The C programming language: Introduction Fall 2003, Jen-Chang Liu.

The C programming language:Introduction

Fall 2003, Jen-Chang Liu

Position of C

Hardwaremachines

MS WindowsUnixLinux

shelldesktop dos

applications

machinelanguage

High-levellanguage

compiler

C, C++Pascal,..

HumanNatural language

Introduction

B -> C C is a general-purpose

programming language Developed on UNIX system by D.

Ritchie Portable, independent of any machine

architecture OS, C compiler and all UNIX application

were written in C

Introduction (cont.)

A relatively small language ANSI C standard in 1988

The first edition of The C Programming Language was usually used as a reference manual of C

American National Standards Institute Develop an unambiguous and machine-

independent definition of C

Tutorial Started by learning by examples… Please try the examples on your own

如何產生可執行程式？

C sourceCodeC 原始程式

Compiler編譯器

ExecutableCode可執行碼

Turbo C 2.01整合式編譯環境

文字檔形式*.c

執行檔形式*.exe

Visual C++

Compiler Unix: cc, gcc(GNU C compiler)

gcc has PC version, you may try to install it Windows: Visual C++, Borland C++

編譯器

aaa.ccompiler aaa.obj

bbb.ccompiler bbb.obj

abc.exe

…

link

Example 1

#include <stdio.h>

main( ) { printf("Hello, world\n"); }

functions

Standard input/output library

printfscanf…

argumentsBody ofmain func

Terminatorof

statement

函式庫

C programming concept

Controlunit

tape

read/write head

…

symbols

state of the machines

Turing machine

printf(“Hello world!”);

Notes Blank words( 空白，換行 ) are irrelevant in C C is well-defined language

with a set of keywords with a set of functions

Case sensitive: error typing is not allowed

inputfunction

output

Example 1 (cont.) “Hello, world\n”

Example 2

Character strings(string constant)

Newline character\t: tab, \b: backspace, \\: \

Program components Function

Contains statements that specify the computing operations to be done

Variable Store values for computing

Outline

Variable and Arithmetic Expression The For Statement Symbolic Constants Character Input and Output Arrays Functions Arguments – Call by Value Character Arrays External Variables and Scope

上課程式實做繳交規定program submission in class

每次上課後將當天實做之程式繳交方式 : email

Subject: work 日期學號例： work1027 93321001 只交上 .c 檔便可

繳交程式必須加上註解 (comments) ，包括姓名，學號程式目的每行的作用

Prog2-1// 劉震昌 , 93xx0xx// 程式目的：測試整數變數並印出#include <stdio.h>void main(void){ int i; /* 變數宣告 declaration */ i=2; /* 設定變數 */ printf(“This is my %dnd C program”, i); // 印出

訊息}

變數宣告與設定宣告：告知編譯器 data type 與變數名稱

Ex. int i;

變數設定

變數名稱可以任取，最好取有意義的文字( 第一個字不可為數字 )

變數型態

i = 2; // 將 i 設成 2

Data types of Variables

int integershort short integerlong long integerchar characterfloat floating pointdouble double-precision

case-sensitivekeywords

Integer withmachine-dependentsizes

Prog4-0 (p. 4-10)

#include <stdio.h>void main(void){ printf(“Size of char : %d\n”, sizeof(char)); printf(“Size of int : %d\n”, sizeof(int)); printf(“Size of short : %d\n”, sizeof(short)); printf(“Size of float : %d\n”, sizeof(float)); printf(“Size of double: %d\n”, sizeof(doubl

e));}

Test the range of (short)int#include <stdio.h>void main(void){

short i;i=32767;printf("i=%d\n", i);i=32768;printf("i=%d\n", i);

}

Variables and arithmetic expression (1.2)

C = (5/9)(F-32) 攝氏 -華氏轉換 Try to output the right table using printf

0 -1720 -640 460 1580 26100 37120 48140 60160 71180 82200 93220 104240 115260 126280 137300 148

F C

0 -1720 -640 460 1580 26100 37120 48140 60160 71180 82200 93220 104240 115260 126280 137300 148

/* temperature version 0 */#include <stdio.h>

main(){ printf(“0 -17\n"); printf(“20 -6\n"); printf(“40 4\n”); /* … */}

0 -1720 -640 460 1580 26100 37120 48140 60160 71180 82200 93220 104240 115260 126280 137300 148


main(){ int F, C;

F = 0; C = 5 * (F-32) / 9; printf("%d %d\n", F, C);

F = F+20; C = 5 * (F-32) / 9; printf("%d %d\n", F, C);

