Are `x = &v` and `*x = v` equivalent?

Given the statement:

int v = 7; 

v has some location in memory. Doing:

x = &v;

will "point" x to the memory location of v, and indeed *x will have the value 7.

However, in this statement:

*x = v;

you are storing the value of v at the address pointed at by x. But x is not pointing at a valid memory address, and so this statement invokes undefined behavior.

So to answer your question, no, the 2 statements are not equivalent.

& means address of.

* means value at.

In x = &v the address of v is assigned to x.

In *x = v - the value at x (the value at the address x) is assigned the value v.

These two are very different statements.

Initially x will contain garbage value. Therefore *x will try to dereference an uninitialised address and will result in an undefined behaviour (in most of the cases a segmentation fault) as *x refers to something which is not initialised. Therefore, *x = v assigns the value in v to the location which is pointed by x.

In the x = &v, x will contain the address of v. From this point x contains the address of v, *x will refer to the value in v. Therefore this statement is correct. Therefore, x = &v assigns the address of v as the value of x.

x = &v modifies x. After the operation x will point to v.

*x = v modifies the object pointed by x. In the example, x doesn't point at anything because the pointer is uninitialised. As such, the behaviour is undefined.