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Network Programming in C第３回「 C Fundamentals ～ command line arguments, pointers, and structures 」
２０ 10 Fall semesterRodney Van Meter
授業 Web ページ SFC-SFShttps://vu.sfc.keio.ac.jp/sfc-sfs/
本講義を「 MY 時間割（仮）」へ登録してください
課題・授業資料などの情報を掲示します 課題は毎回こちらに提出！！ Deadline
Due Tuesday, 10/19 10:59 a.m.
今期の授業スケジュール（予定） 第１回 9/28 ： Introduction / イントロダクション 第２回 10/5 ： C Basics ～ Functions, Variables, Data
Types ・ Makefiles 第３回 10/12 ： C Basics ～ Command Line Arguments
・ Structures ・ Pointers 第４回 10/19 ： C Basics ～ Pointers & Arrays ・ Lists 第５回 10/26 ： file I/O ・ Network Programming 第６回 11/2 ： Network Programming (1) 第７回 11/9 ： Network Programming (2) 第８回 11/16 ： Network Programming （ 3 ） 第９回 11/30 ： Applied Network Programming （ 1 ） 第１０回 12/7 ： Applied Network Programming （２） 第１１回 12/14 ： Work on Projects 第１ 2 回 12/21 ： Work on Projects 第１３回 1/19 ： Final Presentations!!!
今日のお題 Review last week’s homework コマンドライン引数 ポインタの基礎 構造体
課題 1:Homework Prob. 1 浮動小数点のレーンジ(fprange.c)
1. 最小： Write a program that starts withx = 1.0, and divides it by two repeatedly until it becomes zero
2. 最大： Do the same thing getting larger until something happens (what happens?)
3. Do for both “float” and “double”.
課題 2
Makefile を作ってください 一行目は：
all: hello typesizes fortest cline numbers
コマンドライン引数
Command Line Arguments ：argc,argv
What’s a command line argument?% cat –n file% cat file1 file2 file3
Using Arguments main( int argc, char **argv) main( int argc, char *argv[])
argc is number of args argv[0] is command name argv[1] is first argument argv[2] is second argument
Exercise: Using the args#include <stdio.h>int main(int argc, char *argv[]){
int i;for (i=0; i < argc; i++){
printf(“\t%s\n”, argv[i]);}
return 0;
}
Pointers
What’s a Pointer? A pointer is a variable
It points to another variable int i = 10; int j = 20; int *ptr
meaning of * in declaration, says this var
is a pointer to the named typeVariables & addresses
0
Address
ptr
j
i
100
104
108
What’s a Pointer? A pointer is a variable
It points to another variable int i = 10; int j = 20; int *ptr
meaning of * in declaration, says this var
is a pointer to the named typeVariables & addresses
0
Address
ptr
j
i
100
104
108
Pointer Exercise int i = 10; int j = 20; int *ptr = &i;
printf(“&i=%d\n”, &i); printf(“ptr=%d\n”, ptr); printf(“i=%d\n”, i); printf(“*ptr=%d\n”,*ptr); ptr points to variable i
変数とアドレス0
アドレス
ptr
j
i
100
104
108 10
20
108
int x=1, y=5; int z[10]; int *p; p=&x; /* p points to x */ y=*p; /* y gets 1 */ *p = 0; /* x becomes 0 */ p=&z[2]; /* p points to z[2] */
More on Pointers
Pointers in Functions (1)#include <stdio.h>void swap(int x, int y){
int temp;temp = x;x = y;y = temp;
}
int main(){int a = 5;int b = 3;swap (a,b);printf(“a=%d\n”, a);printf(“b=%d\n”,b);return 0;
}
int 型の変数を入れ替えるプログラム
Try this out: does it swap a & b?
Pointers in Functions (2)#include <stdio.h>void swap(int x, int y){
int temp;temp = x;x = y;y = temp;
}
int main(){int a = 5;int b = 3;swap (a,b);printf(“a=%d\n”, a);printf(“b=%d\n”,b);return 0;
}
変数とアドレス0
アドレス
b
a
104
108
5
3
x
y
94
90
交換
Pointers in Functions (3)#include <stdio.h>void swap(int *x, int *y){
int temp;temp = *x;*x = *y;*y = temp;
}
int main(){int a = 5;int b = 3;swap (&a,&b);printf(“a=%d\n”, a);printf(“b=%d\n”,b);return 0;
}
int 型の変数を入れ替えるプログラム
変数とアドレス0
b
a
104
108
108
104
x
y
94
90
*x
*y
Exchange
Data Structures
構造体 (structure) A structure can hold more than one piece
of data, of the same or different types Ex: Holding ID & test score for a student
struct student{int id;int score;
}; Different element types:
struct student2{char name[32];int score;
};
構造体の定義
struct 構造体名 { 型 メンバ名 1; 型 メンバ名 2;
：：
};
使用例
struct student{int id;int score;
};
定義の方法
Using structs as variables
After defining struct, can use when defining variables Instead of type, use struct name
int a;struct student b;
Accessing struct members (1)
Simplest way is using . ( ドット ) 演算子 Ex: struct student’s score value:
struct student b = {70000000,70};printf(“b の点数は %d です \n”, b.score);
Accessing struct members (2)
Can also use a pointer to a struct: -> 演算子 Ex: score in struct student
struct student b = {70000000,70};struct student *c = &b;printf(“b の点数は %d です \n”, c->score);
In place of -> can also use(*pointer).member(*c).score
Using structures: example 1
int main(){
int i;struct student{
int id;int score;
} students[5] = { {70001234,80}, {70204578,70}, {70157384,60}, {70355678,75}, {70207658,49} };
for(i=0;i<5;i++){printf("student id:%d, score:%d\n",
students[i].id, students[i].score);}
}
Definition and initialization together
Structure definition separate from variable declaration & initialization
struct student{int id;int score;
};
int main(){
int i;struct student students[5] = { {70001234,80},
{70204578,70}, {70157384,60}, {70355678,75}, {70207658,49} };
for(i=0;i<5;i++){printf("student id:%d, score:%d\n", students[i].id,
students[i].score);}
}
Using structures: example 2
Initialize at run time instead of compile time
struct student{int id;int score;
};
int main(){
int i;struct student students[5];for(i=0; i<5; i++){
students[i].id = i;students[i].score = i;
}for(i=0;i<5;i++){
printf("student id:%d, score:%d\n", students[i].id, students[i].score);}
}
Using structures: example 3
typedef
typedef struct student{int id;int score;
} STUDENT;STUDENT students[5];
Pointers to structs
struct student st;struct student *sp;sp = &st;sp->id = 7000123;(*sp).score = 23;
sp
7000123
st
23
id
score
printf(“%d\n”, sp->score);
Practice Problem 1 Write a program that accepts as input:
student number login name full name
%./a.out 80749423 kazuhisa “kazuhisa matsuzono”
Number: 7045678Login name: kazuhisaName: kazuhisa matsuzono
Example (Not quite what we’re doing!)
#include <stdio.h>struct student{
char name[20];int math;int jlang;
};void clear(struct student *sei);
int main(){struct student seito;/* clearing data in seito */clear(&seito);
return 0;}
void clear(struct student *sei){strcpy (sei->name, “”);sei->math=0;sei->jlang=0;
}
課題 / Homework Finish Practice #1 Determine size of struct student Determine size of array of struct
student Use memcpy() to copy struct Use assignment to copy struct Use memset() to clear memory
