/ SCHEDULE /06/07(Tue) / Basic of Programming /06/09(Thu) / Fundamental structures /06/14(Tue) / Memory Management /06/1

I117 II I117 PROGRAMMING PRACTICE II SOFTWARE DEVELOPMENT ENV. 1 Research Center for Advanced Computing Infrastructure (RCACI) / Yasuhiro Ohara

I117 II I117 PROGRAMMING PRACTICE II SOFTWARE DEVELOPMENT ENV. 1 Research Center for Advanced Computing Infrastructure (RCACI) / Yasuhiro Ohara [email protected] / SCHEDULE 1. 2011/06/07(Tue) / Basic of

r07.dvi

19 7 ( ) 2019.4.20 1 1.1 (data structure ( (dynamic data structure 1 malloc C free C (garbage collection GC C GC(conservative GC 2 1.2 data next p 3 5 7 9 p 3 5 7 9 p 3 5 7 9 1 1: (single linked list 1

ohp07.dvi

19 7 ( ) 2019.4.20 1 (data structure) ( ) (dynamic data structure) 1 malloc C free 1 (static data structure) 2 (2) C (garbage collection GC) C GC(conservative GC) 2 2 conservative GC 3 data next p 3 5

I117 プログラミング演習II

I117 プログラミング演習 II I117 PROGRAMMING PRACTICE II メモリ管理 1 MEMORY MANAGEMENT 1 情報社会基盤研究センター Research Center for Advanced Computing Infrastructure (RCACI) 小原泰弘 / Yasuhiro Ohara [email protected] スケジュール / SCHEDULE

ohp03.dvi

19 3 ( ) 2019.4.20 CS 1 (comand line arguments) Unix./a.out aa bbb ccc ( ) C main void int main(int argc, char *argv[]) {... 2 (2) argc argv argc ( ) argv (C char ) ( 1) argc 4 argv NULL. / a. o u t \0

r03.dvi

19 ( ) 019.4.0 CS 1 (comand line arguments) Unix./a.out aa bbb ccc ( ) C main void... argc argv argc ( ) argv (C char ) ( 1) argc 4 argv NULL. / a. o u t \0 a a \0 b b b \0 c c c \0 1: // argdemo1.c ---

untitled

C -1 - -2 - concept lecture keywords FILE, fopen, fclose, fscanf, fprintf, EOF, r w a, typedef gifts.dat Yt JZK-3 Jizake tsumeawase 45 BSP-15 Body soap set 3 BT-2 Bath towel set 25 TEA-2 Koutya

r08.dvi

19 8 ( ) 019.4.0 1 1.1 (linked list) ( ) next ( 1) (head) (tail) ( ) top head tail head data next 1: NULL nil ( ) NULL ( NULL ) ( 1 ) (double linked list ) ( ) 1 next 1 prev 1 head cur tail head cur prev

II ( ) prog8-1.c s1542h017%./prog8-1 1 => 35 Hiroshi 2 => 23 Koji 3 => 67 Satoshi 4 => 87 Junko 5 => 64 Ichiro 6 => 89 Mari 7 => 73 D

II 8 2003 11 12 1 6 ( ) prog8-1.c s1542h017%./prog8-1 1 => 35 Hiroshi 2 => 23 Koji 3 => 67 Satoshi 4 => 87 Junko 5 => 64 Ichiro 6 => 89 Mari 7 => 73 Daisuke 8 =>. 73 Daisuke 35 Hiroshi 64 Ichiro 87 Junko

joho07-1.ppt

0xbffffc5c 0xbffffc60 xxxxxxxx xxxxxxxx 00001010 00000000 00000000 00000000 01100011 00000000 00000000 00000000 xxxxxxxx x y 2 func1 func2 double func1(double y) { y = y + 5.0; return y; } double func2(double*

I117 II I117 PROGRAMMING PRACTICE II SCRIPT LANGUAGE 1 Research Center for Advanced Computing Infrastructure (RCACI) / Yasuhiro Ohara

