Computer Programming Language
計算機程式設計
Chapter 2
Introduction to C Program
台大電機系 吳安宇教授
E-mail: [email protected]
Sept. 2002
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Chapter 2 - Introduction to C
Programming
Outline
2.1
Introduction
2.2
A Simple C Program: Printing a Line of Text
2.3
Another Simple C Program: Adding Two Integers
2.4
Memory Concepts
2.5
Arithmetic in C
2.6
Decision Making: Equality and Relational Operators
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2.1
Introduction
• C programming language
– Structured and disciplined approach to program design
• Structured programming
– Introduced in Chapters 3 and 4
– Used throughout the remainder of the book
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1
// Fig. 1.2: fig01_02.cpp
2 // A first program in C++
3 #include <iostream>
Written between /* and */ or following
a //.
• 1. Comments
4
Improve program readability and do not cause the
computer to perform any action.
5 int main()
6 {
7
Comments
preprocessor directive
std::cout << "Welcome to C++!\n";
1. 2. Load
<iostream>
Message to the C++ preprocessor.
• 3. main
Lines beginning with # are preprocessor
directives.
8
9
return 0;
// indicate that program ended successfully
#include <iostream> tells the preprocessor to
include
the contents
theor
filemore
<iostream>,
which
• functions,
3.1 Printone
C++
programs
containofone
of
includes
input/output
as printing to to
which
must
be main operations (such
"Welcome
the screen).
C++\n"
Parenthesis are used to indicate a function
10 }
Welcome to C++!
Prints the string of characters
contained
between
the an integer value.
int means
that main
"returns"
quotation marks.
More in Chapter 3.
return is a way to exit a function
• 3.2 exit (return
from a function.
A left
brace { begins
0) function
The entire line, including
std::cout,
the the
<< body of every
and a right to
braceC++!\n"
} ends it. and
operator,
return 0, in this case,
means the
thatstring "Welcome
semicolon (;), is called a statement.
the program terminatedthe
normally.
• Program Output
All statements must end with a semicolon.
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2.2
1
2
3
4
5
6
7
8
9
10
A Simple C Program: Printing a Line of
Text
/* Fig. 2.1: fig02_01.c
A first program in C */
#include <stdio.h>
int main()
{
printf( "Welcome to C!\n" );
return 0;
}
Welcome to C!
• Comments
– Text surrounded by /* and */ is ignored by computer
– Used to describe program
• #include <stdio.h>
– Preprocessor directive - tells computer to load contents of a certain
file
– <stdio.h> allows standard input/output operations
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2.2
A Simple C Program: Printing a Line of
Text (II)
• int main()
– C++ programs contain one or more functions, exactly
one of which must be main
– Parenthesis used to indicate a function
– int means that main "returns" an integer value
– Braces indicate a block
• The bodies of all functions must be contained in braces
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2.2
A Simple C Program: Printing a Line of
Text (III)
• printf( "Welcome to C!\n" );
– Instructs computer to perform an action
• Specifically, prints string of characters within quotes
– Entire line called a statement
• All statements must end with a semicolon
– \ : escape character
• Indicates that printf should do something out of the
ordinary printed characters.
• \n is the newline character
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Escape Sequence
Escape Sequence
Description
\n
Newline. Position the screen cursor to the
beginning of the next line.
\t
Horizontal tab. Move the screen cursor to the next
tab stop.
\r
Carriage return. Position the screen cursor to the
beginning of the current line; do not advance to the
next line.
\a
Alert. Sound the system bell.
\\
Backslash. Used to print a backslash character.
\"
Double quote. Used to print a double quote
character.
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2.2
A Simple C Program: Printing a Line of
Text (IV)
• return 0;
– A way to exit a function
– return 0, in this case, means that the program terminated
normally
• Right brace }
– Indicates end of main has been reached
• Linker
– When a function is called, linker locates it in the library
– Inserts it into object program
– If function name misspelled, linker will spot error because it
cannot find function in library
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1
/* Fig. 2.5: fig02_05.c
2
3
Outline
Addition program */
#include <stdio.h>
4
5
int main()
6
{
7
1. Initialize variables
int integer1, integer2, sum;
/* declaration */
2. Input
printf( "Enter first integer\n" );
/* prompt */
2.1 Sum
10
scanf( "%d", &integer1 );
/* read an integer */
11
printf( "Enter second integer\n" ); /* prompt */
12
scanf( "%d", &integer2 );
/* read an integer */
13
sum = integer1 + integer2;
/* assignment of sum */
14
printf( "Sum is %d\n", sum );
/* print sum */
8
9
3. Print
15
16
return 0;
