This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
What is the difference between char s[] and char *s in C?
Difference between char *str = “…” and char str[N] = “…”?
I have some code that has had me puzzled.
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[])
{
char* string1 = "this is a test";
char string2[] = "this is a test";
printf("%i, %i\n", sizeof(string1), sizeof(string2));
system("PAUSE");
return 0;
}
When it outputs the size of string1, it prints 4, which is to be expected because the size of a pointer is 4 bytes. But when it prints string2, it outputs 15. I thought that an array was a pointer, so the size of string2 should be the same as string1 right? So why is it that it prints out two different sizes for the same type of data (pointer)?
Arrays are not pointers. Array names decay to pointers to the first element of the array in certain situations: when you pass it to a function, when you assign it to a pointer, etc. But otherwise arrays are arrays - they exist on the stack, have compile-time sizes that can be determined with sizeof, and all that other good stuff.
Arrays and pointers are completely different animals. In most contexts, an expression designating an array is treated as a pointer.
First, a little standard language (n1256):
6.3.2.1 Lvalues, arrays, and function designators
...
3 Except when it is the operand of the sizeof operator or the unary & operator, or is a string literal used to initialize an array, an expression that has type "array of type" is converted to an expression with type "pointer to type" that points to the initial element of the array object and is not an lvalue. If the array object has register storage class, the behavior is undefined.
The string literal "this is a test" is a 15-element array of char. In the declaration
char *string1 = "this is a test";
string1 is being declared as a pointer to char. Per the language above, the type of the expression "this is a test" is converted from char [15] to char *, and the resulting pointer value is assigned to string1.
In the declaration
char string2[] = "this is a test";
something different happens. More standard language:
6.7.8 Initialization
...
14 An array of character type may be initialized by a character string literal, optionally
enclosed in braces. Successive characters of the character string literal (including the
terminating null character if there is room or if the array is of unknown size) initialize the elements of the array.
...
22 If an array of unknown size is initialized, its size is determined by the largest indexed element with an explicit initializer. At the end of its initializer list, the array no longer has incomplete type.
In this case, string2 is being declared as an array of char, its size is computed from the length of the initializer, and the contents of the string literal are copied to the array.
Here's a hypothetical memory map to illustrate what's happening:
Item Address 0x00 0x01 0x02 0x03
---- ------- ---- ---- ---- ----
no name 0x08001230 't' 'h' 'i' 's'
0x08001234 ' ' 'i' 's' ' '
0x08001238 'a' ' ' 't' 'e'
0x0800123C 's' 't' 0
...
string1 0x12340000 0x08 0x00 0x12 0x30
string2 0x12340004 't' 'h' 'i' 's'
0x12340008 ' ' 'i' 's' ' '
0x1234000C 'a' ' ' 't' 'e'
0x1234000F 's' 't' 0
String literals have static extent; that is, the memory for them is set aside at program startup and held until the program terminates. Attempting to modify the contents of a string literal invokes undefined behavior; the underlying platform may or may not allow it, and the standard places no restrictions on the compiler. It's best to act as though literals are always unwritable.
In my memory map above, the address of the string literal is set off somewhat from the addresses of string1 and string2 to illustrate this.
Anyway, you can see that string1, having a pointer type, contains the address of the string literal. string2, being an array type, contains a copy of the contents of the string literal.
Since the size of string2 is known at compile time, sizeof returns the size (number of bytes) in the array.
The %i conversion specifier is not the right one to use for expressions of type size_t. If you're working in C99, use %zu. In C89, you would use %lu and cast the expression to unsigned long:
C89: printf("%lu, %lu\n", (unsigned long) sizeof string1, (unsigned long) sizeof string2);
C99: printf("%zu, %zu\n", sizeof string1, sizeof string2);
Note that sizeof is an operator, not a function call; when the operand is an expression that denotes an object, parentheses aren't necessary (although they don't hurt).
string1 is a pointer, but string2 is an array.
The second line is something like int a[] = { 1, 2, 3}; which defines a to be a length-3 array (via the initializer).
The size of string2 is 15 because the initializer is nul-terminated (so 15 is the length of the string + 1).
