Declare and initialize pointer concisely (i. e. pointer to int)

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String literals are a corner case : they trigger the creation of the literal in static memory, and its access as a char array. Note that the following doesn’t compile, despite 42 being an int literal, because it is not implicitly allocated :

int *p = &42;

In all other cases, you are responsible of allocating the pointed object, be it in automatic or dynamic memory.

int i = 42;
int *p = &i;

Here i is an automatic variable, and p points to it.

int * i;
*i = 42;

You just invoked Undefined Behaviour. i has not been initialized, and is therefore pointing somewhere at random in memory. Then you assigned 42 to this random location, with unpredictable consequences. Bad.

int *i = malloc(sizeof *i);

Here i is initialized to point to a dynamically-allocated block of memory. Don’t forget to free(i) once you’re done with it.

int i = 42, *p = &i;

And here is how you create an automatic variable and a pointer to it as a one-liner. i is the variable, p points to it.

Edit : seems like you really want that variable to be implicitly and anonymously allocated. Well, here’s how you can do it :

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

This thingy is a compound literal. They are anonymous instances with automatic storage duration (or static at file scope), and only exist in C90 and further (but not C++ !). As opposed to string literals, compound literals are mutable, i.e you can modify *p.

Edit 2 : Adding this solution inspired from another answer (which unfortunately provided a wrong explanation) for completeness :

int i[] = {42};

This will allocate a one-element mutable array with automatic storage duration. The name of the array, while not a pointer itself, will decay to a pointer as needed.

Note however that sizeof i will return the “wrong” result, that is the actual size of the array (1 * sizeof(int)) instead of the size of a pointer (sizeof(int*)). That should however rarely be an issue.

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