corrected spelling for the first three chapters

ch02/README.md

@@ -172,7 +172,7 @@ std::cin >> input_value;
 double i = { 3.14 };
 ```
 
-(c): --if you declared 'wage' before, it's right. Otherwhise, you'll get a error:
+(c): --if you declared 'wage' before, it's right. Otherwise, you'll get a error:
     error: use of undeclared identifier 'wage'
 ```cpp
 double wage;
@@ -201,7 +201,7 @@ int main()
 `local_int` is a local variable which is not uninitialized, so it has a undefined value.
 `local_str` is also a local variable which is not uninitialized, but it has a value that is defined by the class. So it is empty string.
 PS: please read P44 in the English version, P40 in Chinese version to get more.
-The note: Uninitialized objects of built-in type defined inside a function body have a undifined value. Objects of class type that we do not explicitly inititalize have a value that is defined by class.
+The note: Uninitialized objects of built-in type defined inside a function body have a undefined value. Objects of class type that we do not explicitly inititalize have a value that is defined by class.
 
 ##Exercise 2.11
 > Explain whether each of the following is a declaration or a
@@ -274,7 +274,7 @@ Yes.It is legal.Printed:
 
 ```
 (a): valid. let d equal 3.14159.
-(b): valid. automatical convert will happen.
+(b): valid. automatic convert will happen.
 (c): valid. but value will be truncated.
 (d): valid. but value will be truncated.
 ```
@@ -309,7 +309,7 @@ p1 = &b;
 
 ####definition:
 
-the pointer is "points to" anyother type.
+the pointer is "points to" any other type.
 
 the reference is "another name" of an **object**.
 
@@ -320,7 +320,7 @@ a pointer is an object in its **own right**.
 2. Once initialized, a reference remains **bound to** its initial object.
 There is **no way** to rebind a reference to refer to a different object.
 a pointer can be **assigned** and **copied**.
-3. a reference always get the object to which the reference was intially bound.
+3. a reference always get the object to which the reference was initially bound.
 a single pointer can point to **several different objects** over its lifetime.
 4. a reference must be initialized.
 a pointer need **not be** initialized at the time it is defined.
@@ -369,7 +369,7 @@ if (*p) // whether the value pointed by p is zero?
 ##Exercise 2.23
 >Given a pointer p, can you determine whether p points to a valid object? If so, how? If not, why not?
 
-No. Because more information needed to determine whether the pointee is valid or not.
+No. Because more information needed to determine whether the pointer is valid or not.
 
 
 ##Exercise 2.24
@@ -531,7 +531,7 @@ decltype((b)) d = a;
 ++d;
 ```
 
-`c` is an int, `d` is a referance of `a`.
+`c` is an int, `d` is a reference of `a`.
 all their value are `4`.
 
 ##Exercise 2.37

ch03/README.md

@@ -2,13 +2,13 @@
 ##Exercise 3.2 : [part1](ex3_2a.cpp) | [part2](ex3_2b.cpp)
 ##Exercise 3.3
 >Explain how whitespace characters are handled in the string
-input operator and in the getline function.
+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,
+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.
@@ -142,7 +142,7 @@ 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 initalized to **zero**. `ia2` is in the function body. so it's
+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.
@@ -151,7 +151,7 @@ You can also use gdb to debug the value when the code is running.
 >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 initalizers.
+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.
@@ -228,7 +228,7 @@ 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 wether *cp is 0 or not. when *cp is not 0, it will print the character until 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'};