/* indicate that program ended successfully */
17 }
Enter first integer
45
Enter second integer
72
Sum is 117
 2000 Prentice Hall, Inc. All rights reserved.
Program Output
2.3
Another Simple C Program: Adding
Two Integers
• As before
– Comments, #include <stdio.h> and main
• int integer1, integer2, sum;
– Declaration of variables
• Variables: locations in memory where a value can be
stored
– int means the variables can hold integers (-1, 3, 0, 47)
– integer1, integer2, sum - variable names (identifiers)
• Identifiers: consist of letters, digits (cannot begin with a
digit), and underscores, case sensitive
• Better be meaningful: e.g. my_student_id, my_age,
rather than id, aa, bb, cc.
– Declarations appear before executable statements
• If not, syntax (compile) error
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2.3
Another Simple C Program: Adding
Two Integers (II)
• scanf( "%d", &integer1 );
– Obtains value from user
• scanf uses standard input (usually keyboard)
– This scanf has two arguments
• %d - indicates data should be a decimal integer
• &integer1 - location in memory to store variable
• & is confusing in beginning - just remember to include it
with the variable name in scanf statements
– It will be discussed later
– User responds to scanf by typing in number, then
pressing the enter (return) key
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2.3
Another Simple C Program: Adding
Two Integers (III)
• = (assignment operator )
– Assigns value to a variable
– Binary operator (has two operands)
sum = variable1 + variable2;
sum gets variable1 + variable2;
– Variable receiving value on left
• printf( "Sum is %d\n", sum );
– Similar to scanf - %d means decimal integer will be printed
• sum specifies what integer will be printed
– Calculations can be performed inside printf statements
printf( "Sum is %d\n", integer1 + integer2 );
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2.4 Memory Concepts
• Variables
– Variable names correspond to locations in the computer's
memory.
– Every variable has a name, a type, a size and a value.
– Whenever a new value is placed into a variable (through
scanf, for example), it replaces (and destroys) previous
value
– Reading variables from memory does not change them
• A visual representation
integer1
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45
Variables define
• 變數命名規則有下列幾項規定：
–
–
–
–
–
只能用大小寫英文字母、1到9的數字和下橫線。
第一個字母必須是字母或數字。
編譯器只認前32個字母。
大小寫字母意義不同 (Case Sensitive, NTU != Ntu)。
不可用c的關鍵字，關鍵字是保留給語言本身使用。
• C語言所內定的資料型態：
資料型態
位元數*
數值範圍
int
依定址長度而定
2 bytes
n/a
-32,768
n/a
32767
Char
float
double
1 bytes
4 bytes
8 bytes
-128
3.4e-38
1.7e-308
127
3.4e38
1.7e308
Void
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Variables define(cont.)
• 修飾資料型態：
const
unsigned
signed
long
short
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如果變數資料的前方有const關鍵字，那麼程式中任
何企圖更改其值的動作，都會被編譯器視為錯誤訊
息。常數修飾子可確保常數值一定，同時提醒任何
讀程式的人不要試圖改變此值。
char
int
long int
int
double
倘若把char,int這兩種變數的負號
去掉，範圍就可以從0開始，儲存
的範圍變成原來的2倍。例：
unsigned char var;
var的數值圍為0~255，記憶體位
元組數為1。
當數值大於int,
double所能儲存
的範圍時，可用
long來加大位元
數
long int為4個位
元組
long double為
10 個位元組
2.5
Arithmetic
• Arithmetic calculations are used in most
programs
– Use * for multiplication and / for division
– Integer division truncates remainder
7 / 5 evaluates to 1
– Modulus operator returns the remainder
7 % 5 evaluates to 2
• Operator precedence
– Some arithmetic operators act before others (i.e.,
multiplication before addition)
• Use parenthesis when needed
– Example: Find the average of three variables a, b and c
• Do not use: a + b + c / 3
• Use: (a + b + c ) / 3
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Operator precedence
• Check z = p * r % q + w / x – y;
• In calculation: z = ((p * r) % q) + (w / x) – y;
• Check y = a * x * x + b * x + c;
• In calculation: y = (a* x * x) + (b * x) + c;
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2.5
•
•
Arithmetic (II)
Arithmetic operators:
C op era tion
Arithm etic
op era tor
Alg eb ra ic
exp ression
C exp ression
Addition
+
f+7
f + 7
Subtraction
p–c
p - c
Multiplication
*
bm
b * m
Division
/
x/y
x / y
Modulus
%
r mod s
r % s
Rules of operator precedence:
Operator(s)
Operation(s)
Order of evaluation (precedence)
()
Parentheses
Evaluated first. If the parentheses are nested, the
expression in the innermost pair is evaluated first. If
there are several pairs of parentheses “on the same level”
(i.e., not nested), they are evaluated left to right.
