Merge pull request fool2fish#31 from wjk20120522/master

Add partial answers of ch01

ch01/1.1/1.1.md

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+# Exercises for Section 1.1
+
+### 1.1.1
+
+What is the difference between a compiler and an interpreter?
+
+#### Answer
+
+A compiler is a program that can read a program in one language - the source language - and translate it into an equivalent program in another language – the target language and report any errors in the source program that it detects during the translation process.
+
+Interpreter directly executes the operations specified in the source program on inputs supplied by the user.
+
+### 1.1.2
+
+What are the advantages of:
+(a) a compiler over an interpreter
+(b) an interpreter over a compiler?
+
+#### Answer
+
+a. The machine-language target program produced by a compiler is usually much faster than an interpreter at mapping inputs to outputs.
+
+b. An interpreter can usually give better error diagnostics than a compiler, because it executes the source program statement by statement.
+
+### 1.1.3
+
+What advantages are there to a language-processing system in which the compiler
+produces assembly language rather than machine language?
+
+#### Answer
+
+The compiler may produce an assembly-language program as its output, because
+assembly language is easier to produce as output and is easier to debug.

ch01/1.6/1.6.md

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+# Exercises for Section 1.6
+
+### 1.6.1
+
+For the block-structured C code of below , indicate the values assigned to w, x, y, and z .
+```
+int w, x, y, z;
+int i = 4; int j = 5;
+{
+  int j = 7;
+  i = 6;
+  w = i + j;
+}
+x = i + j;
+{
+  int i = 8;
+  y = i + j;
+}
+z = i + j;
+```
+
+#### Answer
+
+w = 13, x = 11, y = 13, z = 11.
+
+
+### 1.6.2
+
+Repeat Exercise 1 .6.1 for the code of below .
+```
+int w, x, y, z;
+int i = 3; int j = 4;
+{
+  int i = 5;
+  w = i + j;
+}
+x = i + j;
+{
+  int j = 6;
+  i = 7;
+  y = i + j;
+}
+z = i + j;
+```
+
+#### Answer
+
+w = 9, x = 7, y = 13, z = 11.
+
+
+### 1.6.3
+
+For the block-structured code of Fig. 1 . 14, assuming the usual static scoping of declarations, give the scope for each of the twelve declarations.
+
+#### Answer
+```
+Block B1:
+	declarations:  ->   scope
+		w 				B1-B3-B4
+		x				B1-B2-B4
+		y				B1-B5
+		z				B1-B2-B5
+Block B2:
+	declarations:  ->   scope
+		x				B2-B3
+		z				B2
+Block B3:
+	declarations:  ->   scope
+		w 				B3
+		x				B3
+Block B4:
+	declarations:  ->   scope
+		w 				B4
+		x				B4
+Block B5:
+	declarations:  ->   scope
+		y				B5
+		z				B5
+```
+
+### 1.6.4
+
+What is printed by the fo llowing C code?
+```
+#define a (x + 1)
+int x = 2;
+void b() { x = a; printf("%d\n", x); }
+void c() { int x = 1; printf("%d\n", a); }
+void main () { b(); c(); }
+```
+
+#### Answer
+3
+
+2