C中的函数指针，数组和左值

Question

Let's suppose that we have the following functions (in C):

int sum(int a, int b){  
   return a+b;
}
int diff(int a, int b){
    return a-b;
}

So we know that we can declare an array of funtion pointers in the following way:

int (*test[2]) (int a, int b);
test[0] = sum;
test[1] = diff;

But the following is also valid (but we use heap allocation):

int (**test) (int a, int b) = malloc( 2*sizeof(*test));
test[0] = sum;
test[1] = diff;

So far so good. Now let remember that to declare a (dynamically allocated) array of two integers we can do:

 int* test  = malloc( 2*sizeof(int));

So why we cannot declare an array of function pointers as

int (*test) (int a, int b) = malloc( 2*sizeof(*test)); ?

Is the reason that as test is the same as *test and **test (and so on), malloc( 2*sizeof(*test)) is returning a pointer to a function pointer and therefore it cannot be assigned to (*test) ?

If this supposition is correct, can you explain in detail why we get the compilation error

error: lvalue required as left operand of assignment

when we try to do

int (*test) (int a, int b) = malloc( 2*sizeof(*test));
test=diff; //<--- This is ok.
test+1 = sum; //<--- This is what gives the error!

Disclaimer: I suppose that this is a basic question and the supposition is correct, but I would like a better explanation to have this kind of thing clear one and for all.

---

Edit:

Notice that this is equivalent to

int (*test) (int a, int b) = malloc( 2*sizeof(*test));
*test=*diff; //<--- This is ok.
*(test+1) = *sum; //<--- This is what gives the error!

as this is somewhat more similar to the case:

int *test = malloc(2*sizeof(*test));
*test = 0;
*(test+1) = 1;

Answer 1

So why we cannot declare an array of function pointers as

 int (*test) (int a, int b) = malloc( 2*sizeof(*test)); 

Because test does not point to a function pointer; it is a function pointer. Thus it cannot point to the first element of an array of function pointers.

If you want an array of function pointers use the previous form:

 int (**test) (int a, int b) = malloc( 2*sizeof(*test)); 

Here, *test has function pointer type and thus test can (and does) point to the first element of an array of function pointers. Further:

 error: lvalue required as left operand of assignment 

when we try to do

  int (*test) (int a, int b) = malloc( 2*sizeof(*test)); test=diff; //<--- This is ok. test+1 = sum; //<--- This is what gives the error! 

No matter what type test has, test+1=anything is always invalid C. test+1 can never be an lvalue. I don't see why you would expect this to work.

GCC is also papering over another bug in your program, sizeof(*test) . Since *test has function type, sizeof(*test) is invalid, but GCC silently assigns it the value 1. This results in allocating too little memory for a function pointer, but it doesn't matter anyway because in the following line you throw away the memory you got from malloc and assign something else to test .