An array of unknown size is equivalent to a pointer for sizeof purposes. An array of static size counts as its own type for sizeof purposes, and sizeof reports the size of the storage required for the array. Even though string2 is allocated without an explicit size, the C compiler treats it magically because of the direct initialization by a quoted string and converts it to an array with static size. (Since the memory isn't allocated in any other way, there's nothing else it can do, after all.) Static size arrays are different types from pointers (or dynamic arrays!) for the purpose of sizeof behavior, because that's just how C is.
This seems to be a decent reference on the behaviors of sizeof.
The compiler know that test2 is an array, so it prints out the number of bytes allocated to it(14 letters plus null terminator). Remember that sizeof is a compiler function, so it can know the size of a stack variable.
array is not pointer. Pointer is a variable pointing to a memory location whereas array is starting point of sequential memory allocated
Its because
string1 holds pointer, where pointer has contiguous chars & its
immutable.
string2 is location where your chars sit.
basically C compiler iterprets these 2 differently. beautifully explained here http://c-faq.com/aryptr/aryptr2.html.
Related
I was testing the use of sizeof() for the same String content "abc". my function is like this:
int main(void){
char* pass1 = "abc";
char pass2[] = "abc";
char pass3[4] = "abc";
char pass4[] = "";
scanf("%s", pass4);
printf("sizeof(pass1) is: %lu\n", sizeof(pass1));
printf("sizeof(pass2) is: %lu\n", sizeof(pass2));
printf("sizeof(pass3) is: %lu\n", sizeof(pass3));
printf("sizeof(pass4) is: %lu\n", sizeof(pass4));
return 0;
}
I input "abc" for pass4, the output is like this:
sizeof(pass1) is: 8
sizeof(pass2) is: 4
sizeof(pass3) is: 4
sizeof(pass4) is: 1
I was expecting all 4s. I thought the 4 above string definitions are the same.
why sizeof(pass1) returns 8? Why sizeof(pass4) is 1?
When you take sizeof on a pointer type, you'll get the size in bytes of the memory address. In this case 8 is the size of the address (in bytes). sizeof on statically allocated read only strings in C will return the actual size in bytes of the string including the null byte.
sizeof gives the size of its operand. To understand the results you are seeing, you need to understand what pass1, pass2, pass3, and pass4 actually are.
pass1 is a pointer to char (i.e. a char *) so sizeof pass1 gives the size of a pointer (a variable which contains a memory address, not an array). That is 8 with your compiler. The size of a pointer is implementation defined, so this may give different results with different compilers. The fact you have initialised pass1 so it points at the first character of a string literal "abc" does not change the fact that pass1 is declared as a pointer, not an array.
pass2 is an array initialised using the literal "abc" which - by convention - is represented in C using an array of four characters (the three letters 'a' to 'c', plus an additional character with value zero ('\0').
pass3 is also an array of four char, since it is declared that way char pass3[4] = <etc>. If you had done char pass3[4] = "abcdef", you would still find that sizeof pass3 is 4 (and the 4 elements of pass3 will be 'a' to 'd' (with other character 'e', 'f', and '\0' in the string literal "abcdef" not used to initialise pass3).
Since both pass2 and pass3 are arrays of four characters, their size is 4 (in general, the size of an array is the size of the array element multiplied by number of elements). The standard defines sizeof char to be 1, and 1*4 has a value 4.
pass4 is initialised using the literal "". That string literal is represented using a single char with value '\0' (and no characters before it, since none are between the double quotes). So pass4 has size 1 for the same reason that pass2 has size 4.
In this declaration
char* pass1 = "abc";
in the right side used as an initializer there is a character array that has size equal to 4. String literals have types of character arrays. Take into account that string literals include the terminating zero. You can check this the following way
printf("sizeof( \"abc\") is: %zu\n", sizeof( "abc"));
In the declaration the array is used to initialize a pointer. Used as initializer of a pointer the array is implicitly converted to the pointer to its first element.
Thus the pointer pass1 points to the first element of the string literal "abc". The size of the pointer itself in your system is equal to 8 bytes.
