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A few questions regarding C strings:
both char* and char[] are pointers?
I've learned about pointers and I can tell that char* is a pointer, but why is it automatically a string and not just a char pointer that points to 1 char; why can it hold strings?
Why, unlike other pointers, when you assign a new value to the char* pointer you are actually allocating new space in memory to store the new value and, unlike other pointers, you just replace the value stored in the memory address the pointer is pointing at?
A pointer is not a string.
A string is a constant object having type array of char and, also, it has the property that the last element of the array is the null character '\0' which, in turn, is an int value (converted to char type) having the integer value 0.
char* is a pointer, but char[] is not. The type char[] is not a "real" type, but an incomplete type. The C language is specified in such a way that, in the moment that you define a concrete variable (object) having array of char type, the size of the array is well determined in some way or another. Thus, none variable has type char[] because this is not a type (for a given object).
However, automatically every object having type array of N objects of type char is promoted to char *, that is, a pointer to char pointing to the initial object of the array.
On the other hand, this promotion is not always performed. For example, the operator sizeof() will give different results for char* than for an array of N chars. In the former case, the size of a pointer to char is given (which is in general the same amount for every pointer...), and in the last case gives you the value N, that is, the size of the array.
The behaviour is differente when you declare function arguments as char* and char[]. Since the function cannot know the size of the array, you can think of both declarations as equivalent.
Actually, you are right here: char * is a pointer to just 1 character object. However, it can be used to access strings, as I will explain you now: In the paragraph 1. I showed you that the strings are considered objects in memory having type array of N chars for some N. This value N is big enough to allow an ending null character (as all "string" is supposed to be in C).
So, what's the deal here?
The key point to understand this issues is the concept of object (in memory).
When you have a string or, more generally, an array of char, this means that you have figured out some manner to hold an array object in memory.
This object determines a portion of RAM memory that you can access safely, because C has assigned enough memory for it.
Thus, when you point to the first byte of this object with a char* variable, actually you have guaranteed access to all the adjacent elements to the "right" of that memory place, because those places are well defined by C as having the bytes of the array above.
Briefly: the adjacent (to the right) bytes of the byte pointed by a char* variable can be accessed, they are valid places to access, so the pointer can be "iterated" to walk through these bytes, up to the end of the string, without "risks", since all the bytes in an array are contiguous well defined positions in memory.
This is a complicated question, but it reveals that you are not understanding the relationship between pointers, arrays, and string literals in C.
A pointer is just a variable pointing to a position in memory.
A pòinter to char points to just 1 object having type char.
If the adjacent bytes of the pointed position correspond to an array of chars, they will be accessible by the pointer, so the pointer can "walk on" the memory bytes occupied by the array object.
A string literal is considered as an array of char object, which implictely add an ending byte with value 0 (the null character).
In any case, an array of T object has a well defined "size".
A string literal has an additional property: it's a constant object.
Try to fit and gather these concepts in your mind to figure out what's going on.
And ask me for clarification.
ADDITIONAL REMARKS:
Consider the following piece of code:
#include <stdio.h>
int main(void)
{
char *s1 = "not modifiable";
char s2[] = "modifiable";
printf("%s ---- %s\n\n", s1, s2);
printf("Size of array s2: %d\n\n", (int)sizeof(s2));
s2[1] = '0', s2[3] = s2[5] = '1', s2[4] = '7',
s2[6] = '4', s2[7] = '8', s2[9] = '3';
printf("New value of s2: %s\n\n",s2);
//s1[0] = 'X'; // Attempting to modify s1
}
In the definition and initialization of s1 we have the string literal "not modifiable", which has constant content and constant address. Its address is assigned to the pointer s1 as initialization.
Any attempt to modify the bytes of the string will give some kind of error, because the array content is read-only.
In the definition and initialization of s2, we have the string literal "modifiable", which has, again, constant content and constant address. However, what happens now is that, as part of the initialization, the content of the string is copied to the array of char s2. The size of the array s2 is not specified (the declaration char s2[] gives an incomplete type), but after initialization the size of the array is well determined and defined as the exact size of the copied string (plus 1 character used to hold the null character, or end-of-string mark).
So, the string literal "modifiable" is used to initialize the bytes of the array s2, which is modifiable.
The right manner to do that is by changing a character at the time.
