In C, why can't an integer value be assigned t

Question

I've been looking through the site but haven't found an answer to this one yet.

It is easiest (for me at least) to explain this question with an example.

I don't understand why this is valid:

#include <stdio.h>

int main(int argc, char* argv[])
{
  char *mystr = "hello";
}

But this produces a compiler warning ("initialization makes pointer from integer without a cast"):

#include <stdio.h>

int main(int argc, char* argv[])
{
  int *myint = 5;
}

My understanding of the first program is that creates a variable called mystr of type pointer-to-char, the value of which is the address of the first char ('h') of the string literal "hello". In other words with this initialization you not only get the pointer, but also define the object ("hello" in this case) which the pointer points to.

Why, then, does int *myint = 5; seemingly not achieve something analogous to this, i.e. create a variable called myint of type pointer-to-int, the value of which is the address of the value '5'? Why doesn't this initialization both give me the pointer and also define the object which the pointer points to?

Answer 1

The problem is that you are trying to assign the address 5 to the pointer. Here you are not dereferencing the pointer, you are declaring it as a pointer and initializing it to the value 5 (as an address which surely is not what you intend to do). You could do the following.

#include <stdio.h>

int main(int argc, char* argv[])
{
  int *myint, b;
  b = 5;
  myint = &b;
}

Answer 2

In fact, you can do so using a compound literal, a feature added to the language by the 1999 ISO C standard.

A string literal is of type char[N], where N is the length of the string plus 1. Like any array expression, it's implicitly converted, in most but not all contexts, to a pointer to the array's first element. So this:

char *mystr = "hello";

assigns to the pointer mystr the address of the initial element of an array whose contents are "hello" (followed by a terminating '\0' null character). Incidentally, it's safer to write:

const char *mystr = "hello";

There are no such implicit conversions for integers -- but you can do this:

int *ptr = &(int){42};

(int){42} is a compound literal, which creates an anonymous int object initialized to 42; & takes the address of that object.

But be careful: The array created by a string literal always has static storage duration, but the object created by a compound literal can have either static or automatic storage duration, depending on where it appears. That means that if the value of ptr is returned from a function, the object with the value 42 will cease to exist while the pointer still points to it.

As for:

int *myint = 5;

that attempts to assign the value 5 to an object of type int*. (Strictly speaking it's an initialization rather than an assignment, but the effect is the same). Since there's no implicit conversion from int to int* (other than the special case of 0 being treated as a null pointer constant), this is invalid.

Answer 3

"In C, why can't an integer value be assigned to an int* the same way a string value can be assigned to a char*?"

Because it's not even a similar situation, let alone "the same way".

A string literal is an array of chars which – being an array – can be implicitly converted to a pointer to its first element. Said pointer is a char *.

But an int is not either a pointer in itself, nor an array, nor anything else implicitly convertible to a pointer. These two scenarios just don't have anything in common.

Answer 4

When you do char* mystr = "foo";, the compiler will create the string "foo" in a special read-only portion of your executable, and effectively rewrite the statement as char* mystr = address_of_that_string;

The same is not implemented for any other type, including integers. int* myint = 5; will set myint to point to address 5.

Answer 5

The first program is actually creating a pointer to a string which is an array of chars. The array would actually be a pointer itself in this case. In your second program you're assigning a scalar integer value to a pointer.

You should try this instead:

int main(int argc, char* argv[])
{
    int temp = 5;
    int * my_int_pointer = &temp;
}

Answer 6

i'll split my answer to two parts:

1st, why char* str = "hello"; is valid:

char* str declare a space for a pointer (number that represents a memory address on the current architecture)

when you write "hello" you actually fill the stack with 6 bytes of data
(don't forget the null termination) lets say at address 0x1000 - 0x1005.

str="hello" assigns the start address of that 5 bytes (0x1000) to the *str

so what we have is :
1. str, which takes 4 bytes in memory, holds the number 0x1000 (points to the first char only!)
2. 6 bytes 'h' 'e' 'l' 'l' 'o' '\0'

2st, why int* ptr = 0x105A4DD9; isn't valid:

well, this is not entirely true!
as said before, a Pointer is a number that represent an address,
so why cant i assign that number ?

it is not common because mostly you extract addresses of data and not enter the address manually.
but you can if you need !!!...
because it isn't something that is commonly done,
the compiler want to make sure you do so in propose, and not by mistake and forces you to CAST your data as
int* ptr = (int*)0x105A4DD9;
(used mostly for Memory mapped hardware resources)

Hope this clear things out.
Cheers