In these declerations
char pass2[] = "abc";
char pass3[4] = "abc";
the string literal is used to initialize arrays. In this case each element of the arrays is initialized by the corresponding element of the string literal. All other elements of the arrays are zero initialized. If the size of an array is not specified then it is calculated from the number of initializers.
So in this declaration
char pass2[] = "abc";
the array pass2 will have 4 elements because the sgtring literal provides four initializers.
In this declaration
char pass3[4] = "abc";
there is explicitly specified that the array has 4 elements.
Thus the both arrays has size equal to 4 and the pointer declared first has size of 8 bytes.
I have a two part question:
Understand output from sizeof
Understand how strings are stored in variables (e.g. bits and ram)
Question 1
I'm trying to understand the output from the following piece of C code.
printf("a: %ld\n", sizeof("a")); // 2
printf("abc: %ld\n", sizeof("abc")); // 4
It always seems to be one larger than the actual number of characters specified.
The docs suggest that the returned value represents the size of the object (in this case a string) in bytes. So if the size of a gives us back 2 bytes, then I'm curious how a represents 16 bits of information.
If I look at the binary representation of the ASCII character a I can see it is 01100001. But that's only showing 3 bits out of 1 byte being used.
Question 2
Also, how do large strings get stored into a variable in C? Am I right in thinking that they have to be stored within an array, like so:
char my_string[5] = "hello";
Interestingly when I have some code like:
char my_string = "hello";
printf("my_string: %s\n", my_string);
I get two compiler errors:
- incompatible pointer to integer conversion initializing 'char' with an expression of type 'char [6]'
- format specifies type 'char *' but the argument has type 'char'
...which I don't understand. Firstly it states the type is presumed to be a size of [6] when there's only 5 characters. Secondly the mention of a pointer here seems odd to me? Why does printf expect a pointer and why does not specifying the length of the variable/array result in a pointer to integer error?
By the way I seemingly can set the length of the variable/array to 5 rather than 6 and it'll work as I'd expect it to char my_string[5] = "hello";.
I'm probably just missing something very basic/fundamental about how bits and strings work in C.
Any help understanding this would be appreciated.
The first part of the question is due to the way strings are stored in C. Strings in C are nothing more than a series of characters (char) with a \0 added at the end, which is the reason you're seeing a +1 when you do sizeof. Notice in your second part if you were to say char my_string[4] = "hello"; you'd also get a compiler error saying there wasn't enough size for this string. That's also related to this.
Now onto the second part, strings themselves are a series of characters. However, you don't store every character by themselves in a variable. You instead have a pointer to these series of characters that will allow you to access them from some part of memory. Additional information regarding pointers and strings in C can be found here: Pointer to a String in C
In C, a string is a sequence of character values followed by a zero valued terminator. For example, the string "hello" is the sequence of character values {'h', 'e', 'l', 'l', 'o', 0 }1. Strings (including string literals) are stored as arrays of char (or wchar_t for wide-character strings). To account for the terminator, the size of the array must always be one greater than the number of characters in the string:
char greeting[6] = "hello";
The storage for greeting will look like
+---+
greeting: |'h'| greeting[0]
+---+
|'e'| greeting[1]
+---+
|'l'| greeting[2]
+---+
|'l'| greeting[3]
+---+
|'o'| greeting[4]
+---+
| 0 | greeting[5]
+---+
Storage for a string literal is largely the same2:
+---+
"hello": |'h'| "hello"[0]
+---+
|'e'| "hello"[1]
+---+
|'l'| "hello"[2]
+---+
|'l'| "hello"[3]
+---+
|'o'| "hello"[4]
+---+
| 0 | "hello"[5]
+---+
Yes, you can apply the subscript operator [] to a string literal just like any other array expression.
Except when it is the operand of the sizeof or unary & operators, or is a string literal used to initialize a character array in a declaration, an expression of type "N-element of T" will be converted ("decay") to an expression of type "pointer to T", and the value of the expression will be the address of the first element of the array. So, the string literal "hello" is an expression of type "6-element array of char". If I pass that literal as an argument to a function like
printf( "%s\n", "hello" );
then both of the string literal expressions "%s" and "hello" are converted from "4-element array of char"3 and "6-element array of char" to "pointer to char", so what printf receives are pointer values, not array values.