I117 II I117 PROGRAMMING PRACTICE II SCRIPT LANGUAGE 1 Research Center for Advanced Computing Infrastructure (RCACI) / Yasuhiro Ohara [email protected] / SCHEDULE 1. 2011/06/07(Tue) / Basic of Programming

ex14.dvi

1,, 0, b (b b 2 b ) n k n = n j b j, (0 n j b 1), n =(n k n k 1...n 1 n 0 ) b, n j j j +1, 0,...,b 1 (digit). b b, n b 1 ñ, ñ = k (b 1 n j )b j b N, n b n, n = b N n, n =ñ+1 b N, n m n + m (mod b N ),

untitled

II yacc 005 : 1, 1 1 1 %{ int lineno=0; 3 int wordno=0; 4 int charno=0; 5 6 %} 7 8 %% 9 [ \t]+ { charno+=strlen(yytext); } 10 "\n" { lineno++; charno++; } 11 [^ \t\n]+ { wordno++; charno+=strlen(yytext);}

Introduction Purpose This training course demonstrates the use of the High-performance Embedded Workshop (HEW), a key tool for developing software for

Introduction Purpose This training course demonstrates the use of the High-performance Embedded Workshop (HEW), a key tool for developing software for embedded systems that use microcontrollers (MCUs)

Microsoft Word - Cプログラミング演習(12)

第 12 回 (7/9) 4. いくつかのトピック (5)main 関数の引数を利用したファイル処理 main 関数は, 起動する環境から引数を受け取ることができる 例えば 次に示すように,main 関数に引数を用いたプログラムを作成する 01 /* sample */ 02 /* main 関数の引数 */ 03 #include 04 05 main(int argc, char

program.dvi

2001.06.19 1 programming semi ver.1.0 2001.06.19 1 GA SA 2 A 2.1 valuename = value value name = valuename # ; Fig. 1 #-----GA parameter popsize = 200 mutation rate = 0.01 crossover rate = 1.0 generation

新・明解Ｃ言語 ポインタ完全攻略

2 1-1 1-1 /* 1-1 */ 1 int n = 100; int *p = &n; printf(" n %d\n", n); /* n int */ printf("*&n %d\n", *&n); /* *&n int */ printf(" p %p\n", p); /* p int * */ printf("&*p %p\n", &*p); /* &*p int * */ printf("sizeof(n)

1 C STL(1) C C C libc C C C++ STL(Standard Template Library ) libc libc C++ C STL libc STL iostream Algorithm libc STL string vector l

C/C++ 2007 6 18 1 C STL(1) 2 1.1............................................... 2 1.2 stdio................................................ 3 1.3.......................................... 10 2 11 2.1 sizeof......................................

3.1 stdio.h iostream List.2 using namespace std C printf ( ) %d %f %s %d C++ cout cout List.2 Hello World! cout << float a = 1.2f; int b = 3; cout <<

C++ C C++ 1 C++ C++ C C++ C C++? C C++ C *.c *.cpp C cpp VC C++ 2 C++ C++ C++ [1], C++,,1999 [2],,,2001 [3], ( )( ),,2001 [4] B.W. /D.M.,, C,,1989 C Web [5], http://kumei.ne.jp/c_lang/ 3 Hello World Hello

1-4 int a; std::cin >> a; std::cout << "a = " << a << std::endl; C++( 1-4 ) stdio.h iostream iostream.h C++ include.h 1-4 scanf() std::cin >>

1 C++ 1.1 C C++ C++ C C C++ 1.1.1 C printf() scanf() C++ C hello world printf() 1-1 #include printf( "hello world\n" ); C++ 1-2 std::cout

1 # include < stdio.h> 2 # include < string.h> 3 4 int main (){ 5 char str [222]; 6 scanf ("%s", str ); 7 int n= strlen ( str ); 8 for ( int i=n -2; i