/* … */}

/*algorithm */F=0 -> CF+20 -> CF+20 -> C…

Arithmetic expression C = 5*(F-32)/9;

C = 5/9*(F-32);

Expressions are evaluated in this direction one-by-one

=0Because their type is integer,division will truncate the remainder

Equation in C F = F + 20;

F 無解？

F + 20 Temp

F

More about printf printf("%d\t%d\n", F, C);

Variables should be properly initialized C does not have build-in input or output

Call the function printf and scanf

The current value of the variable will be printed on display

More about printf Format numeric output

%d

tight output

0

20

40

60

120

%3d

0

20

40

60

120

Right justified

Course break

http://alg.csie.ncnu.edu.tw/course/cprogram/introduction.htm




0 -1720 -640 460 1580 26100 37120 48140 60160 71180 82200 93220 104240 115260 126280 137300 148


main(){ float F, C;

F = 0.0; C = 5.0 * (F-32.0) / 9.0; printf(“%f %f\n", F, C);

F = F+20.0; C = 5.0 * (F-32.0) / 9.0; printf("%f %f\n", F, C);

/* … */}

/*algorithm */F=0 -> CF+20 -> CF+20 -> C…

Floating point C = 5.0/9.0 *(F-32.0);

(F-32) : integer will convert to float for computation

C = 5/9 *(F-32.0); What will happen?

printf("%3.0f\t%6.1f\n", F, C);

Evaluateorder

整數位數小數位數

Floating point (cont.) Many print-out form

others

%f%6f%.2f%6.2f

%o otcal%x hexadecimal%c character%s character string%% %

流程控制： while loop 迴圈

Testcondition

Enter loop

Yes (non-0)

ExecuteLoop body

noexit

F=0F=F+20F=F+20…

F=F+20

0 -1720 -640 460 1580 26100 37120 48140 60160 71180 82200 93220 104240 115260 126280 137300 148

F=0

F=F+20

F<=300

while loop 迴圈

F = 0;while (F <= 300){ F = F+20;}

Testcondition

Enter loop

Yes(non-0)

ExecuteLoop body

noexit

/* C Language */

F=0

F<=300

F=F+20

#include <stdio.h>

/* temperature version 3.0 */

main(){ int fahr, celsius; int lower, upper, step;

lower = 0; /* lower limit of temperature table */ upper = 300; /* upper limit */ step = 20; /* step size */

fahr = lower; while (fahr <= upper){ celsius = 5*(fahr-32)/9; printf("%d\t%d\n", fahr, celsius); fahr = fahr + step; }}

Variable Declarationint = integer

Variableassignment

whileloop

#include <stdio.h>

/* print F-C table */

main(){ float fahr, celsius; int lower, upper, step;

lower = 0; /* lower limit of temperature table */ upper = 300; /* upper limit */ step = 20; /* step size */

fahr = lower; while (fahr <= upper){ celsius = 5.0/9.0 *(fahr-32.0); printf("%3.0f\t%6.1f\n", fahr, celsius); fahr = fahr + step; }}

floating pointversion of ex3.

Floating point costant

Exercise 1 Print the following numbers using while

loops11235813213455

Outline


for loop

for(initial ; loop test ; increment){ Loop body… … }

Testcondition

Enter loop

yes

ExecuteLoop body

noexit

initial value

ExecuteLoop increment

F=0

F=F+20;

F<=300

for(F=0; F<=300; F=F+20){ printf(“%d\n”, F);}

for statement

#include <stdio.h>

/* print F-S table */main(){ int fahr;

for(fahr=0; fahr <= 300; fahr = fahr+20) printf("%3d %6.1f\n", fahr, (5.0/9.0)*(fahr-32));}

initial increment

exit looptest

Type conversion

Exercise 2 Print the following numbers using for

loops11235813213455

Review of HW#2

*** *** **** ***** ****** ******* ******** ********* **********

#include<stdio.h>

main(){

int i,j;for(i=1;i<=10;i++){ /* 第一層迴圈

* 增加直行 */for(j=1;j<=i;j++){

printf("*"); /* 第二層迴圈 * 增加橫的 */

}printf("\n");

}}

Review of HW#2

*** *** **** ***** ****** ******* ******** ********* **********

#include<stdio.h>main(){ int i,k; for(i=0;i<10;i++){ for(k=0;k<i;k++){ printf("*"); } printf("\n"); }}

Review of HW#2

*** *** **** ***** ****** ******* ******** ********* **********

#include <stdio.h>

main (){ int A, B; for(A=1; A<=10; A= A+1){ B=0; printf("\n"); while(B<A){ printf("*"); B=B+1; } }}