You've already seen two exceptions to the conversion rule. You saw it in your code when you used the sizeof operator and got a value one more than you expected. sizeof evaluates to the number of bytes required to store the operand. Because of the zero terminator, it takes N+1 bytes to store an N-character string.
The second exception is the declaration of the greeting array above; since I'm using the string literal to initialize the array, the literal is not converted to a pointer value first. Note the you can write that declaration as
char greeting[] = "hello";
In that case, the size of the array is taken from the size of the initializer.
The third exception occurs when the array expression is the operand of the unary & operator. Instead of evaluating to a pointer to a pointer to char (char **), the expression &greeting evaluates to type "pointer to 6-element array of char", or char (*)[6].
The length of a string is the number of characters before to zero terminator. All the standard library functions that deal with strings expect to see that terminator. The size of the array to store that string must be at least one greater than the maximum length of the string you intend to store.
Sometimes you'll see people write '\0' instead of a naked 0 to represent a string terminator; they mean the same thing.
Storage for string literals is allocated at program startup and held until the program terminates. String literals may be stored in a read-only memory segment; attempting to modify the contents of a string literal results in undefined behavior.
'\n' counts as a single character.
This question already has answers here:
How to declare strings in C [duplicate]
(4 answers)
Closed 8 years ago.
A few weeks ago I started learning the programming language C. I have knowledge in web technologies like HMTL/CSS, Javscript, PHP, and basic server administration, but C is confusing me. To my understanding, the C language does not have a data type for strings, just characters, however I may be wrong.
I have heard there are two ways of declaring a string. What is the difference between these two lines of declaring a string:
a.) char stringName[];
b.) char *stringName;
I get that char stringName[]; is an array of characters. However, the second line confuses me. To my understanding the second line makes a pointer variable. Aren't pointer variables supposed to be the memory address of another variable?
In the C language, a "string" is, as you say, an array of char. Most string functions built into the C spec expect the string to be "NUL terminated", meaning the last char of the string is a 0. Not the code representing the numeral zero, but the actual value of 0.
For example, if you're platform uses ASCII, then the following "string" is "ABC":
char myString[4] = {65, 66, 67, 0};
When you use the char varName[] = "foo" syntax, you're allocating the string on the stack (or if its in a global space, you're allocating it globally, but not dynamically.)
Memory management in C is more manual than in many other langauges you may have experience with. In particular, there is the concept of a "pointer".
char *myString = "ABC"; /* Points to a string somewhere in memory, the compiler puts somewhere. */
Now, a char * is "an address that points to a char or char array". Notice the "or" in that statement, it is important for you, the programmer, to know what the case is.
It's important to also ensure that any string operations you perform don't exceed the amount of memory you've allocated to a pointer.
char myString[5];
strcpy(myString, "12345"); /* copy "12345" into myString.
* On no! I've forgot space for my nul terminator and
* have overwritten some memory I don't own. */
"12345" is actually 6 characters long (don't forget the 0 at the end), but I've only reserved 5 characters. This is what's called a "buffer overflow", and is the cause of many serious bugs.
The other difference between "[]" and "*", is that one is creating an array (as you guessed). The other one is not reserving any space (other than the space to hold the pointer itself.) That means that until you point it somewhere that you know is valid, the value of the pointer should not be used, for either reading or writing.
Another point (made by someone in the comment)
You cannot pass an array as a parameter to a function in C. When you try, it gets converted to a pointer automatically. This is why we pass around pointers to strings rather than the strings themselves
In C, a string is a sequence of character values followed by a 0-valued byte1 . All the library functions that deal with strings use the 0 terminator to identify the end of the string. Strings are stored as arrays of char, but not all arrays of char contain strings.
For example, the string "hello" is represented as the character sequence {'h', 'e', 'l', 'l', 'o', 0}2 To store the string, you need a 6-element array of char - 5 characters plus the 0 terminator:
char greeting[6] = "hello";
or
char greeting[] = "hello";
In the second case, the size of the array is computed from the size of the string used to initialize it (counting the 0 terminator). In both cases, you're creating a 6-element array of char and copying the contents of the string literal to it. Unless the array is declared at file scope (oustide of any function) or with the static keyword, it only exists for the duration of the block in which is was declared.