ABC066 / ARC077 writer: nuip 2017 7 1 For International Readers: English editorial starts from page 8. A : ringring a + b b + c a + c a, b, c a + b + c 1 # include < stdio.h> 2 3 int main (){ 4 int a,

mstrcpy char *mstrcpy(const char *src); mstrcpy malloc (main free ) stdio.h fgets char *fgets(char *s, int size, FILE *stream); s size ( )

2008 3 10 1 mstrcpy char *mstrcpy(const char *src); mstrcpy malloc (main free ) stdio.h fgets char *fgets(char *s, int size, FILE *stream); s size ( ) stream FILE ( man ) 40 ( ) %./a.out String : test

Ｃ言語によるアルゴリズムとデータ構造

Algorithms and Data Structures in C 4 algorithm List - /* */ #include List - int main(void) { int a, b, c; int max; /* */ Ÿ 3Ÿ 2Ÿ 3 printf(""); printf(""); printf(""); scanf("%d", &a); scanf("%d",

tuat1.dvi

( 1 ) http://ist.ksc.kwansei.ac.jp/ tutimura/ 2012 6 23 ( 1 ) 1 / 58 C ( 1 ) 2 / 58 2008 9 2002 2005 T E X ptetex3, ptexlive pt E X UTF-8 xdvi-jp 3 ( 1 ) 3 / 58 ( 1 ) 4 / 58 C,... ( 1 ) 5 / 58 6/23( )

untitled

II 4 Yacc Lex 2005 : 0 1 Yacc 20 Lex 1 20 traverse 1 %% 2 [0-9]+ { yylval.val = atoi((char*)yytext); return NUM; 3 "+" { return + ; 4 "*" { return * ; 5 "-" { return - ; 6 "/" { return / ; 7 [ \t] { /*

file"a" file"b" fp = fopen("a", "r"); while(fgets(line, BUFSIZ, fp)) {... fclose(fp); fp = fopen("b", "r"); while(fgets(line, BUFSIZ, fp)) {... fclose

I117 9 2 School of Information Science, Japan Advanced Institute of Science and Technology file"a" file"b" fp = fopen("a", "r"); while(fgets(line, BUFSIZ, fp)) {... fclose(fp); fp = fopen("b", "r"); while(fgets(line,

void hash1_init(int *array) int i; for (i = 0; i < HASHSIZE; i++) array[i] = EMPTY; /* i EMPTY */ void hash1_insert(int *array, int n) if (n < 0 n >=

II 14 2018 7 26 : : [email protected] 14,, 8 2 12:00 1 O(1) n O(n) O(log n) O(1) 32 : 1G int 4 250 M 2.5 int 21 2 0 100 0 100 #include #define HASHSIZE 100 /* */ #define NOTFOUND 0

lexex.dvi

(2018, c ) http://istksckwanseiacjp/ ishiura/cpl/ 4 41 1 mini-c lexc,, 2 testlexc, lexc mini-c 1 ( ) mini-c ( ) (int, char, if, else, while, return 6 ) ( ) (+, -, *, /, %, &, =, ==,!=, >, >=,

1 1.1 C 2 1 double a[ ][ ]; 1 3x x3 ( ) malloc() malloc 2 #include <stdio.h> #include

1 1.1 C 2 1 double a[ ][ ]; 1 3x3 0 1 3x3 ( ) 0.240 0.143 0.339 0.191 0.341 0.477 0.412 0.003 0.921 1.2 malloc() malloc 2 #include #include #include enum LENGTH = 10 ; int

ex01.dvi

,. 0. 0.0. C () /******************************* * $Id: ex_0_0.c,v.2 2006-04-0 3:37:00+09 naito Exp $ * * 0. 0.0 *******************************/ #include int main(int argc, char **argv) double

memo

数理情報工学演習第一 C プログラミング演習 ( 第 5 回 ) 2015/05/11 DEPARTMENT OF MATHEMATICAL INFORMATICS 1 今日の内容 : プロトタイプ宣言 ヘッダーファイル, プログラムの分割 課題 : 疎行列 2 プロトタイプ宣言 3 C 言語では, 関数や変数は使用する前 ( ソースの上のほう ) に定義されている必要がある. double sub(int

