer to array, not a pointer to char
printf("%s",c);// you are printing a array of chars, but you are sending a char
you need do this:
int sizeofstring = 200; // max size of buffer
char *c = malloc(sizeof(char))*sizeofstring; //almost equals to declare char c[200]
scanf("%s",c);
printf("%s",c);
question.3 how many bytes are being allocated to the pointer c? when i
try to input a string, it takes just a word and not the whole string.
why so ?
In your code, you only are allocating 1 byte because sizeof(char) = 1byte = 8bit, you need allocate sizeof(char)*N, were N is your "string" size.
char a[]="this is an array of characters"; // declaration type 1
char *b="this is an array of characters";// declaration type 2
Here you are declaring two variables, a and b, and initializing them. "this is an array of characters" is a string literal, which in C has type array of char. a has type array of char. In this specific case, the array does not get converted to a pointer, and a gets initialized with the array "this is an array of characters". b has type pointer to char, the array gets converted to a pointer, and b gets initialized with a pointer to the array "this is an array of characters".
printf("%s",*b); // gives a segmentation fault
printf("%s",b); // displays the string
In an expression, *b dereferences the pointer b, so it evaluates to the char pointed by b, i.e: T. This is not an address (which is what "%s" is expecting), so you get undefined behavior, most probably a crash (but don't try to do this on embedded systems, you could get mysterious behaviour and corrupted data, which is worse than a crash). In the second case, %s expects a pointer to a char, gets it, and can proceed to do its thing.
char *d=malloc(sizeof(char)); // 1)
scanf("%s",d); // 2)
printf("%s",d);// 3)
In C, sizeof returns the size in bytes of an object (= region of storage). In C, a char is defined to be the same as a byte, which has at least 8 bits, but can have more (but some standards put additional restrictions, e.g: POSIX requires 8-bit bytes, i.e: octets). So, you are allocating 1 byte. When you call scanf(), it writes in the memory pointed to by d without restraint, overwriting everything in sight. scanf() allows maximum field widths, so:
Allocate more memory, at least enough for what you want + 1 terminating ASCII NUL.
Tell scanf() to stop, e.g: scanf("%19s") for a maximum 19 characters (you'll need 20 bytes to store that, counting the terminating ASCII NUL).
And last (if markdown lets me):
char c=malloc(sizeof(char)); // 4)
scanf("%c",c); // 5)
printf("%c",c);// 6)
c is not a pointer, so you are trying to store an address where you shouldn't. In scanf, "%c" expects a pointer to char, which should point to an object (=region of storage) with enough space for the specified field width, 1 by default. Since c is not a pointer, the above may crash in some platforms (and cause worse things on others).
I see several problems in your code.
Question 1: The difference is:
a gets allocated in writable memory, the so-called data segment. Here you can read and write as much as you want. sizeof a is the length of the string plus 1, the so-called string terminator (just a null byte).
b, however, is just a pointer to a string which is located in the rodata. That means, in a data area which is read only. sizeof b is whatever is the pointer size on your system, maybe 4 or 8 on a PC or 2 on many embedded systems.
Question 2: The printf() format wants a pointer to a string. With *b, you dereferene the pointer you have and give it the first byte of data, which is a t (ASCII 84 or something like that). The callee, however, treats it as a pointer, dereferences it and BAM.
With b, however, everything goes fine, as it is exactly the right call.
Question 3: malloc(sizeof(char)) allocates exactly one byte. sizeof(char) is 1 by definition, so the call is effectively malloc(1). The input just takes a word because %s is defined that way.
Question 4:
char c=malloc(sizeof(char)); // 4)
shound give you a warning: malloc() returns a pointer which you try to put into a char. ITYM char *...
As you continue, you give that pointer to scanf(), which receives e.g. instead of 0x80043214 a mere 0x14, interprets it as a pointer and BAM again.
The correct way would be
char * c=malloc(1024);
scanf("%1024s", c);
printf("%s", c);
Why? Well, you want to read a string. 1 byte is too small, better allocate more.
In scanf() you should take care that you don't allow reading more than your buffer can hold - thus the limitation in the format specifier.
and on printing, you should use %s, because you want the whole string to be printed and not only the first character. (At least, I suppose so.)
Ad Q1: The first is an array of chars with a fixed pointer a pointing to it. sizeof(a) will return something like 20 (strlen(a)+1). Trying to assign something to a (like a = b) will fail, since a is fixed.