The string literal "hello" is also stored in a 6-element array of char, but it's stored in such a way that it is allocated when the program is loaded into memory and held until the program terminates3, and is visible throughout the program. When you write
char *greeting = "hello";
you are assigning the address of the first element of the array that contains the string literal to the pointer variable greeting.
As always, a picture is worth a thousand words. Here's a simple little program:
#include <string.h>
#include <stdio.h>
#include <ctype.h>
int main( void )
{
char greeting[] = "hello"; // greeting contains a *copy* of the string "hello";
// size is taken from the length of the string plus the
// 0 terminator
char *greetingPtr = "hello"; // greetingPtr contains the *address* of the
// string literal "hello"
printf( "size of greeting array: %zu\n", sizeof greeting );
printf( "length of greeting string: %zu\n", strlen( greeting ) );
printf( "size of greetingPtr variable: %zu\n", sizeof greetingPtr );
printf( "address of string literal \"hello\": %p\n", (void * ) "hello" );
printf( "address of greeting array: %p\n", (void * ) greeting );
printf( "address of greetingPtr: %p\n", (void * ) &greetingPtr );
printf( "content of greetingPtr: %p\n", (void * ) greetingPtr );
printf( "greeting: %s\n", greeting );
printf( "greetingPtr: %s\n", greetingPtr );
return 0;
}
And here's the output:
size of greeting array: 6
length of greeting string: 5
size of greetingPtr variable: 8
address of string literal "hello": 0x4007f8
address of greeting array: 0x7fff59079cf0
address of greetingPtr: 0x7fff59079ce8
content of greetingPtr: 0x4007f8
greeting: hello
greetingPtr: hello
Note the difference between sizeof and strlen - strlen counts all the characters up to (but not including) the 0 terminator.
So here's what things look like in memory:
Item Address 0x00 0x01 0x02 0x03
---- ------- ---- ---- ---- ----
"hello" 0x4007f8 'h' 'e' 'l' 'l'
0x4007fc 'o' 0x00 ??? ???
...
greetingPtr 0x7fff59079ce8 0x00 0x00 0x00 0x00
0x7fff59879cec 0x00 0x40 0x7f 0xf8
greeting 0x7fff59079cf0 'h' 'e' 'l' 'l'
0x7fff59079cf4 'o' 0x00 ??? ???
The string literal "hello" is stored at a vary low address (on my system, this corresponds to the .rodata section of the executable, which is for static, constant data). The variables greeting and greetingPtr are stored at much higher addresses, corresponding to the stack on my system. As you can see, greetingPtr stores the address of the string literal "hello", while greeting stores a copy of the string contents.
Here's where things can get kind of confusing. Let's look at the following print statements:
printf( "greeting: %s\n", greeting );
printf( "greetingPtr: %s\n", greetingPtr );
greeting is a 6-element array of char, and greetingPtr is a pointer to char, yet we're passing them both to printf in exactly the same way, and the string is being printed out correctly; how can that work?
Unless it is the operand of the sizeof or unary & operators, or is a string literal used to initialize another array in a declaration, an expression of type "N-element array of T" will be converted ("decay") to an expression of type "pointer to T", and the value of the expression will be the address of the first element of the array.
In the printf call, the expression greeting has type "6-element array of char"; since it isn't the operand of the sizeof or unary & operators, it is converted ("decays") to an expression of type "pointer to char" (char *), and the address of the first element is actually passed to printf. IOW, it behaves exactly like the greetingPtr expression in the next printf call4.
The %s conversion specifer tells printf that its corresponding argument has type char *, and that it it should print out the character values starting from that address until it sees the 0 terminator.
Hope that helps a bit.
1. Often referred to as the NUL terminator; this should not be confused with the NULL pointer constant, which is also 0-valued but used in a different context.
2. You'll also see the terminating 0-valued byte written as '\0'. The leading backslash "escapes" the value, so instead of being treated as the character '0' (ASCII 48), it's treated as the value 0 (ASCII 0)).
