##Exercise 4.1
What is the value returned by 5 + 10 * 20/2?
105
##Exercise 4.2
Using Table 4.12 (p. 166), parenthesize the following expressions to indicate the order in which the operands are grouped:
* vec.begin() //=> *(vec.begin())
* vec.begin() + 1 //=> (*(vec.begin())) + 1##Exercise 4.3
Order of evaluation for most of the binary operators is left undefined to give the compiler opportunities for optimization. This strategy presents a trade-off between efficient code generation and potential pitfalls in the use of the language by the programmer. Do you consider that an acceptable trade-off? Why or why not?
Yes, I think it necessary to hold the trade-off. Because the speed is always the biggest advantage of C++. Sometimes, we need the compiler's features for efficient work. But if you are not a expert. I have to advice you do not touch the undefined behaviors.
For an instance, cout << i << ++i <<endl should never appear in your code.
##Exercise 4.4
Parenthesize the following expression to show how it is evaluated. Test your answer by compiling the expression (without parentheses) and printing its result.
12 / 3 * 4 + 5 * 15 + 24 % 4 / 2
// parenthesize
((12/3)*4) + (5*15) + ((24%4)/2)
// 16 + 75 + 0 = 91
// print: 91##Exercise 4.5
Determine the result of the following expressions.
-30 * 3 + 21 / 5 // -90+4 = -86
-30 + 3 * 21 / 5 // -30+63/5 = -30+12 = -18
30 / 3 * 21 % 5 // 10*21%5 = 210%5 = 0
-30 / 3 * 21 % 4 // -10*21%4 = -210%4 = -2##Exercise 4.6
Write an expression to determine whether an int value is even or odd.
i%2 == 0 ? "even" : "odd"##Exercise 4.7
What does overflow mean? Show three expressions that will overflow.
short svalue = 32767; ++svalue; // -32768
unsigned uivalue = 0; --uivalue; // 4294967295
unsigned short usvalue = 65535; ++usvalue; // 0##Exercise 4.8
Explain when operands are evaluated in the logical AND, logical OR, and equality operators.
logical AND : true only if both its operands evaluated to true;
logical OR : true if either of its operands evaluated to true;
equality operators : true only if its operands are equal.
##Exercise 4.9
Explain the behavior of the condition in the following if:
const char *cp = "Hello World";
if (cp && *cp)cp is a pointer to const char *, and it's not a nullptr. true.
*cp is a const char: 'H', and it is explicit a nonzero value. true.
true && true = true.
##Exercise 4.10
Write the condition for a while loop that would read ints from the standard input and stop when the value read is equal to 42.
int i = 0;
while(cin >> i && i != 42)##Exercise 4.11
Write an expression that tests four values, a, b, c, and d, and ensures that a is greater than b, which is greater than c, which is greater than d.
a>b && b>c && c>d##Exercise 4.12
Assuming i, j, and k are all ints, explain what i != j < k means.
if j<k is true, means i != 1, else means i != 0.
##Exercise 4.13
What are the values of i and d after each assignment?
int i; double d;
d = i = 3.5; // i = 3, d = 3.0
i = d = 3.5; // d = 3.5, i = 3##Exercise 4.14
Explain what happens in each of the if tests:
if (42 = i) // complie error: expression is not assignable
if (i = 42) // true.##Exercise 4.15
The following assignment is illegal. Why? How would you correct it?
double dval; int ival; int *pi;
dval = ival = pi = 0;
// pi is a pointer to int.
// can not assign to 'int' from type 'int *'
// correct it:
dval = ival = 0;
pi = 0;##Exercise 4.16
Although the following are legal, they probably do not behave as the programmer expects. Why? Rewrite the expressions as you think they should be.
if (p = getPtr() != 0)
if (i = 1024)
// why? always true. use an assigment as a condition.
// correct it
if ((p=getPtr()) != 0)
if (i == 1024)##Exercise 4.17
Explain the difference between prefix and postfix increment.
The postfix operators increment(or decrement) the operand but yield a copy of the original, unchanged value as its result.
The prefix operators return the object itself as an lvalue.
The postfix operators return a copy of the object's original value as an rvalue.
##Exercise 4.18
What would happen if the while loop on page 148 that prints the elements from a vector used the prefix increment operator?