& & a a * * ptr p int a ; int *a ; int a ; int a int *a

int a = 123; a 123 :100 a 123 int *ptr = & a; a ptr ptr a 100 a 123 200 *ptr 200 a & & a a * * ptr p --------------------------------------------------------------------------------------------- int a

超初心者用

3 1999 10 13 1. 2. hello.c printf( Hello, world! n ); cc hello.c a.out./a.out Hello, world printf( Hello, world! n ); 2 Hello, world printf n printf 3. ( ) int num; num = 100; num 100 100 num int num num

BW BW

Induced Sorting BW 11T2042B 2015 3 23 1 1 1.1................................ 1 1.2................................... 1 2 BW 1 2.1..................................... 2 2.2 BW.................................

Introduction Purpose This training course describes the configuration and session features of the High-performance Embedded Workshop (HEW), a key tool

Introduction Purpose This training course describes the configuration and session features of the High-performance Embedded Workshop (HEW), a key tool for developing software for embedded systems that

,,,,., C Java,,.,,.,., ,,.,, i

24 Development of the programming s learning tool for children be derived from maze 1130353 2013 3 1 ,,,,., C Java,,.,,.,., 1 6 1 2.,,.,, i Abstract Development of the programming s learning tool for children

PowerPoint プレゼンテーション

プログラミング応用演習 第 4 回再帰的構造体 プログラミングを 余談 : 教えることの難しさ 丁寧に説明しないと分かってもらえない 説明すると 小難しくなる学生が目指すべきところプログラム例を説明されて理解できる違うやり方でも良いので自力で解決できる おっけー 動けば良い という意識でプログラミング 正しく動くことのチェックは必要 解答例と自分のやり方との比較が勉強になる 今日のお題 再帰的構造体

yacc.dvi

2017 c 8 Yacc Mini-C C/C++, yacc, Mini-C, run,, Mini-C 81 Yacc Yacc, 1, 2 ( ), while ::= "while" "(" ")" while yacc 1: st while : lex KW WHILE lex LPAREN expression lex RPAREN statement 2: 3: $$ = new

(K&R 2.9) ~, &,, >>, << 2. (K&R 5.7) 3. (K&R 5.9) 4. (K&R 5.10) (argc argv atoi(), atof() ) 5. (K&R 7.5) (K&R 7.6) - FILE, stdin, stdout, std

3 3 1. 1 (K&R.9) ~, &,, >>,

ex01.dvi

,. 0. 0.0. C () /******************************* * $Id: ex_0_0.c,v.2 2006-04-0 3:37:00+09 naito Exp $ * * 0. 0.0 *******************************/ #include int main(int argc, char **argv) { double

programmingII2019-v01

II 2019 2Q A 6/11 6/18 6/25 7/2 7/9 7/16 7/23 B 6/12 6/19 6/24 7/3 7/10 7/17 7/24 x = 0 dv(t) dt = g Z t2 t 1 dv(t) dt dt = Z t2 t 1 gdt g v(t 2 ) = v(t 1 ) + g(t 2 t 1 ) v v(t) x g(t 2 t 1 ) t 1 t 2

WinHPC ppt

MPI.NET C# 2 2009 1 20 MPI.NET MPI.NET C# MPI.NET C# MPI MPI.NET 1 1 MPI.NET C# Hello World MPI.NET.NET Framework.NET C# API C# Microsoft.NET java.net (Visual Basic.NET Visual C++) C# class Helloworld

