Assignment 1 Tips

1. Assignment 1 Tips Advanced OOPCS 440/540Spring 2014Kenneth Chiu

2. structs • struct A { char c; double x;int array[1];}; • Can you do this? • A a1, a2;…a1 = a2; // Assignment.if (a1 == a2) { … }; // Equality.

3. Function Pointers • Syntax is like this: • int foo(double x);int (*fp)(double x) = foo;i = (*fp)(3.14); • Can reassign: • int foo2(double x);int foo3(double x);int foo4(int x);fp = foo2;fp = foo3;fp = 0;fp = foo4; • Can be combined with structs to look like methods: • structFoo { ... int (*foo2)(Foo *, double x);};...// Assuming that the function pointer has been// properly initialized, it can be used as below.f->foo2(f, 3.14); • Why is f being passed to foo2()?

4. String Literal Concatenation • Any problems with the below? • printf(“A long, long, long, long,” “long, long sentence.\n”); • Two string literals right next to each other are automatically concatenated at compile-time. • Handy for long strings.

5. Macros • Can be used for token replacement: • #define PI 3.14159area = PI*r*r; • #define GREETING “Hola!”printf(“%s\n”, GREETING);printf(GREETING); printf(“\n”); • #define FOREVER while (1) {FOREVER printf(“Stuck in a loop...\n”); } • #define TIMES *#define GETS =a GETS 3 TIMES 2; • Can have parameters: • #define alloc(type) (type *) malloc(sizeof(type))structFoo *f = alloc(Foo); • #define SQR(x) x*xarea = PI*SQR(r);a2 = PI*SQR(2+5);

6. Consider: • #define M(t) \ int foo_##t() { \printf(“%s\n”, “A_” #t); \ } • What does M(int) do? • The backslash is a continuation character for long macros. • The ## splices two preprocessor tokens together into one. • The # operator stringifies the following token by putting “” around it. • Recall that two string literals that are right next to each other are automatically joined by the compiler into one string literal. • Result is: • int foo_int() { printf(“%s\n”, “A_int”); }

7. Containers and Iterators • A container holds values. • An iterator is essentially a “pointer” (or “cursor”) into a container. • Every container has a way to obtain an iterator pointing to the first element. • Every container has a special iterator value, the “end” value. It represents one past the last element of the container. • What happens if you decrement the “begin” iterator? How about increment the “end” iterator?

8. Example (assume std): • list<int> l1;l1.push_back(1);l1.push_back(2);for (list<int>::iterator it = l1.begin(); it != l1.end(); ++it) {cout << *it << endl;} • list<MyClass *> l2;l2.push_back(new MyClass(“A”));l2.push_back(new MyClass(“B”));for (list<MyClass *>::iterator it = l2.begin(); it != l2.end(); ++it) {cout << *it << endl;}

9. // C++11list<int> l1{1, 2};for (auto it = l1.begin(); it != l1.end(); ++it) {cout << *it << endl;}// Same as above.for (auto i: l1) {cout << i << endl;}list<MyClass*> l2{new MyClass(“A”), new MyClass(“B”)};for (auto p: l2) {cout << p << endl;}

10. Why Doesn’t End Iterator in the Assignment Point to the Last Element? • Glib answer: That’s the way the C++ standard library does it. • Keep in mind: • Iterators in C++ are modeled after pointers. • Idiomatic iteration over arrays in C/C++: • for (inti = 0; i < n; i++) { a[i] = … ;} • Common way to iterate using pointers: • int *begin = new int[n], *end = begin + n;for (int *p = begin; p < end; p++) { *p = …;} • But we could do: • for (inti = 0; i <= n - 1; i++) { a[i] = …;} • int *begin = new int[n], *end = begin + n – 1;for (int *p = begin; p <= end; p++) { *p = …;}

11. To iterate over a region of length n in the middle: • for (inti = pos; i < pos + n; i++) { …} • for (int *p = pos; p < pos + n; p++) { …} • for (int i = pos; i <= pos + n - 1; i++) { …} • for (int *p = pos; p <= pos + n - 1; p++) { …} • [Show end.cpp]

12. Arrays and String Literals • What does this print? • #include <stdio.h>int main() {  char *p;  char array[20];printf("%zd, %zd, %zd\n",sizeof p, sizeof array, sizeof "hello");}

13. String Literals • What’s wrong with this? • void foo(char *);...foo(“hello”);

14. What’s a Double-Ended Queue?