For more handy ways of modifying and assigning strings, it has to be used the standard header <string.h>.
char *s is a pointer, char s[] is an array of characters. Ex.
char *s = "hello";
char c[] = "world";
s = c; //Legal
c = address of some other string //Illegal
char *s is not a string; it points to an address. Ex
char c[] = "hello";
char *s = &c[3];
Assigning a pointer is not creating memory; you are pointing to memory. Ex.
char *s = "hello";
In this example when you type "hello" you are creating special memory to hold the string "hello" but that has nothing to do with the pointer, the pointer simply points to that spot.
version 1
char *cad1="hell";
char *cad2="home";
int j;
cad2=cad1;
for (j=0;j<4;j++){
printf("%c",cad1[j]);
}
cad2[0]='a';
for (j=0;j<4;j++){
printf("%c",cad2[j]);
}
version 2
char cad1[]="hell";
char cad2[]="home";
int j;
cad2=cad1;
for (j=0;j<4;j++){
printf("%c",cad1[j]);
}
cad2[0]='a';
for (j=0;j<4;j++){
printf("%c",cad2[j]);
}
version 3
char cad1[]="hell";
char *cad2="home";
int j;
cad2=cad1;
for (j=0;j<4;j++){
printf("%c",cad1[j]);
}
cad2[0]='a';
for (j=0;j<4;j++){
printf("%c",cad2[j]);
}
The question that I have is why version 1 hangs the dev c++?, version 2 says there is an incompatible assignment in cad2=cad1? and why version 3 works normal?
When you declare pointer like,
char *cad1="hell";
"hell" is called as constant string literal and so may be stored in read-only memory. - compiler is free to choose whatever it likes.
But when you declare it as,
char cad2[]="hell";
"hell" is stored as array member. ie, it will be stored as,
cad[0] = 'h', cad[1] = 'e', cad[2] = 'l', cad[3] = 'l', cad[4] = '\0'
C doesn't guarantee any defined behavior for changing constant literals. It may crash hang or spoil other valid data. Its called as undefined behavior.
Since you are changing cad1 which is pointing to constant literal your application hangs.
In version 2, both cad1 and cad2 are of array type. Direct array assignments in C is illegal. So you got error. Refer this link for all details as mentioned by others.
To answer why version 3 works,
cad1 is an array and cad2 is pointer here. By the statement cad2 = cad1 you made cad2 to point the memory which can be modified ( still, size is restricted). So changing cad1 and cad2 are same as they point same modifiable memory.
In version 1, cad2 is equal to cad1 which points to the constant string "hell". Later on, you attempt to modify that constant string, which is unpredictable. Version 3, in contrast, has cad1 declared as a char array, so you get a non-constant copy of the string, so modifying it will work.
For version 2, it's probably because both are arrays (not pointers), so I'm sure there's some issues there.
If cad is declared as char* cad="hell"; then that is a string literal (of length 4 plus 1 for a null terminator) and any attempt to modify a string literal is undefined behaviour. Anything could happen.
char cad[]="home"; will allocate 5 chars on the stack, cad[4] is '\0' - the null terminator; used by many string functions in C in modelling a set of chars as a string to mark the string end. You are free to modify these data although changing cad[4] will cause you trouble when using C string library functions as you will have removed their stopping condition.
Throughout your code you have cad2=cad1; Note that this does not copy the string, just the pointer; use strcpy in the C standard library to copy strings.
Really you should write const char* cad="hell";. Newer c++ compilers will insist on it.
why version 1 hangs the dev c++?
Read comments:
char *cad1="hell"; // pointer to constant string "hell"
char *cad2="home";
cad2=cad1; // now cad2 points to constant string "hell" too
cad2[0]='a'; // modifying of constant string causes undefined behaviour.
version 2 says there is an incompatible assignment in cad2=cad1?
Read comments:
char cad2[]="home"; // cad2 is array
cad2=cad1; // error because you can not assign to arrays in C.
why version 3 works normal?
Read comments:
char cad1[]="hell"; // cad1 is array
char *cad2="home";
cad2=cad1; // now cad2 point to first element of array cad1
cad2[0]='a'; // you can modify arrays in C
Note that you can not assign to arrays but you can modify them by copying or assigning values to their elements.
I get the error message "error: incompatible types when assigning to type ‘char[10]’ from type ‘char *’ ", when trying to initialize name in the following structure:
#include <stdio.h>
struct info
{
char name[10];
int years;
};
int main(void){
struct info b;
b.name = "Michael";
b.years = 19;
return 0
;}
Since you're just trying to initialize your struct, you could probably get away with:
struct info b = {"Michael", 19};
C makes a distinction between initialization and assignment (even though they use the same operator). Arrays cannot be assigned, (memcpy or similar must be used instead) but data can come into existence with a certain value, which the initializer specifies.
Use this instead:
strcpy(b.name,"Michael");
Arrays are not pointers in C. You need to strcpy to copy a string into another location.
Let's see what happens when the main() runs.
struct info b;
By this you initialize the struct, including the name in it.
name is a char array (as Jay pointed out, it's different from a pointer), and now it is assigned a chunk of memory with some random stuff in it.