The second is a pointer pointing to an array of char and hence is the sizeof(b) usually 4 on 32-bit or 8 on 64-bit. Assigning something to b will work, since the pointer can take a new value.
Of course, *a or *b work on both.
Ad Q2: printf() with the %s argument takes a pointer to a char (those are the "strings" in C). Hence, printf("%s", *b) will crash, since the "pointer" used by printf() will contain the byte value of *b.
What you could do, is printf("%c", *b), but that would only print the first character.
Ad Q3: sizeof(char) is 1 (by definition), hence you allocate 1 byte. The scanf will most likely read more than one byte (remember that each string will be terminated by a null character occupying one char). Hence the scanf will trash memory, likely to cause memory sometime later on.
Ad 4: Maybe that's the trashed memory.
Both declaration are the same.
b point to the first byte so when you say *b it's the first character.
printf("%s", *b)
Will fail as %s accepts a pointer to a string.
char is one byte.
I'm a little bit confused about something. I was under the impression that the correct way of reading a C string with scanf() went along the lines of
(never mind the possible buffer overflow, it's just a simple example)
char string[256];
scanf( "%s" , string );
However, the following seems to work too,
scanf( "%s" , &string );
Is this just my compiler (gcc), pure luck, or something else?
An array "decays" into a pointer to its first element, so scanf("%s", string) is equivalent to scanf("%s", &string[0]). On the other hand, scanf("%s", &string) passes a pointer-to-char[256], but it points to the same place.
Then scanf, when processing the tail of its argument list, will try to pull out a char *. That's the Right Thing when you've passed in string or &string[0], but when you've passed in &string you're depending on something that the language standard doesn't guarantee, namely that the pointers &string and &string[0] -- pointers to objects of different types and sizes that start at the same place -- are represented the same way.
I don't believe I've ever encountered a system on which that doesn't work, and in practice you're probably safe. None the less, it's wrong, and it could fail on some platforms. (Hypothetical example: a "debugging" implementation that includes type information with every pointer. I think the C implementation on the Symbolics "Lisp Machines" did something like this.)
I think that this below is accurate and it may help.
Feel free to correct it if you find any errors. I'm new at C.
char str[]
array of values of type char, with its own address in memory
array of values of type char, with its own address in memory
as many consecutive addresses as elements in the array
including termination null character '\0' &str, &str[0] and str, all three represent the same location in memory which is address of the first element of the array str
char *strPtr = &str[0]; //declaration and initialization
alternatively, you can split this in two:
char *strPtr; strPtr = &str[0];
strPtr is a pointer to a char
strPtr points at array str
strPtr is a variable with its own address in memory
strPtr is a variable that stores value of address &str[0]
strPtr own address in memory is different from the memory address that it stores (address of array in memory a.k.a &str[0])
&strPtr represents the address of strPtr itself
I think that you could declare a pointer to a pointer as:
char **vPtr = &strPtr;
declares and initializes with address of strPtr pointer
Alternatively you could split in two:
char **vPtr;
*vPtr = &strPtr
*vPtr points at strPtr pointer
*vPtr is a variable with its own address in memory
*vPtr is a variable that stores value of address &strPtr
final comment: you can not do str++, str address is a const, but
you can do strPtr++
Man, pointers continue to give me trouble. I thought I understood the concept.(Basically, that you would use *ptr when you want to manipulate the actual memory saved at the location that ptr points to. You would just use ptr if you would like to move that pointer by doing things such as ptr++ or ptr--.) So, if that is the case, if you use the asterisk to manipulate the files that the pointer is pointing to, how does this work:
char *MallocAndCopy( char *line ) {
char *pointer;
if ( (pointer = malloc( strlen(line)+1 )) == NULL ) {
printf( "Out of memory!!! Goodbye!\n" );
exit( 0 );
}
strcpy( pointer, line );
return pointer;
}
malloc returns a pointer, so I understand why the "pointer" in the if condition does not utilize the asterisk. However, in the strcpy function, it is sending the CONTENTS of line to the CONTENTS of pointer. Shouldn't it be:
strcpy( *pointer, *line);
????? Or is my understanding of pointers correct and that is just the way that the strcpy function works?
A C-style string is an array of character bytes. When you pass an array around as a pointer to the array type, the pointer contains the address of the first element of the array.
The strcpy function takes the pointer to the first char of the source array (which is the start of the string) and the pointer to the first char in the destination array, and iterates over the source until it reaches a '\0' character, which terminates the string.
