I am learning C++ from Primer 5th edition and I am at Returning a Pointer to an Array. The declaration of this function is:
int (*func(int i))[10];
Read: What does sizeof(&array) return? to understand diffrence between array name and address of array.
Q1 I want to know the difference between:
In your code:
int *func(){
static int a[]={1,2,3};
return a;
}
you are returning address of first element. Actually type of a is int[3] that decays into int*. Important is
You stores address into int* p and can assess elements of array as p[i].
Whereas if your function would be int int (*func())[3] then you return &a, and assign to int(*p)[3] and can access (*p)[i].
Note: type of &a is int(*)[3].
Q2 How i can make this function call work, because in the book, there isn't any concrete example.
like:
int (*func())[3]{
static int a[]={1,2,3};
return &a;
}
And main():
int main(){
int i=0;
int(*p)[3] = func();
for(i=0; i<3; i++)
printf(" %d\n", (*p)[i]);
return 0;
}
You can check second version of code working id Ideone
Q1 I want to know the difference between:
As you are interested to know diffrence between two so now compare two different declarations of p in two versions of code:
1) : int* p; and we access array elements as p[i] that is equals to *(p + i).
2) : int (*p)[i] and we access array elements as (*p)[i] that is equals to *((*p) + i) or just = *(*p + i). ( I added () around *p to access array element because precedence of [] operator is higher then * So simple *p[i] means defense to the array elements).
Edit:
An addition information other then return type:
In both kind of functions we returns address that is of a static variable (array), and a static object life is till program not terminates. So access the array outsize func() is not a problem.
Consider if you returns address of simple array (or variable) that is not static (and dynamically allocated) then it introduce as Undefined behavior in your code that can crash.