*, /, or %
Multiplication Division Evaluated second. If there are several, they re
Modulus
evaluated left to right.
+ or -
Addition
Subtraction
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Evaluated last. If there are several, they are
evaluated left to right.
2.6
Decision Making: Equality and
Relational Operators
• Executable statements
– Perform actions (calculations, input/output of data)
– Perform decisions
• May want to print "pass" or "fail" given the value of a test grade
• if control structure
– Simple version in this section, more detail later
– If a condition is true, then the body of the if statement
executed
• 0 is false, non-zero is true
– Control always resumes after the if structure
• Keywords
– Special words reserved for C
– Cannot be used as identifiers or variable names
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2.6
Decision Making: Equality and
Relational Operators (II)
Standard algebraic equality
operator or
relational operator
C++ equality
or relational
operator
Example
of C++
condition
Meaning of
C++ condition
>
>
x>y
<
<
x<y
x is greater than y
x is less than y
_
>
>=
x >= y
_
<
<=
x <= y
x is greater than or equal to
y
x is less than or equal to y
==
!=
x == y
x != y
x is equal to y
x is not equal to y
Relational operators
Equality operators
=
=
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2.6
Decision Making: Equality and
Relational Operators (III)
Keyw o rd s
auto
double
int
struct
break
case
char
const
continue
default
do
else
enum
extern
float
for
goto
if
long
register
return
short
signed
sizeof
static
switch
typedef
union
unsigned
void
volatile
while
C program 2002
1
/* Fig. 2.13: fig02_13.c
Outline
2
Using if statements, relational
3
operators, and equality operators */
4
#include <stdio.h>
5
6
int main()
7
{
1. Declare variables
int num1, num2;
2. Prompt and get Input
10
printf( "Enter two integers, and I will tell you\n" );
2.1 if statements
11
printf( "the relationships they satisfy: " );
12
scanf( "%d%d", &num1, &num2
8
9
);
/* read two integers */
3. Print
13
14
15
if ( num1 == num2 )
printf( "%d is equal to %d\n", num1, num2 );
16
17
18
if ( num1 != num2 )
printf( "%d is not equal to %d\n", num1, num2 );
19
20
21
if ( num1 < num2 )
printf( "%d is less than %d\n", num1, num2 );
22
23
24
if ( num1 > num2 )
printf( "%d is greater than %d\n", num1, num2 );
25
26
27
if ( num1 <= num2 )
printf( "%d is less than or equal to %d\n",
Inc. Allnum2
rights reserved.
28  2000 Prentice Hall,num1,
);
29
30
if ( num1 >= num2 )
31
Outline
printf( "%d is greater than or equal to %d\n",
32
num1, num2 );
33
34
return 0;
3.1 Exit main
/* indicate program ended successfully */
35 }
Enter two integers, and I will tell you
the relationships they satisfy: 3 7
3 is not equal to 7
3 is less than 7
3 is less than or equal to 7
Enter two integers, and I will tell you
the relationships they satisfy: 22 12
22 is not equal to 12
22 is greater than 12
22 is greater than or equal to 12
 2000 Prentice Hall, Inc. All rights reserved.
Program Output
Homework #1
• Practice Turbo C Compiler
• Run the Program of Fig. 2.5 on Page 31: Addition
Program
– Try input 55, 89
– Try input 88, -100
– Try input 590100, 600200
• Change the Addition Program to Division Program
– Sum = integer1 + integer2; /* old statement */
–
Quotient = integer1 / integer2; /* 1st new statement */
–
Remainder = integer1 % integer2; /* 2nd new statement */
– Modify printf() to show the computed results: Quotient and Remainder.
– Try above three inputs and other kinds of inputs
• Due next Friday at 1:10pm: In one floppy disk (label
your name) with printed results.
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