バイオプログラミング第 1 榊原康文 佐藤健吾 慶應義塾大学理工学部生命情報学科

バイオプログラミング第 1 榊原康文 佐藤健吾 慶應義塾大学理工学部生命情報学科 ポインタ変数の扱い方 1 ポインタ変数の宣言 int *p; double *q; 2 ポインタ変数へのアドレスの代入 int *p; と宣言した時,p がポインタ変数 int x; と普通に宣言した変数に対して, p = &x; は x のアドレスのポインタ変数 p への代入 ポインタ変数の扱い方 3 間接参照 (

pptx

iphone 2010 8 18 C [email protected] C Hello, World! Hello World hello.c! printf( Hello, World!\n );! os> ls! hello.c! os> cc hello.c o hello! os> ls! hello!!hello.c! os>./hello! Hello, World!! os>! os>

HARK Designer Documentation 0.5.0 HARK support team 2013 08 13 Contents 1 3 2 5 2.1.......................................... 5 2.2.............................................. 5 2.3 1: HARK Designer.................................

I. Backus-Naur BNF S + S S * S S x S +, *, x BNF S (parse tree) : * x + x x S * S x + S S S x x (1) * x x * x (2) * + x x x (3) + x * x + x x (4) * *

2015 2015 07 30 10:30 12:00 I. I VI II. III. IV. a d V. VI. 80 100 60 1 I. Backus-Naur BNF S + S S * S S x S +, *, x BNF S (parse tree) : * x + x x S * S x + S S S x x (1) * x x * x (2) * + x x x (3) +

XcalableMP入門

XcalableMP 1 HPC-Phys@, 2018 8 22 XcalableMP XMP XMP Lattice QCD!2 XMP MPI MPI!3 XMP 1/2 PCXMP MPI Fortran CCoarray C++ MPIMPI XMP OpenMP http://xcalablemp.org!4 XMP 2/2 SPMD (Single Program Multiple Data)

Page 1 of 6 B (The World of Mathematics) November 20, 2006 Final Exam 2006 Division: ID#: Name: 1. p, q, r (Let p, q, r are propositions. ) (10pts) (a

Page 1 of 6 B (The World of Mathematics) November 0, 006 Final Exam 006 Division: ID#: Name: 1. p, q, r (Let p, q, r are propositions. ) (a) (Decide whether the following holds by completing the truth

解きながら学ぶＣ言語

printf 2-5 37 52 537 52 printf("%d\n", 5 + 37); 5370 source program source file.c ex00.c 0 comment %d d 0 decimal -2 -p.6 3-2 5 37 5 37-22 537 537-22 printf("537%d\n", 5-37); function function call ( )argument,

comment.dvi

( ) (sample1.c) (sample1.c) 2 2 Nearest Neighbor 1 (2D-class1.dat) 2 (2D-class2.dat) (2D-test.dat) 3 Nearest Neighbor Nearest Neighbor ( 1) 2 1: NN 1 (sample1.c) /* -----------------------------------------------------------------

# let st1 = {name = "Taro Yamada"; id = };; val st1 : student = {name="taro Yamada"; id=123456} { 1 = 1 ;...; n = n } # let string_of_student {n

II 6 / : 2001 11 21 (OCaml ) 1 (field) name id type # type student = {name : string; id : int};; type student = { name : string; id : int; } student {} type = { 1 : 1 ;...; n : n } { 1 = 1 ;...; n = n

£Ã¥×¥í¥°¥é¥ß¥ó¥°ÆþÌç (2018) - Â裵²ó ¨¡ À©¸æ¹½Â¤¡§¾ò·ïʬ´ô ¨¡

(2018) 2018 5 17 0 0 if switch if if ( ) if ( 0) if ( ) if ( 0) if ( ) (0) if ( 0) if ( ) (0) ( ) ; if else if ( ) 1 else 2 if else ( 0) 1 if ( ) 1 else 2 if else ( 0) 1 if ( ) 1 else 2 (0) 2 if else

新版明解Ｃ言語 実践編

2 List - "max.h" a, b max List - max "max.h" #define max(a, b) ((a) > (b)? (a) : (b)) max List -2 List -2 max #include "max.h" int x, y; printf("x"); printf("y"); scanf("%d", &x); scanf("%d", &y); printf("max(x,