That's also why when you call malloc, the size that you pass to it is strlen(line)+1, because you need to allocate one more byte for the termination character (as far as I know).
char* strcpy(char *destination, const char *source) is the signature for strcpy. It expects to get pointers as arguments. In your instance, pointer is already a pointer, and so is line.
strcpy will take the pointers source and destination, and copy the underlying bytes pointed at from source to destination until it either hits the \0 (NULL) byte in the string pointed to by source, or if it segmentation faults because it never encounters that byte (unterminated string) and just reads off into the abyss.
If you used *pointer, you would actually be dereferencing the pointer and getting the char at the address the pointer points to.
The signature of strcpy is
char * strcpy ( char * destination, const char * source );
You are sending in pointer which is char*, and line which is also char*. Thus you are matching the signature exactly as expected. Sending in *pointer or *line would be sending 'the values these pointers point to' - which would be wrong.
scrcpy takes one pointer, and writes the contents it is pointing to to the location pointed by the another pointer. It is like rewriting cell contents in the table. You don't necessarily have to cut and paste a part of the table, you can rewrite the numbers in it. In that case pointer is just a hint which cells you have to touch.
Look at the declaration of strpcy().
char *strcpy(
char *strDestination,
const char *strSource
);
You need to pass a pointer. So you don't dereference pointer and line. Because that would pass a char.
The * is inside strcpy.
Applying the * operator is called dereferencing, and is exactly the same as applying the [0].
strlen and strcpy need to know where the strings start so that they can access all of their elements, not just the first.
The type of pointer is char * while the type of *pointer is char, meaning a single character with no concept of neighboring characters.
By writing *pointer, you obtain the character that it points to (which happens to be the first character of the string that pointer represents). Passing it to strcpy involves creating a copy of that character. It's impossible for strcpy to know where it should be writing, unless you give it a pointer to that data.
In a similar fashion, by providing *line you're only giving strcpy a copy of the first character of your string.
So, you're basically saying:
I have two strings. I'm not giving them to you. The first letter of one is C, the first letter of the other is ยต. Copy the contents of the first to the second
I see why you expect to pass *pointer and *line to strcpy. But remember that pointer points to the location in memory where the content is stored and where there is strlen(line)+1 bytes of memory is reserved and allocated. Thus what strcpy is exactly doing is to copy strlen(line) bytes of memory starting from the address line into corresponding locations, starting from the address of pointer. If you pass *pointer and *line, strcpy will have access to only the first char and not the rest of strlen(line)-1. Hope that clarifies your confusion about the pointers and why we need to pass pointers to strcpy.
Your basic understanding of strcpy is correct. The function indeed copies the contents of line over the contents of pointer. However, you have to pass pointers to the function and not the data itself. This is done for several reasons.
First, the data that is being copied is a string (that is, an array) and not a regular variable. When passing an array to a function, you can't cram the entire array into a single function argument so instead you have to pass a pointer to the beginning of the array. This is a limitation of array passing in C and is not specific to strcpy.
Also, passing data to a function provides the function with a copy of the variable's content. If you directly passed data (or de-referenced pointers) to strcpy, the function would be working with copies and not the original data. It would be unable to write data to the original string.
Essentially, by handing pointers to strcpy you are telling it the location of the source data and of the destination. The function handles all of the de-referencing internally. In human language, the call to strcpy can be thought of as "take the string that starts at memory address line and write a copy it starting at memory address pointer". To communicate memory addresses, you use pointers.
Consider the strcpy code as:
char * strcpy(char * dst, char* src)
{
int i = 0 ;
for (;dst[i]=source[i++];);
return dst ;
}
\0(NULL) is the end of both the string and the for loop.
Your function is basically what strdup does. The function is defined on many platform, if it isn't you can define this way:
char * my_strdup(const char* s)
{
size_t len = strlen(s);
char *r = malloc(len+1);
return r ? memcpy(r, s, len+1) : NULL;
}
memcpy is usually faster than strcpy for longer strings.
I'm learning C right now and got a bit confused with character arrays - strings.
char name[15]="Fortran";
No problem with this - its an array that can hold (up to?) 15 chars
char name[]="Fortran";
C counts the number of characters for me so I don't have to - neat!
char* name;
Okay. What now? All I know is that this can hold an big number of characters that are assigned later (e.g.: via user input), but
Why do they call this a char pointer? I know of pointers as references to variables
Is this an "excuse"? Does this find any other use than in char*?
What is this actually? Is it a pointer? How do you use it correctly?
thanks in advance,
lamas
I think this can be explained this way, since a picture is worth a thousand words...