It will print from the second element and will dereference the v.end() at last.(It's undefined and very dangerous)
##Exercise 4.19
Given that ptr points to an int, that vec is a vector, and that ival is an int, explain the behavior of each of these expressions. Which, if any, are likely to be incorrect? Why? How might each be corrected?
ptr != 0 && *ptr++ // check ptr is not a nullptr. and check the pointer value.
ival++ && ival // check ival and ival+1 whether equal zero.
vec[ival++] <= vec[ival] // incorrect. It is an **undefined behavior.**
// correct:
vec[ival] <= vec[ival+1]##Exercise 4.20
Assuming that iter is a vector::iterator, indicate which, if any, of the following expressions are legal. Explain the behavior of the legal expressions and why those that aren’t legal are in error.
*iter++; // return *iter, then ++iter.
(*iter)++; // illegal, *iter is a string, cannot increment value.
*iter.empty() // illegal, iter should use '->' to indicate whether empty.
iter->empty(); // indicate the iter' value whether empty.
++*iter; // illegal, string have not increment.
iter++->empty(); // return iter->empty(), then ++iter.##Exercise 4.21 ##Exercise 4.22 ##Exercise 4.23
The following expression fails to compile due to operator precedence. Using Table 4.12 (p. 166), explain why it fails. How would you fix it?
string s = "word";
string pl = s + s[s.size() - 1] == 's' ? "" : "s" ;Operator Precedence: ?: < +
Fix it:
string pl = s + (s[s.size() - 1] == 's' ? "" : "s") ;##Exercise 4.24
Our program that distinguished between high pass, pass, and fail depended on the fact that the conditional operator is right associative. Describe how that operator would be evaluated if the operator were left associative.
if the operator were left associative.
finalgrade = (grade > 90) ? "high pass" : (grade < 60) ? "fail" : "pass";would same as :
finalgrade = ((grade > 90) ? "high pass" : (grade < 60)) ? "fail" : "pass";if grade > 90, first conditional operator's result is high pass. so the finalgrade is always fail.
It's contradictory obviously.
##Exercise 4.25
What is the value of ~'q' << 6 on a machine with 32-bit ints and 8 bit chars, that uses Latin-1 character set in which 'q' has the bit pattern 01110001?
The final value in decimal representation is -7296.
The bitwise NOT operator (~) yields us the Ones' Complement of 0000 0000 0000 0000 0000 0000 0111 0001, which is 1111 1111 1111 1111 1111 1111 1000 1110. The value of 1111 1111 1111 1111 1111 1111 1000 1110 in decimal form is -114. This may come as a surprise to some as the unsigned value of said binary sequence is 4294967182. The most significant bit (the left-most bit, commonly referred to as the sign bit) is however "turned on", or 1, which siginifies a negatation operation on that particular bit. The value of that particular bit is then -2147483648.
We then shift the bits 6 digits to the left, which yields us 1111 1111 1111 1111 1110 0011 1000 0000. Overflowing bits were discarded. The decimal representation of the binary sequence is -7296.
##Exercise 4.26
In our grading example in this section, what would happen if we used unsigned int as the type for quiz1?
no different in most situation. unsigned int have the same size as unsigned long on most machine. But the second one could make sure that it have at least 32 bits on any machine.
##Exercise 4.27
What is the result of each of these expressions?
unsigned long ul1 = 3, ul2 = 7;
ul1 & ul2 // == 3
ul1 | ul2 // == 7
ul1 && ul2 // == true
ul1 || ul2 // == ture##Exercise 4.28 ##Exercise 4.29
Predict the output of the following code and explain your reasoning. Now run the program. Is the output what you expected? If not, figure out why.
int x[10]; int *p = x;
cout << sizeof(x)/sizeof(*x) << endl;
cout << sizeof(p)/sizeof(*p) << endl;The first result is 10. It returns the number of elements in x. But the second result depends on your machine. It would be 2 on the 64-bit machine and 1 on the 32-bit machine. Because of the size of pointer is different on various machines.
reference: Why the size of a pointer is 4bytes in C++
##Exercise 4.30
Using Table 4.12 (p. 166), parenthesize the following expressions to match the default evaluation:
sizeof x + y // (sizeof x)+y . "sizeof" has higher precedence than binary `+`.
sizeof p->mem[i] // sizeof(p->mem[i])
sizeof a < b // sizeof(a) < b
sizeof f() //If `f()` returns `void`, this statement is undefined, otherwise it returns the size of return type. reference: sizeof operator
##Exercise 4.31
The program in this section used the prefix increment and decrement operators. Explain why we used prefix and not postfix. What changes would have to be made to use the postfix versions? Rewrite the program using postfix operators.
postfix will copy itself as return, then increment or decrement. prefix will increment or decrement first, and return itself. so prefix is more effective in this program.(reduce one copy space.)
