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int *p; // *p has type int
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Here *p is a declarator, which is the part of a definition that defines a single variable Even though the * and the p are part of a single declarator, most C++ programmers write this definition as
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int* p; // p has type int*
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to emphasize the notion that p has a particular type (ie, int*) These two usages are equivalent because spaces around the * are neutral However, the latter usage conceals a pitfall that is so important that it deserves special
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int* p, q; // What does this definition mean
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defines p as an object of type "pointer to int" and q as an object of type int This example is easier to understand if we view it this way:
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int *p, q; // *p and q have type int
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or, for that matter, this way:
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int (*p), q; // (*p) and q have type int
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Still better, we can make our intentions crystal clear by writing
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int* p; int q; // *p has type int // q has type int
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We now know enough to write a simple program that uses pointers:
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int main() { int x = 5; // p points to x int* p = &x; cout << "x = " << x << endl; // change the value of x through p *p = 6; cout << "x = " << x << endl; return 0; }
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The output of this program will be
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x = 5 x = 6
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Immediately after we have defined p, the state of our variables is
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It should be no surprise that x is 5 when we execute the first output expression The next statement changes the value of x to 6 by executing *p = 6 Remember, once p contains the address of x, *p and x are two different ways of referring to the same object Thus, the x is 6 when the second output expression is executed It may be useful to think of a pointer to an object as an iterator that refers to the only element of a "container" that contains that object and nothing else
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In 622/113, we saw a program that passed a function as an argument to another function, and noted that there was slightly more going on there than met the eye The truth is that functions are not objects, and there is no way to copy or assign them, or to pass them as arguments directly In particular, there is no way for a program to create or modify a function only the compiler can do that All that a program can ever do with a function is call it or take its address Nevertheless, we can call a function with another function as an argument, as we did when we passed median_analysis as an argument to write_analysis in 622/112 What happens is that the compiler quietly translates such calls so as to use pointers to functions instead of using functions directly Pointers to functions behave similarly to any other pointers Once you have dereferenced such a pointer, however, all you can do with the resulting function is call it or take the function's address yet again Declarators for pointers to functions resemble other declarators For example, just as we wrote
int *p;
to say that *p has type int, thereby implying that p is a pointer, we might write
int (*fp)(int);
to say that if we dereference fp, and call it with an int argument, the result has type int By implication, fp is a pointer to a function that takes an int argument and returns an int result Because all that you can do with a function is to take its address or call it, any use of a function that is not a call is assumed to be taking its address, even without an explicit & Similarly, you can call a pointer to a function without dereferencing the pointer explicitly So, for example, if we have a function whose type matches fp, such as
int next(int n) { return n + 1; }
then we can make fp point to next by writing either of the following two statements:
// these two statements are equivalent fp = &next; fp = next;
Similarly, if we have an int variable named i, we can use fp to call next, and thereby increment i, by writing either
// these two statements are equivalent i = (*fp)(i); i = fp(i);
Finally, if we write a function that looks like it takes another function as a parameter, the compiler quietly translates the parameter to be a pointer to a function instead So, for example, in the write_analysis function in 622/113, the parameter that we wrote as
double analysis(const vector<Student_info>&)