1.ppt

/* * Program name: hello.c */ #include int main() { printf( hello, world\n ); return 0; /* * Program name: Hello.java */ import java.io.*; class Hello { public static void main(string[] arg)

: CR (0x0d) LF (0x0a) line separator CR Mac LF UNIX CR+LF MS-DOS WINDOWS Japan Advanced Institute of Science and Technology

I117 8 1 School of Information Science, Japan Advanced Institute of Science and Technology : CR (0x0d) LF (0x0a) line separator CR Mac LF UNIX CR+LF MS-DOS WINDOWS Japan Advanced Institute of Science and

Minimum C Minimum C Minimum C BNF T okenseq W hite Any D

6 2019 5 14 6.1 Minimum C....................... 6 1 6.2....................................... 6 7 6.1 Minimum C Minimum C BNF T okenseq W hite Any Digit ::= 0 1 2... 9. Number ::= Digit Digit. Alphabet

{:from => Java, :to => Ruby } Java Ruby KAKUTANI Shintaro; Eiwa System Management, Inc.; a strong Ruby proponent http://kakutani.com http://www.amazon.co.jp/o/asin/4873113202/kakutani-22 http://www.amazon.co.jp/o/asin/477413256x/kakutani-22

cpp1.dvi

2017 c 1 C++ (1) C C++, C++, C 11, 12 13 (1) 14 (2) 11 1 n C++ //, [List 11] 1: #include // C 2: 3: int main(void) { 4: std::cout

エラー処理・分割コンパイル・コマンドライン引数

L10(2017-12-05 Tue) : Time-stamp: 2017-12-17 Sun 11:59 JST hig. recv/send http://hig3.net ( ) L10 (2017) 1 / 21 IP I swallow.math.ryukoku.ac.jp:13 = 133.83.83.6:13 = : IP ( = ) (well-known ports), :. :,.

Microsoft Word - j201drills27.doc

Drill 1: Giving and Receiving (Part 1) Directions: Describe each picture using the verb of giving and the verb of receiving. (1) (2) (3) (4) 1 (5) (6) Drill 2: Giving and Receiving (Part 1) Directions:

1) OOP 2) ( ) 3.2) printf Number3-2.cpp #include <stdio.h> class Number Number(); // ~Number(); // void setnumber(float n); float getnumber();

: : :0757230G :2008/07/18 2008/08/17 1) OOP 2) ( ) 3.2) printf Number3-2.cpp #include class Number Number(); // ~Number(); // void setnumber(float n); float getnumber(); private: float num; ;

Sport and the Media: The Close Relationship between Sport and Broadcasting SUDO, Haruo1) Abstract This report tries to demonstrate the relationship be

Sport and the Media: The Close Relationship between Sport and Broadcasting SUDO, Haruo1) Abstract This report tries to demonstrate the relationship between broadcasting and sport (major sport and professional

AtCoder Regular Contest 073 Editorial Kohei Morita(yosupo) A: Shiritori if python3 a, b, c = input().split() if a[len(a)-1] == b[0] and b[len(

AtCoder Regular Contest 073 Editorial Kohei Morita(yosupo) 29 4 29 A: Shiritori if python3 a, b, c = input().split() if a[len(a)-1] == b[0] and b[len(b)-1] == c[0]: print( YES ) else: print( NO ) 1 B:

Microsoft Word - j201drills27.doc

Drill 1: Giving and Receiving (Part 1) [Due date: ] Directions: Describe each picture using the verb of giving and the verb of receiving. E.g.) (1) (2) (3) (4) 1 (5) (6) Drill 2: Giving and Receiving (Part

第1回 プログラミング演習3 センサーアプリケーション

C プログラミング - ポインタなんて恐くない! - 藤田悟 [email protected] 目標 C 言語プログラムとメモリ ポインタの関係を深く理解する C 言語プログラムは メモリを素のまま利用できます これが原因のエラーが多く発生します メモリマップをよく頭にいれて ポインタの動きを理解できれば C 言語もこわくありません 1. ポインタ入門編 ディレクトリの作成と移動 mkdir