##Exercise 3.1 : part1 | part2 ##Exercise 3.2 : part1 | part2 ##Exercise 3.3

Explain how whitespace characters are handled in the string input operator and in the getline function.

The getline function takes an input stream and a string. This function reads the given stream up to and including the first newline and stores what it read—not including the newline—in its string argument. After getline sees a newline, even if it is the first character in the input, it stops reading and returns. If the first character in the input is a newline, then the resulting string is the empty string.

getline function whitespace handling, do not ignore the beginning of the line blank characters read characters until it encounters a line break, read to termination and discard newline (line breaks removed from the input stream but is not stored in the string object).

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##Exercise 3.4 : part1 | part2 ##Exercise 3.5 : part1 | part2 ##Exercise 3.6 ##Exercise 3.7

What would happen if you define the loop control variable in the previous exercise as type char? Predict the results and then change your program to use a char to see if you were right.

No different.

auto& c : str

We use auto to let the compiler determine the type of c. which in this case will be char&.

##Exercise 3.8 ##Exercise 3.9

What does the following program do? Is it valid? If not, why not?

string s;
cout << s[0] << endl;

invalid in theory, but the compiler passes.

s is empty, so s[0] is undefined. But the compiler always define s[0] with \0, so you can use s[0] in the cout.

##Exercise 3.10 ##Exercise 3.11

Is the following range for legal? If so, what is the type of c?

const string s = "Keep out!";
for (auto &c : s){/*... */}

When you don't change c's value, it's legal, else it's illegal.

For example:

cout << c;  // legal.
c = 'X';    // illegal.

The type of c is const char&. read-only variable is not assignable.

##Exercise 3.12

Which, if any, of the following vector definitions are in error? For those that are legal, explain what the definition does. For those that are not legal, explain why they are illegal.

vector<vector<int>> ivec;         // legal(c++11), vectors.
vector<string> svec = ivec;       // illegal, different type.
vector<string> svec(10, "null");  // legal, vector have 10 strings: "null".

##Exercise 3.13

How many elements are there in each of the following vectors? What are the values of the elements?

vector<int> v1;         // size:0,  no values.
vector<int> v2(10);     // size:10, value:0
vector<int> v3(10, 42); // size:10, value:42
vector<int> v4{10};     // size:1,  value:10
vector<int> v5{10, 42}; // size:2,  value:10, 42
vector<string> v6{10};  // size:10, value:""
vector<string> v7{10, "hi"};  // size:10, value:"hi"

##Exercise 3.14 ##Exercise 3.15 ##Exercise 3.16 ##Exercise 3.17 ##Exercise 3.18 ##Exercise 3.19 ##Exercise 3.20 ##Exercise 3.21 (Generics Version) ##Exercise 3.22 ##Exercise 3.23 ##Exercise 3.24 ##Exercise 3.25 ##Exercise 3.26

In the binary search program on page 112, why did we write mid=beg+(end-beg)/2; instead of mid=(beg+end) /2;?

Because the iterator of vector don't define the + operator between the two iterators. beg + end is illegal.

We can only use the subtraction between the two iterators.

##Exercise 3.27

Assuming txt_size is a function that takes no arguments and returns an int value, which of the following definitions are illegal? Explain why.

unsigned buf_size = 1024;

int ia[buf_size];   // illegal, The dimension value must be a constant expression.
int ia[4 * 7 - 14]; // legal
int ia[txt_size()]; // illegal, The dimension value must be a constant expression.
char st[11] = "fundamental";  // illegal, the string's size is 12.

##Exercise 3.28

What are the values in the following arrays?

string sa[10];
int ia[10];

int main() {
  string sa2[10];
  int ia2[10];
}

please see 2.2.1. Variable Definitions -> Default Initialization.

std::string isn't a build-in type. The initializer will set it empty. ia and ia2 are build-type. But ia isn't in the function body, so it will be initialized to zero. ia2 is in the function body. so it's value is undefined.

You can also use gdb to debug the value when the code is running.

##Exercise 3.29:

List some of the drawbacks of using an array instead of a vector.

1. can't add elements to an array.
2. can't use auto to deduce the type from a list of initializers.
3. no arrays of references.
4. can't use template and iterator.
5. vector have lots of useful methods and algorithms.

##Exercise 3.30

Identify the indexing errors in the following code:

constexpr size_t array_size = 10;
int ia[array_size];
for (size_t ix = 1; ix <= array_size; ++ix)
      ia[ix] = ix;

The size of ia is 10, so the index of value should less than 10. ix cannot equal the array_size.

##Exercise 3.31 ##Exercise 3.32 ##Exercise 3.33

What would happen if we did not initialize the scores array in the program on page 116?

If we did not initialize the scores array. the array is undefined. the value will be Unknown.

Look like this:

##Exercise 3.34

Given that p1 and p2 point to elements in the same array, what does the following code do? Are there values of p1 or p2 that make this code illegal?

p1 += p2 - p1;

we assume p1 and p2 point to an array arr. so p1 = &arr[0]; and p2 = &arr[0]. p2 - p1 is the distance of arr[0] to arr[0], and must be zero. so p1 += 0; can not change the p1's point.

p1 += p2 - p1; same as p1 = p2;. If p2 and p1 are legal, this code always legal.

##Exercise 3.35 ##Exercise 3.36 ##Exercise 3.37

What does the following program do?

const char ca[] = {'h', 'e', 'l', 'l', 'o'};
const char *cp = ca;
while (*cp) {
    cout << *cp << endl;
    ++cp;
}

Print all the elements of the array.

---

WARNING!!!!
When we use a string, the compiler put it in the section .rodata, the code uses C-style character string without adding a '\0' in the end of ca.
So, when we code like this:

const char ca[] = {'h', 'e', 'l', 'l', 'o'};
const char s[] = "world";
const char *cp = ca;
while (*cp) {
cout << *cp;
++cp;
}

The code will print "helloworld" when you run it. because the character list in the .rodata like this:

h e l l o w o r l d \0  

While(*cp) judge weather *cp is 0 or not. when *cp is not 0, it will print the character until 0.
When you change the code like this:

const char ca[] = {'h', 'e', 'l', 'l', 'o', '\0'};  

the character list in the .rodata:

h e l l o \0 w o r l d \0  

The program will run correctly. So when using C-style character string, be careful!!

---

see .rodata, you can use this command:

hexdump -C a.out

##Exercise 3.38

In this section, we noted that it was not only illegal but meaningless to try to add two pointers. Why would adding two pointers be meaningless?

Because Subtracting two points gives a logically explainable result - the offset in memory between two points. Similarly, you can subtract or add an integral number to/from a pointer, which means "move the pointer up or down". Adding a pointer to a pointer is something which is hard to explain. The result is meaningless.

---

References:

• Why can't I add pointers

##Exercise 3.39 ##Exercise 3.40 ##Exercise 3.41 ##Exercise 3.42 ##Exercise 3.43 ##Exercise 3.44 ##Exercise 3.45