We use prefix and not postfix, just because of the Advice: Use Postfix Operators only When Necessary on §4.5. Increment and Decrement Operators.
Advice: Use Postfix Operators only When Necessary
Readers from a C background might be surprised that we use the prefix increment in the programs we've written. The reason is simple: The prefix version avoids unnecessary work. It increments the value and returns the incremented version.The postfix operator must store the original value so that it can return the unincremented value as its result. If we don’t need the unincremented value, there’s no need for the extra work done by the postfix operator.
For ints and pointers, the compiler can optimize away this extra work. For more complicated iterator types, this extra work potentially might be more costly. By habitually using the prefix versions, we do not have to worry about whether the performance difference matters. Moreover—and perhaps more importantly—we can express the intent of our programs more directly.
So, it's just a good habits. And there are no changes if we have to be made to use the postfix versions. Rewrite:
for(vector<int>::size_type ix = 0; ix != ivec.size(); ix++, cnt--)
ivec[ix] = cnt; This is not an appropriate example to discuss the difference of prefix and postfix. Look at the section Built-in comma operator on this page.
reference: Usage of the Built-in Comma Operator
##Exercise 4.32
Explain the following loop.
constexpr int size = 5;
int ia[size] = {1,2,3,4,5};
for (int *ptr = ia, ix = 0;
ix != size && ptr != ia+size;
++ix, ++ptr) { /* ... */ }ptr and ix have the same function. The former use a pointer, and the latter use the index of array. we use the loop to through the array.(just choose one from ptr and ix)
##Exercise 4.33
Using Table 4.12 (p. 166) explain what the following expression does:
someValue ? ++x, ++y : --x, --yBecause of the most lowest precedence of the comma operator, the expression is same as:
(someValue ? ++x, ++y : --x), --yIf someValue is true, then ++x, and the result is y, if someValue is false, then --x, and the result is --y. so it is also same as:
someValue ? (++x,y) : (--x,--y);Even though the result has nothing to do with x, the evaluation of someValue does effect the operation on x.
##Exercise 4.34
Given the variable definitions in this section, explain what conversions take place in the following expressions: (a) if (fval) (b) dval = fval + ival; (c) dval + ival * cval; Remember that you may need to consider the associativity of the operators.
if (fval) // fval converted to bool
dval = fval + ival; // ival converted to fval, then the result of fval add ival converted to double.
dval + ival * cval; // cval converted to int, then that int and ival converted to double.##Exercise 4.35
Given the following definitions,
char cval; int ival; unsigned int ui; float fval; double dval;identify the implicit type conversions, if any, taking place:
cval = 'a' + 3; // 'a' promoted to int, then the result of ('a' + 3)(int) converted to char.
fval = ui - ival * 1.0; // ival promoted to double, ui also promoted to double. then that double converted to float.
dval = ui * fval; // ui promoted to float. then that float converted to double.
cval = ival + fval + dval; // ival converted to float, then that float and fval converted to double. At last, that double converted to char(by truncation).##Exercise 4.36
Assuming i is an int and d is a double write the expression i *= d so that it does integral, rather than floating-point, multiplication.
i *= static_cast<int>(d);##Exercise 4.37
Rewrite each of the following old-style casts to use a named cast:
int i; double d; const string *ps; char *pc; void *pv;
pv = (void*)ps; // pv = const_cast<string*>(ps); or pv = static_cast<void*>(const_cast<string*>(ps));
i = int(*pc); // i = static_cast<int>(*pc);
pv = &d; // pv = static_cast<void*>(&d);
pc = (char*)pv; // pc = reinterpret_cast<char*>(pv);##Exercise 4.38
Explain the following expression:
double slope = static_cast<double>(j/i);j/i is an int(by truncation), then converted to double and assigned to slope.