warning: assignment makes integer from pointer without a cast

When I declare a char * to a fixed开发者_如何学Go string and reuse the pointer to point to another string

/* initial declaration */
char *src = "abcdefghijklmnop";
.....

/* I get the   "warning: assignment makes integer from pointer without a cast" */
*src ="anotherstring";

I tried to recast the pointer but no success.

The expression *src refers to the first character in the string, not the whole string. To reassign src to point to a different string tgt, use src = tgt;.

When you write the statement

*src = "anotherstring";

the compiler sees the constant string "abcdefghijklmnop" like an array. Imagine you had written the following code instead:

char otherstring[14] = "anotherstring";
...
*src = otherstring;

Now, it's a bit clearer what is going on. The left-hand side, *src, refers to a char (since src is of type pointer-to-char) whereas the right-hand side, otherstring, refers to a pointer.

This isn't strictly forbidden because you may want to store the address that a pointer points to. However, an explicit cast is normally used in that case (which isn't too common of a case). The compiler is throwing up a red flag because your code is likely not doing what you think it is.

It appears to me that you are trying to assign a string. Strings in C aren't data types like they are in C++ and are instead implemented with char arrays. You can't directly assign values to a string like you are trying to do. Instead, you need to use functions like strncpy and friends from <string.h> and use char arrays instead of char pointers. If you merely want the pointer to point to a different static string, then drop the *.

The warning comes from the fact that you're dereferencing src in the assignment. The expression *src has type char, which is an integral type. The expression "anotherstring" has type char [14], which in this particular context is implicitly converted to type char *, and its value is the address of the first character in the array. So, you wind up trying to assign a pointer value to an integral type, hence the warning. Drop the * from *src, and it should work as expected:

src = "anotherstring";

since the type of src is char *.

What Jeremiah said, plus the compiler issues the warning because the production:

*src ="anotherstring";

says: take the address of "anotherstring" -- "anotherstring" IS a char pointer -- and store that pointer indirect through src (*src = ... ) into the first char of the string "abcdef..." The warning might be baffling because there is nowhere in your code any mention of any integer: the warning seems nonsensical. But, out of sight behind the curtain, is the rule that "int" and "char" are synonymous in terms of storage: both occupy the same number of bits. The compiler doesn't differentiate when it issues the warning that you are storing into an integer. Which, BTW, is perfectly OK and legal but probably not exactly what you want in this code.

-- pete

structure of node:-

struct node {
  char key[10];
  struct node *left, *right;
};

Also it may arrive when you are doing something like this

struct node *newNode(char item[10]) {
  struct node *temp = (struct node *)malloc(sizeof(struct node));
  temp->key[10] = item;
  temp->left = temp->right = NULL;
  return temp;
}

It will show warning in line 3 and all of your code will be messed up. Good news is:-

you have to prove that it is an array and it can be simply done by

temp->key[10] = item[10];

As simple as that...