Does tmp = malloc(sizeof(x)); the same automatically as tmp = malloc(sizeof(x)); *tmp = x;?

More specifically, is malloc instantly initialising my variable or is it just allocating memory and I have to initialise it myself?

CodePudding user response：

No.

The memory returned by malloc isn't initialized.

Quoting cppreference,

Allocates size bytes of uninitialized storage.

CodePudding user response：

No. Even if malloc weren't defined to be uninitialized storage, sizeof(x) has nothing to do with the runtime value of x (it's just a friendly way to find the size of a variable's underlying type). There's no runtime use of the value of x at all in the code you provided.

CodePudding user response：

The operator sizeof yields the size in bytes of its operand.

That is for example if the identifier x denotes a name of a variable of the type int then the expression sizeof( x ) yields the value 4 provided that for objects of the type int the compiler reserves 4 bytes.

So this call

tmp = malloc(sizeof(x));

will be equivalent to the call

tmp = malloc(sizeof(int));

that in turn is equivalent to

tmp = malloc( 4 );

This statement just tries to allocate dynamically 4 bytes. The allocated memory is uninitialized.

Moreover the expression sizeof( x ) if x is not a variable length array is evaluated at compile time before the program begins its execution.

You could initialize it with zeroes the following way

tmp = calloc( 1, sizeof( x ) );

CodePudding user response：

No, the function malloc() simply allocates memory on the heap. It doesn't initialize the pointer.