Review of HW#2

*** *** **** ***** ****** ******* ******** ********* **********

#include<stdio.h>

int main (){

int a,b;

for(a=0;a<10;a++){for(b=1;b<=a;b++)

printf("*", b);printf("*\n" ,a);

}return 0;

}

Review of HW#2

*** *** **** ***** ****** ******* ******** ********* **********

#include <stdio.h>

int main(){

int a,b; a=10; b=11;

while(a>=0){ while(b<11){

printf("*");b=b+1;

} printf("\n"); b=a; a=a-1; } }

Outline


Symbol constants

#include <stdio.h>

#define LOWER 0#define UPPER 300#define STEP 20

/* print F-S table */main(){ int fahr;

for(fahr=LOWER; fahr <= UPPER; fahr = fahr+STEP) printf("%3d %6.1f\n", fahr, (5.0/9.0)*(fahr-32));}

Symbol name(Meaningful, easy to read)

Symbol value

Replace this symbol nameat compile time

Outline Variable and Arithmetic Expression The For Statement Symbolic Constants Character Input and Output Arrays Functions Arguments – Call by Value Character Arrays External Variables and Scope

Character I/O A text stream is a sequence of characters getchar() getch() getche() getc()

putchar(c) putch(c,stdout) putc(c,stdout)

I/O devices are also takenas files

輸入stdin

輸出stdout

Example: File copying

#include <stdio.h>

/* echo, version 1 */main(){ int c;

c=getchar(); while( c != EOF ){ putchar(c); c = getchar(); }}

not equal to

End Of File

A constant definedin stdio.hNOT the same asany char values

Example: File copying EOF

Print the value of EOF

End of file

OS Keyboardterminationsignal(ctrl-Z)

#include <stdio.h>

main(){ printf("EOF = %d\n", EOF);}

Exercise Modify your homework#2

1. Use symbol constant 2. Read keyboard input to decide the level of output

*** *** **** ***** ****** ******* ******** ********* **********

Example: File copyingAssignment

c= getchar(); Assignment is an expression and has a valu

e, which is the value of the left hand side after assignment.

#include <stdio.h>main(){ int c;

while( (c=getchar()) != EOF ){ putchar(c); }}

Precedence of = and !=

Example: Character counting

#include <stdio.h>

main(){ long nc; /* number of character */

nc = 0; while(getchar() != EOF) ++nc; printf("%ld\n",nc);}

Good namingConventionFor variables

nc = nc+1;

32-bitinteger

Example : Character counting 2

#include <stdio.h>

main(){ double nc;

for(nc = 0; getchar()!= EOF; ++nc) ; printf("%.0f\n",nc);}

null statement

Increaserange

Example: Line counting

#include <stdio.h>

/* count lines */main(){ int c, nl;

nl = 0; while( (c=getchar()) != EOF) if(c == '\n') ++nl; printf("%d\n", nl);}

Test condition

is equal to

條件測試characterconstant

ASCII valueof input char.

“\n” ?

if statement

if( expression ){ statement 1; }else{ statement 2; }

TestYES

Statement 1

NO

IF

statement 3;Statement 2

Statement 3

else

Exercise #2 Modify the previous program, such that

it counts the number of occurrences of ‘{’ and ‘}’ in a file

Example: Word counting

#include <stdio.h>#define IN 1#defin OUT 2

/* count lines, word, characters */main(){ int c, nl, nw, nc, state;

state = OUT; nl = nw = nc = 0; while( (c=getchar()) != EOF){ ++nc; if(c == '\n') ++nl; if(c == ' ' || c == '\n' || c == '\t;) state = OUT; else if(state == OUT){ state = IN; ++nw; } } printf("%d %d %d\n", nl, nw, nc);}

Word: separate by space,Tab, and newline

record whether nowis in a word or not

nl=(nw=(nc=0));OR

OR: ||AND: &&

Evaluate from left to right

Outline


Example: Count digits Store occurrence of each digit in an

array

#include <stdio.h>

/* count digits, white space, and others */main(){ int c, i, nwhite, nother; int ndigit[10];

/* initialize */

Declaration of array

ndigit[0]ndigit[1]ndigit[2]ndigit[3]ndigit[4]

ndigit[9]??????