We'll start off with char name[] = "Fortran", which is an array of chars, the length is known at compile time, 7 to be exact, right? Wrong! it is 8, since a '\0' is a nul terminating character, all strings have to have that.
char name[] = "Fortran";
+======+ +-+-+-+-+-+-+-+--+
|0x1234| |F|o|r|t|r|a|n|\0|
+======+ +-+-+-+-+-+-+-+--+
At link time, the compiler and linker gave the symbol name a memory address of 0x1234.
Using the subscript operator, i.e. name[1] for example, the compiler knows how to calculate where in memory is the character at offset, 0x1234 + 1 = 0x1235, and it is indeed 'o'. That is simple enough, furthermore, with the ANSI C standard, the size of a char data type is 1 byte, which can explain how the runtime can obtain the value of this semantic name[cnt++], assuming cnt is an integer and has a value of 3 for example, the runtime steps up by one automatically, and counting from zero, the value of the offset is 't'. This is simple so far so good.
What happens if name[12] was executed? Well, the code will either crash, or you will get garbage, since the boundary of the array is from index/offset 0 (0x1234) up to 8 (0x123B). Anything after that does not belong to name variable, that would be called a buffer overflow!
The address of name in memory is 0x1234, as in the example, if you were to do this:
printf("The address of name is %p\n", &name);
Output would be:
The address of name is 0x00001234
For the sake of brevity and keeping with the example, the memory addresses are 32bit, hence you see the extra 0's. Fair enough? Right, let's move on.
Now on to pointers...
char *name is a pointer to type of char....
Edit:
And we initialize it to NULL as shown Thanks Dan for pointing out the little error...
char *name = (char*)NULL;
+======+ +======+
|0x5678| -> |0x0000| -> NULL
+======+ +======+
At compile/link time, the name does not point to anything, but has a compile/link time address for the symbol name (0x5678), in fact it is NULL, the pointer address of name is unknown hence 0x0000.
Now, remember, this is crucial, the address of the symbol is known at compile/link time, but the pointer address is unknown, when dealing with pointers of any type
Suppose we do this:
name = (char *)malloc((20 * sizeof(char)) + 1);
strcpy(name, "Fortran");
We called malloc to allocate a memory block for 20 bytes, no, it is not 21, the reason I added 1 on to the size is for the '\0' nul terminating character. Suppose at runtime, the address given was 0x9876,
char *name;
+======+ +======+ +-+-+-+-+-+-+-+--+
|0x5678| -> |0x9876| -> |F|o|r|t|r|a|n|\0|
+======+ +======+ +-+-+-+-+-+-+-+--+
So when you do this:
printf("The address of name is %p\n", name);
printf("The address of name is %p\n", &name);
Output would be:
The address of name is 0x00005678
The address of name is 0x00009876
Now, this is where the illusion that 'arrays and pointers are the same comes into play here'
When we do this:
char ch = name[1];
What happens at runtime is this:
The address of symbol name is looked up
Fetch the memory address of that symbol, i.e. 0x5678.
At that address, contains another address, a pointer address to memory and fetch it, i.e. 0x9876
Get the offset based on the subscript value of 1 and add it onto the pointer address, i.e. 0x9877 to retrieve the value at that memory address, i.e. 'o' and is assigned to ch.
That above is crucial to understanding this distinction, the difference between arrays and pointers is how the runtime fetches the data, with pointers, there is an extra indirection of fetching.
Remember, an array of type T will always decay into a pointer of the first element of type T.
When we do this:
char ch = *(name + 5);
The address of symbol name is looked up
Fetch the memory address of that symbol, i.e. 0x5678.
At that address, contains another address, a pointer address to memory and fetch it, i.e. 0x9876
Get the offset based on the value of 5 and add it onto the pointer address, i.e. 0x987A to retrieve the value at that memory address, i.e. 'r' and is assigned to ch.
Incidentally, you can also do that to the array of chars also...
Further more, by using subscript operators in the context of an array i.e. char name[] = "..."; and name[subscript_value] is really the same as *(name + subscript_value).
i.e.
name[3] is the same as *(name + 3)
And since the expression *(name + subscript_value) is commutative, that is in the reverse,
*(subscript_value + name) is the same as *(name + subscript_value)
Hence, this explains why in one of the answers above you can write it like this (despite it, the practice is not recommended even though it is quite legitimate!)
3[name]
Ok, how do I get the value of the pointer?