Next
b.name = "Michael";
By this you try to assign a string literal to your char array.
Normally, you assign string literal to char array in the declaration like this char a[] = "hello";
It will copy the string literal to a chunk of memory and let a "sit aside the memory". However in your case, b.name is already sitting aside another chunk of memory, and you cannot change the sitting by just do an assignment, because an array is not a pointer.
So, if you want to modify b.name, you either use strcpy as Vaughn Cato said, or you do that character by charater, like
b.name[0] = 'M';
b.name[1] = 'i';
...
which essentially is the same thing as 'strcpy'
if you do char name[10] you are staticaly allocating the memory whose base address can be accessed by the name variable. You cannot chage the address pointer to a statically allocated array. i.e you cannot add a different address to name variable.
In you do b.name = "Micheal", you are doing just that. You are trying to add the base address of "Micheal" string to name variable.
So to do that as #Vaughn said, copy the "Micheal" to the memory pointed by b.name. or in the structure, change char name[10] to char *name
The problem is that name's address is necessarily inside the memory for the info object containing it, while the string literal is in some distinct area of memory. The compiler makes sure their locations are sensible given where they're mentioned in your program, and they can't be moved or merged arbitrarily. Either:
name should be changed to a const char* and pointed at the literal (here, const char* as string literals must not be modified), or
the content from the literal strcpy()-ied into name (but you'll need to be careful to make sure name is always large enough for the text and terminating ASCII NUL/0 character), or
name changed to a char* and assigned to a strdup() heap-allocated copy of the literal (which ensures you can modify name afterwards, and that the memory controlled by name will be large enough to store a copy of the literal, but you will need to free(name) when you've finished with it).
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Closed 11 years ago.
Possible Duplicate:
What is the difference between char s[] and char *s in C?
Do these statements about pointers have the same effect?
All this time I thought that whenever I need to copy a string(either literal or in a variable) I need to use strcpy(). However I recently found out this:
char a[]="test";
and this
char *a="test";
From what I understand the second type is unsafe and will print garbage in some cases. Is that correct? What made me even more curious is why the following doesn't work:
char a[5];
a="test";
or this
char a[];
a="test";
but this works however
char *a;
a="test";
I would be greatful if someone could clear up things a bit.
char a[]="test";
This declares and initializes an array of size 5 with the contents of "test".
char *a="test";
This declares and initializes a pointer to the literal "test". Attempting to modify a literal through a is undefined behavior (and probably results in the garbage you are seeing). It's not unsafe, it just can't be modified since literals are immutable.
char a[5];
a="test";
This fails even when both a and "test" have the exact same type, just as any other attempt to copy arrays thorugh assignment.
char a[];
a="test";
This declares an array of unknown size. The declaration should be completed before being used.
char *a;
a="test";
This works just fine since "test" decays to a pointer to the literal's first element. Attempting to modify its contents is still undefined behavior.
Let's examine case by case:
char a[]="test";
This tells the compiler to allocate 5 bytes on the stack, put 't' 'e' 's' 't' and '\0' on it. Then the variable a points to where 't' was written and you have a pointer pointing to a valid location with 5 available spaces. (That is if you view a as a pointer. In truth, the compiler still treats a as a single custom type that consists of 5 chars. In an extreme case, you can imagine it something like struct { char a, b, c, d, e; } a;)
char *a="test";
"test" (which like I said is basically 't' 'e' 's' 't' and '\0') is stored somewhere in your program, say a "literal's area", and a is pointing to it. That area is not yours to modify but only to read. a by itself doesn't have any specific memory (I am not talking about the 4/8 bytes of pointer value).
char a[5];
a = "test";
You are telling the compiler to copy the contents of one string over to another one. This is not a simple operation. In the case of char a[] = "test"; it was rather simple because it was just 5 pushes on the stack. In this case however it is a loop that needs to copy 1 by 1.
Defining char a[];, well I don't think that's even possible, is it? You are asking for a to be an array of a size that would be determined when initialized. When there is no initialization, it's just doesn't make sense.
char *a;
a = "test";
You are defining a as a pointer to arrays of char. When you assign it to "test", a just points to it, it doesn't have any specific memory for it though, exactly like the case of char *a = "test";
Like I said, assigning arrays (whether null-terminated arrays of char (string) or any other array) is a non-trivial task that the compiler doesn't do for you, that is why you have functions for it.
Do not confuse assignment and initialisation in C, they are different.
In C a string is not a data type, it is a convention, utilising an array and a nul terminator. Like any array, when you assign it, it's name resolves as a mere pointer. You can assign a pointer, but that is not the same as assigning a string.
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