3. In practice, space is set aside for it in the generated binary file, often in a section marked read-only; attempting to modify the contents of a string literal invokes undefined behavior.
4. This is also why the declaration of greeting copies the string contents to the array, while the declaration of greetingPtr copies the address of the first element of the string. The string literal "hello" is also an array expression. In the first declaration, since it's being used to initialize another array in a declaration, the contents of the array are copied. In the second declaration, the target is a pointer, not an array, so the expression is converted from an array type to a pointer type, and the resulting pointer value is copied to the variable.
In C (and in C++), arrays and pointers are represented similarly; an array is represented by the address of the first element in the array (which is sufficient to gain access to the other elements, since elements are contiguous in memory within an array). This also means that an array does not, by itself, indicate where it ends, and thus you need some way of identifying the end of the array, either by passing around the length as a separate variable or by using some convention (such as that there is a sentinel value that is placed in the last position of the array to indicate the end of the array). For strings, the latter is the common convention, with '\0' (the NUL character) indicating the end of the string.
#include <stdio.h>
main()
{
char * ptr;
ptr = "hello";
printf("%p %s" ,"hello",ptr );
getchar();
}
Hi, I am trying to understand clearly how can arrays get assign in to pointers. I notice when you assign an array of chars to a pointer of chars ptr="hello"; the array decays to the pointer, but in this case I am assigning a char of arrays that are not inside a variable and not a variable containing them ", does this way of assignment take a memory address specially for "Hello" (what obviously is happening) , and is it possible to modify the value of each element in "Hello" wich are contained in the memory address where this array is stored. As a comparison, is it fine for me to assign a pointer with an array for example of ints something as vague as thisint_ptr = 5,3,4,3; and the values 5,3,4,3 get located in a memory address as "Hello" did. And if not why is it possible only with strings? Thanks in advanced.
"hello" is a string literal. It is a nameless non-modifiable object of type char [6]. It is an array, and it behaves the same way any other array does. The fact that it is nameless does not really change anything. You can use it with [] operator for example, as in "hello"[3] and so on. Just like any other array, it can and will decay to pointer in most contexts.
You cannot modify the contents of a string literal because it is non-modifiable by definition. It can be physically stored in read-only memory. It can overlap other string literals, if they contain common sub-sequences of characters.
Similar functionality exists for other array types through compound literal syntax
int *p = (int []) { 1, 2, 3, 4, 5 };
In this case the right-hand side is a nameless object of type int [5], which decays to int * pointer. Compound literals are modifiable though, meaning that you can do p[3] = 8 and thus replace 4 with 8.
You can also use compound literal syntax with char arrays and do
char *p = (char []) { "hello" };
In this case the right-hand side is a modifiable nameless object of type char [6].
The first thing you should do is read section 6 of the comp.lang.c FAQ.
The string literal "hello" is an expression of type char[6] (5 characters for "hello" plus one for the terminating '\0'). It refers to an anonymous array object with static storage duration, initialized at program startup to contain those 6 character values.
In most contexts, an expression of array type is implicitly converted a pointer to the first element of the array; the exceptions are:
When it's the argument of sizeof (sizeof "hello" yields 6, not the size of a pointer);
When it's the argument of _Alignof (a new feature in C11);
When it's the argument of unary & (&arr yields the address of the entire array, not of its first element; same memory location, different type); and
When it's a string literal in an initializer used to initialize an array object (char s[6] = "hello"; copies the whole array, not just a pointer).
None of these exceptions apply to your code:
char *ptr;
ptr = "hello";
So the expression "hello" is converted to ("decays" to) a pointer to the first element ('h') of that anonymous array object I mentioned above.
So *ptr == 'h', and you can advance ptr through memory to access the other characters: 'e', 'l', 'l', 'o', and '\0'. This is what printf() does when you give it a "%s" format.
That anonymous array object, associated with the string literal, is read-only, but not const. What that means is that any attempt to modify that array, or any of its elements, has undefined behavior (because the standard explicitly says so) -- but the compiler won't necessarily warn you about it. (C++ makes string literals const; doing the same thing in C would have broken existing code that was written before const was added to the language.) So no, you can't modify the elements of "hello" -- or at least you shouldn't try. And to make the compiler warn you if you try, you should declare the pointer as const:
const char *ptr; /* pointer to const char, not const pointer to char */
ptr = "hello";
(gcc has an option, -Wwrite-strings, that causes it to treat string literals as const. This will cause it to warn about some C code that's legal as far as the standard is concerned, but such code should probably be modified to use const.)