That is what the * is used for,
Suppose the pointer name has that pointer memory address of 0x9878, again, referring to the above example, this is how it is achieved:
char ch = *name;
This means, obtain the value that is pointed to by the memory address of 0x9878, now ch will have the value of 'r'. This is called dereferencing. We just dereferenced a name pointer to obtain the value and assign it to ch.
Also, the compiler knows that a sizeof(char) is 1, hence you can do pointer increment/decrement operations like this
*name++;
*name--;
The pointer automatically steps up/down as a result by one.
When we do this, assuming the pointer memory address of 0x9878:
char ch = *name++;
What is the value of *name and what is the address, the answer is, the *name will now contain 't' and assign it to ch, and the pointer memory address is 0x9879.
This where you have to be careful also, in the same principle and spirit as to what was stated earlier in relation to the memory boundaries in the very first part (see 'What happens if name[12] was executed' in the above) the results will be the same, i.e. code crashes and burns!
Now, what happens if we deallocate the block of memory pointed to by name by calling the C function free with name as the parameter, i.e. free(name):
+======+ +======+
|0x5678| -> |0x0000| -> NULL
+======+ +======+
Yes, the block of memory is freed up and handed back to the runtime environment for use by another upcoming code execution of malloc.
Now, this is where the common notation of Segmentation fault comes into play, since name does not point to anything, what happens when we dereference it i.e.
char ch = *name;
Yes, the code will crash and burn with a 'Segmentation fault', this is common under Unix/Linux. Under windows, a dialog box will appear along the lines of 'Unrecoverable error' or 'An error has occurred with the application, do you wish to send the report to Microsoft?'....if the pointer has not been mallocd and any attempt to dereference it, is guaranteed to crash and burn.
Also: remember this, for every malloc there is a corresponding free, if there is no corresponding free, you have a memory leak in which memory is allocated but not freed up.
And there you have it, that is how pointers work and how arrays are different to pointers, if you are reading a textbook that says they are the same, tear out that page and rip it up! :)
I hope this is of help to you in understanding pointers.
That is a pointer. Which means it is a variable that holds an address in memory. It "points" to another variable.
It actually cannot - by itself - hold large amounts of characters. By itself, it can hold only one address in memory. If you assign characters to it at creation it will allocate space for those characters, and then point to that address. You can do it like this:
char* name = "Mr. Anderson";
That is actually pretty much the same as this:
char name[] = "Mr. Anderson";
The place where character pointers come in handy is dynamic memory. You can assign a string of any length to a char pointer at any time in the program by doing something like this:
char *name;
name = malloc(256*sizeof(char));
strcpy(name, "This is less than 256 characters, so this is fine.");
Alternately, you can assign to it using the strdup() function, like this:
char *name;
name = strdup("This can be as long or short as I want. The function will allocate enough space for the string and assign return a pointer to it. Which then gets assigned to name");
If you use a character pointer this way - and assign memory to it, you have to free the memory contained in name before reassigning it. Like this:
if(name)
free(name);
name = 0;
Make sure to check that name is, in fact, a valid point before trying to free its memory. That's what the if statement does.
The reason you see character pointers get used a whole lot in C is because they allow you to reassign the string with a string of a different size. Static character arrays don't do that. They're also easier to pass around.
Also, character pointers are handy because they can be used to point to different statically allocated character arrays. Like this:
char *name;
char joe[] = "joe";
char bob[] = "bob";
name = joe;
printf("%s", name);
name = bob;
printf("%s", name);
This is what often happens when you pass a statically allocated array to a function taking a character pointer. For instance:
void strcpy(char *str1, char *str2);
If you then pass that:
char buffer[256];
strcpy(buffer, "This is a string, less than 256 characters.");
It will manipulate both of those through str1 and str2 which are just pointers that point to where buffer and the string literal are stored in memory.
Something to keep in mind when working in a function. If you have a function that returns a character pointer, don't return a pointer to a static character array allocated in the function. It will go out of scope and you'll have issues. Repeat, don't do this:
char *myFunc() {
char myBuf[64];
strcpy(myBuf, "hi");
return myBuf;
}
That won't work. You have to use a pointer and allocate memory (like shown earlier) in that case. The memory allocated will persist then, even when you pass out of the functions scope. Just don't forget to free it as previously mentioned.
This ended up a bit more encyclopedic than I'd intended, hope its helpful.
Editted to remove C++ code. I mix the two so often, I sometimes forget.
char* name is just a pointer. Somewhere along the line memory has to be allocated and the address of that memory stored in name.
It could point to a single byte of memory and be a "true" pointer to a single char.
It could point to a contiguous area of memory which holds a number of characters.