16 bits

/* initialize */ nwhite = nother = 0; for(i=0; i<10; ++i) ndigit[i] = 0; while((c = getchar() != EOF) if(c >= '0' && c <= '9') ++ndigit[ c-'0' ]; else if(c == ' ' || c == '\n' || c == '\t') ++nwhite; else ++nother; printf("digit = "); for(i=0; i<10; ++i) printf(" %d", ndigit[i]); printf(", white space = %d, other = %d\n", nwhite, nother); }

in the interval [0,9]

Multi-way decision

if( expression ){ statement 1; }else if { statement 2; }…else{ statement N;}

statement K;

TestYES

Statement 1

NO

IF

Statement K

TestYES

Statement 2

else if

else if…

Test

else

NO

Statement N

YESStatement N-1

…

Outline


Functions Function, subroutine, procedure printf, getchar, putchar, …

input

output

…body…(hidden from user)

Example: Power Power(2,3) -> 23

#include <stdio.h>

int power(int m, int n);

main(){ int i; for(i=0; i<10; ++i) printf("%d %d %d\n", i, power(2,i), power(-3,i)); return 0;}

function prototypec.f. function definitionint power(int, int);

int power(int base, int n){ int i, p; p = 1; for(i=1; i<=n; ++i) p = p*base; return p;}

Return-type function-name(parameter declarations…){Declarationsstatements}

Samefile?

These variable names are localto this function

Functions (cont.)

power(int base, int n)

return p;

…body…(hidden from user)

power(2,i)

arguments(formal arguments)

parameters(actual arguments)

Early definition

int power();…int power(int, int)int base, n;{ int i, p; p = 1; for(i=1; i<=n; ++i) p = p*base; return p;}

? Compiler cannotcheck the correctnessof the parameters of function calls

Outline


Call by value

Store-program concept Program is data

Data segment

Program segment

…

…

memory

Data segment

Program segment

main

power

…

address

2i

Program segment

…

…

memory

Program segment

main

power

…

address

power(2,i)

basen

Call by value

Call by value - example

int power(int base, int n){ int p;

for(p=1; n>0; --n) p = p * base; return p;}

int power(int base, int n){ int i, p; p = 1; for(i=1; i<=n; ++i) p = p*base; return p;}

Outline


Character arrays Character arrays = strings

“Hello\n”

H e l l o \n \0

Null character

H e l l o \n \0s

結尾字元char s[10];

S[0]S[1]S[2]S[3] …

… …

S[9]

Character arrays: example Read text lines and print the longest

while (there’s another line) if (it’s longer than the previous longest) save it save its lengthprint longest line

Algorithm outline

#include <stdio.h>#define MAXLINE 1000

int getline(char line[], int maxline);void copy(char to[], char from[]);

main(){ int len; int max; char line[MAXLINE]; char longest[MAXLINE]; max = 0; while((len=getline(line, MAXLINE)) > 0) if(len > max){ max = len; copy(longest, line); } if(max > 0) printf("%s", longest); return 0;}

0 if no input

int getline(char s[], int lim){ int c, i; for(i=0; i<lim-1 && (c=getchar())!=EOF && c!='\n'; ++i) s[i] = c; if(c == '\n'){ s[i]=c; ++i; } s[i]='\0'; return i;}

void copy(char to[], char from[]){ int i i=0; while((to[i] = from[i]) != '\0') ++i;}

left to right evaluation

when i=0?

lineMAXLINE

…

…

memory

Program segment

main

getline

…

address

slim

Call by valuefor char arrays

int getline(line, MAXLINE)…

…

char line[MAXLINE]

line[0]…line[MAXLINE-1]

Outline


Scope of variables Local variables in a function

Automatic variables

int getline(char s[], int lim){ int c, i; for(i=0; i<lim-1 && (c=getchar())!=EOF && c!='\n'; ++i) s[i] = c;

…}

code segment

Data segment

enter

Allocatewhen entering

#include <stdio.h>#define MAXLINE 1000

int max;char line[MAXLINE];char longest[MAXLINE];

int getline(void);void copy(void);

main(){ int len; extern int max; extern char longest[]; max = 0; while((len=getline()) > 0) if(len > max){ max = len; copy(); } if(max > 0) printf("%s", longest); return 0;}

int getline(void){ int c, i; extern char line[]; for(i=0; i<lim-1 && (c=getchar())!=EOF && c!='\n'; ++i) s[i] = c; if(c == '\n'){ s[i]=c; ++i; } s[i]='\0'; return i;}

void copy(void){ int i extern char line[], longest[]; i=0; while((to[i] = from[i]) != '\0') ++i;}

External var.definition

External var.declaration

External variables Advantage

Less communication of variables in functions calls

Disadvantage Variables can be changed in unexpected

way Functions lose its generality (must live with

the external variables…)

External variables (cont.) Write them in a header file

extern int max;extern char line[];extern char longest[];

ggyy.h

#include <ggyy.h>

int getline(void){ int c, i;

for(i=0; …}

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The C programming language: Introduction Fall 2003, Jen-Chang Liu.

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