#include <stdio.h>
main()
{
char * ptr;
ptr = "hello";
//instead of above tow lines you can write char *ptr = "hello"
printf("%p %s" ,"hello",ptr );
getchar();
}
Here you have assigned string literal "hello" to ptr it means string literal is stored in read only memory so you can't modify it. If you declare char ptr[] = "hello";, then you can modify the array.
Say what?
Your code allocates 6 bytes of memory and initializes it with the values 'h', 'e', 'l', 'l', 'o', and '\0'.
It then allocates a pointer (number of bytes for the pointer depends on implementation) and sets the pointer's value to the start of the 5 bytes mentioned previously.
You can modify the values of an array using syntax such as ptr[1] = 'a'.
Syntactically, strings are a special case. Since C doesn't have a specific string type to speak of, it does offer some shortcuts to declaring them and such. But you can easily create the same type of structure as you did for a string using int, even if the syntax must be a bit different.
I am trying to learn the basics, I would think that declaring a char[] and assigning a string to it would work.
thanks
int size = 100;
char str[size];
str = "\x80\xbb\x00\xcd";
gives error "incompatible types in assignment". what's wrong?
thanks
You can use a string literal to initialize an array of char, but you can't assign an array of char (any more than you can assign any other array). OTOH, you can assign a pointer, so the following would be allowed:
char *str;
str = "\x80\xbb\x00\xcd";
This is actually one of the most difficult parts of learning a programming language.... str is an array, that is, a part of memory (size times a char, so size chars) that has been reserved and labeled as str. str[0] is the first character, str[1] the second... str[size-1] is the last one. str itself, without specifiying any character, is a pointer to the memory zone that was created when you did
char str[size]
As Jerry so clearly said, in C you can not initialize arrays that way. You need to copy from one array to other, so you can do something like this
strncpy(str, "\x80\xbb\x00\xcd", size); /* Copy up to size characters */
str[size-1]='\0'; /* Make sure that the string is null terminated for small values of size */
Summarizing: It's very important to make a difference between pointers, memory areas and array.
Good luck - I am pretty sure that in less time than you imagine you will be mastering these concepts :)
A char-array can be implicitely cast to a char* when used as Rvalue, but not when used as Lvalue - that's why the assignment won't work.
You cannot assign array contents using the =operator. That's just a fact of the C language design. You can initialize an array in the declaration, such as
char str[size] = "\x80\xbb\x00\xcd";
but that's a different operation from an assignment. And note that in this case, and extra '\0' will be added to the end of the string.
The "incompatible types" warning comes from how array expressions are treated by the language. First of all, string literals are stored as arrays of char with static extent (meaning they exist over the lifetime of the program). So the type of the string literal "\x80\xbb\x00\xcd" is "4 5-element array of char". However, in most circumstances, an expression of array type will implicitly be converted ("decay") from type "N-element array of T" to "pointer to T", and the value of the expression will be the address of the first element in the array. So, when you wrote the statement
str = "\x80\xbb\x00\xcd";
the type of the literal was implicitly converted from "4 5-element array of char" to "pointer to char", but the target of the assignment is type "100-element array of char", and the types are not compatible (above and beyond the fact that an array expression cannot be the target of the = operator).
To copy the contents of one array to another you would have to use a library function like memcpy, memmove, strcpy, etc. Also, for strcpy to function properly, the source string must be 0-terminated.
Edit per R's comment below, I've struck out the more dumbass sections of my answer.
To assign a String Literal to the str Array you can use a the String copy function strcpy.
char a[100] = "\x80\xbb\x00\xcd"; OR char a[] = "\x80\xbb\x00\xcd";
str is the name of an array. The name of an array is the address of the 0th element. Therefore, str is a pointer constant. You cannot change the value of a pointer constant, just like you cannot change a constant (you can't do 6 = 5, for example).