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call trap(square,1,2,100) would return 1.41421. The Trapezoidal Rule returns the sum of the areas of the n trapezoids that would approximate the area under the graph of f. For example, if n = 5, then it would return the following, where h = (b a)/5, the width of each h -- [ f ( a ) + 2f ( a + h ) + 2f ( a + 2h ) + 2f ( a + 3h ) + 2f ( a + 4h ) + f ( b ) ] 2 trapezoid.
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7.1 7.2 7.3 Apply the address operator & to the variable &x. Apply the dereference operator * to the variable *p. The declaration int n1=n; defines n1 to be a clone of n; it is a separate object that has the same value as n. The declaration int& n2=n; defines n2 to be a synonym of n; it is the same object as n, with the same address. The declaration int& r = n; declares r to be a reference (alias) for the int variable n. The assignment p = &n; assigns the address of n to the pointer p. The declaration int* q = p; declares q to be a pointer (memory address) pointing to the same int to which p points. The assignment n = *p; assigns to n the int to which p points. a. True: &x == x and &y == y because &x and &y are synonyms for x and y, respectively; so if (x == y) then they all have the same value. b. False: different objects can have the same value, but different objects have different addresses. a. A dangling pointer is a pointer that has not been initialized. It is dangerous because it could be pointing to unallocated memory, or inaccessible memory. b. If a pointer pointing to unallocated memory is dereferenced, it could change the value of some unidentified variable. If a pointer pointing to inaccessible memory is dereferenced, the program will probably crash (i.e., terminate abruptly). c. Initialize pointers when they are declared. You cannot have a reference to a constant; it s address is not accessible. The reference operator & cannot be applied to a constant. The variable p has type char, while the expression &c has type pointer to char. To initialize p to &c, p would have to be declared as type char*. The declaration is invalid because the expression &&n is illegal. The reference operator & can be applied only to objects (variables and class instances). But &n is not an object, it is only a reference. References do not have addresses, so &&n does not exist. Static binding is when memory is allocated at compile time, as with the array declaration: double a[400]; Dynamic binding is when memory is allocated at run time, by means of the new operator: double* p; p = new double[400]; The variable p has type char*, while the expression c has type char. To initialize p to c, p would have the same type as c: either both char or both char*. The only problem is that the array name a is a constant pointer, so it cannot be incremented. The following modified code would be okay: short a[32]; short* p = a; for (int i = 0; i < 32; i++) *p++ = i*i; a. m = 46 b. n = 44 c. &m = 0x3fffd00
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d. *p = 46 e. r = 46 f. *q = 46 a. mutable lvalue; b. not an lvalue; c. immutable lvalue; d. immutable lvalue; e. mutable lvalue; f. immutable lvalue; g. mutable lvalue if return type is a non-local reference; otherwise not an lvalue; h. mutable lvalue; i. mutable lvalue; j. mutable lvalue, unless p points to a constant, in which case *p is an immutable lvalue; k. mutable lvalue; l. immutable lvalue; The pointers p and q have different types: p is pointer to float while q is pointer to short. It is an error to assign the address in one pointer type to a different pointer type. It is an error to add two pointers. Test it to see if it is NULL. In particular, you should never try to dereference it. p is a pointer to a pointer to a pointer to a pointer to a double. It could be used to represent a four-dimensional array. The value of p is the same as the address of x: 0x3fffd1c. The value of q depends upon sizeof(double). If objects of type double occupy 8 bytes, then an offset of 8(5) = 40 is added to p to give q the hexadecimal value 0x3fffd44. The only expressions among these six that are illegal are p + q and n - q. The name of an array is a variable that contains the address of the first element of the array. This address cannot be changed, so the array name is actually a constant pointer. In the following code that adds all the elements of the array a, each increment of the pointer p locates the next element: const SIZE = 3; short a[SIZE] = {22, 33, 44}; short* end = a + SIZE; // adds SIZE*sizeof(short) = 6 to a for (short* p = a; p < end; p++) sum += *p; The value a[i] returned by the subscripting operator [] is the value stored at the address computed from the expression a + i. In that expression, a is a pointer to its base type T and i is an int, so the offset i*sizeof(T) is added to the address a. The same evaluation would be made from the expression i + a which is what would be used for i[a]. The declaration double * f(); declares f to be a function that returns a pointer to double. The declaration double (* f)(); declares *f to be a pointer to a function that returns a double. a. float a[8]; b. float* a[8]; c. float (* a)[8]; d. float* (* a)[8]; e. float f(); f. float* f(); g. float (* f)(); h. float* (* f)();
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