c string basics, why unassigned?

Question

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

Answer 1

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";

Answer 2

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 :)

Answer 3

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.

Answer 4

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.

Answer 5

To assign a String Literal to the str Array you can use a the String copy function strcpy.

Answer 6

char a[100] = "\x80\xbb\x00\xcd"; OR char a[] = "\x80\xbb\x00\xcd";

Answer 7

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).