If those characters happen to end with a null terminator, low and behold you have a pointer to a string.
char *name, on it's own, can't hold any characters. This is important.
char *name just declares that name is a pointer (that is, a variable whose value is an address) that will be used to store the address of one or more characters at some point later in the program. It does not, however, allocate any space in memory to actually hold those characters, nor does it guarantee that name even contains a valid address. In the same way, if you have a declaration like int number there is no way to know what the value of number is until you explicitly set it.
Just like after declaring the value of an integer, you might later set its value (number = 42), after declaring a pointer to char, you might later set its value to be a valid memory address that contains a character -- or sequence of characters -- that you are interested in.
It is confusing indeed. The important thing to understand and distinguish is that char name[] declares array and char* name declares pointer. The two are different animals.
However, array in C can be implicitly converted to pointer to its first element. This gives you ability to perform pointer arithmetic and iterate through array elements (it does not matter elements of what type, char or not). As #which mentioned, you can use both, indexing operator or pointer arithmetic to access array elements. In fact, indexing operator is just a syntactic sugar (another representation of the same expression) for pointer arithmetic.
It is important to distinguish difference between array and pointer to first element of array. It is possible to query size of array declared as char name[15] using sizeof operator:
char name[15] = { 0 };
size_t s = sizeof(name);
assert(s == 15);
but if you apply sizeof to char* name you will get size of pointer on your platform (i.e. 4 bytes):
char* name = 0;
size_t s = sizeof(name);
assert(s == 4); // assuming pointer is 4-bytes long on your compiler/machine
Also, the two forms of definitions of arrays of char elements are equivalent:
char letters1[5] = { 'a', 'b', 'c', 'd', '\0' };
char letters2[5] = "abcd"; /* 5th element implicitly gets value of 0 */
The dual nature of arrays, the implicit conversion of array to pointer to its first element, in C (and also C++) language, pointer can be used as iterator to walk through array elements:
/ *skip to 'd' letter */
char* it = letters1;
for (int i = 0; i < 3; i++)
it++;
In C a string is actually just an array of characters, as you can see by the definition. However, superficially, any array is just a pointer to its first element, see below for the subtle intricacies. There is no range checking in C, the range you supply in the variable declaration has only meaning for the memory allocation for the variable.
a[x] is the same as *(a + x), i.e. dereference of the pointer a incremented by x.
if you used the following:
char foo[] = "foobar";
char bar = *foo;
bar will be set to 'f'
To stave of confusion and avoid misleading people, some extra words on the more intricate difference between pointers and arrays, thanks avakar:
In some cases a pointer is actually semantically different from an array, a (non-exhaustive) list of examples:
//sizeof
sizeof(char*) != sizeof(char[10])
//lvalues
char foo[] = "foobar";
char bar[] = "baz";
char* p;
foo = bar; // compile error, array is not an lvalue
p = bar; //just fine p now points to the array contents of bar
// multidimensional arrays
int baz[2][2];
int* q = baz; //compile error, multidimensional arrays can not decay into pointer
int* r = baz[0]; //just fine, r now points to the first element of the first "row" of baz
int x = baz[1][1];
int y = r[1][1]; //compile error, don't know dimensions of array, so subscripting is not possible
int z = r[1]: //just fine, z now holds the second element of the first "row" of baz
And finally a fun bit of trivia; since a[x] is equivalent to *(a + x) you can actually use e.g. '3[a]' to access the fourth element of array a. I.e. the following is perfectly legal code, and will print 'b' the fourth character of string foo.
#include <stdio.h>
int main(int argc, char** argv) {
char foo[] = "foobar";
printf("%c\n", 3[foo]);
return 0;
}
One is an actual array object and the other is a reference or pointer to such an array object.
The thing that can be confusing is that both have the address of the first character in them, but only because one address is the first character and the other address is a word in memory that contains the address of the character.
The difference can be seen in the value of &name. In the first two cases it is the same value as just name, but in the third case it is a different type called pointer to pointer to char, or **char, and it is the address of the pointer itself. That is, it is a double-indirect pointer.
#include <stdio.h>
char name1[] = "fortran";
char *name2 = "fortran";
int main(void) {
printf("%lx\n%lx %s\n", (long)name1, (long)&name1, name1);
printf("%lx\n%lx %s\n", (long)name2, (long)&name2, name2);
return 0;
}
Ross-Harveys-MacBook-Pro:so ross$ ./a.out
100001068
100001068 fortran
100000f58
100001070 fortran