Programming Language B

Transcription

1 Programming Language B Takako Nemoto (JAIST) 17 December Takako Nemoto (JAIST) 17 December 1 / 17

2 A tip for the last homework 1 Do Exercise 9-5 ( 9-5) in p.249 (in the latest edtion). The type of the second argument of str_char could be changed. Make a function int str_char (const char char s[], int c){... s.t. if a string s contains the character c, then it returns the smallest index of c; otherwise, it returns 1. Make also the function main so that we can input the string and the character you want to search and see the result. Takako Nemoto (JAIST) 17 December 2 / 17

3 Error sample 1 % c is a conversion specifier for single character. The reason this program does not work seems the Enter key at the end of the input of the first string is recognized as the second character input. To avoid it, we can use scanf("%*c%c", &search);. The first %*c means ignore the input. Takako Nemoto (JAIST) 17 December 3 / 17

4 Error sample 2 The reason the second program does not work seems the length of the array search. Since, as we studied last time, the length of the string in the system is always the length of string+'\0'. To avoid the problem, we should change the length of search more than 2. Takako Nemoto (JAIST) 17 December 4 / 17

5 How should we do? #include <stdio.h> /*10-1.c*/ void sum_diff(int n1, int n2, int sum, int diff){ sum = n1 + n2; diff = (n1 > n2)? n1 - n2 : n2 - n1; int na, nb; int wa = 0, sa = 0; puts("input two integers: "); printf("integer 1: "); scanf("%d", &na); printf("integer 2: "); scanf("%d", &nb); sum_diff(na, nb, wa, sa); printf("the sum is %d and the difference is %d.\n", wa, sa); return 0; Takako Nemoto (JAIST) 17 December 5 / 17

6 Why it does not work well? First of all, the function can return only one value. (This is why we set void for the return value of sum_diff.) The function sum_diff calls only the value of na, nb, wa, sa. It does not access the memory location for them, so they cannot rewrite the value of them. Such call of arguments are called call-by-value. Takako Nemoto (JAIST) 17 December 6 / 17

7 Address operator & #include <stdio.h> int n; double x; int a[3]; printf("the address of n: %p\n", &n); printf("the address of x: %p\n", &x); printf("the address of a[0]: %p\n", &a[0]); printf("the address of a[1]: %p\n", &a[1]); printf("the address of a[2]: %p\n", &a[2]); return 0; Takako Nemoto (JAIST) 17 December 7 / 17

8 Address operator & #include <stdio.h> int n; double x; int a[3]; Unary & operator is called address operator. By applying it to an object, it returns the address in the memory location. The conversion specifier %p for addresses stands for pointer. printf("the address of n: %p\n", &n); printf("the address of x: %p\n", &x); printf("the address of a[0]: %p\n", &a[0]); printf("the address of a[1]: %p\n", &a[1]); printf("the address of a[2]: %p\n", &a[2]); return 0; Takako Nemoto (JAIST) 17 December 7 / 17

9 We have already used the address operator. Recall that we used & often in scanf. int n; scanf("%d", &n); This means Install the scaned value into the address of n. Takako Nemoto (JAIST) 17 December 8 / 17

10 Usage of pointers #include <stdio.h> int sato = 178; int sanaka = 175; int masaki = 179; int *isako, *hiroko; isako = &sato; hiroko = &masaki; printf("*isako = %d\n", *isako); printf("*hiroko = %d\n", *hiroko); isako = &sanaka; *hiroko = 180; printf("sato = %d\n", sato); printf("sanaka = %d\n", sanaka); printf("masaki = %d\n", masaki); printf("*isako = %d\n", *isako); printf("*hiroko = %d\n", *hiroko); return 0; Takako Nemoto (JAIST) 17 December 9 / 17

11 Usage of pointers #include <stdio.h> int sato = 178; int sanaka = 175; int masaki = 179; int *isako, *hiroko; isako = &sato; hiroko = &masaki; printf("*isako = %d\n", *isako); printf("*hiroko = %d\n", *hiroko); isako = &sanaka; *hiroko = 180; printf("sato = %d\n", sato); printf("sanaka = %d\n", sanaka); printf("masaki = %d\n", masaki); printf("*isako = %d\n", *isako); printf("*hiroko = %d\n", *hiroko); Variables declared with * is called pointer. In this example, isako and hiroko are pointers. Their type are called pointer type to int-type object, pointer type to int, or int * type. If you declare them as int *isako, hiroko then hiroko has int type. return 0; Takako Nemoto (JAIST) 17 December 9 / 17

12 Usage of pointers #include <stdio.h> int sato = 178; int sanaka = 175; int masaki = 179; int *isako, *hiroko; isako = &sato; hiroko = &masaki; printf("*isako = %d\n", *isako); printf("*hiroko = %d\n", *hiroko); isako = &sanaka; *hiroko = 180; printf("sato = %d\n", sato); printf("sanaka = %d\n", sanaka); printf("masaki = %d\n", masaki); printf("*isako = %d\n", *isako); printf("*hiroko = %d\n", *hiroko); Pointers has the value of addresses of the objects with declared type. When the address of x is assigned into a pointer p, we say p points x. In this example, isako points sato and hiroko points first masaki and changed to sanaka. Unary * operator is applied to pointers and takes the object the pointer points. If the pointer p points the object x, then *p is an alias (alternative name) for x. E.g., *hiroko = 180; has the same meaning masaki = 180;. return 0; Takako Nemoto (JAIST) 17 December 10 / 17

13 Pointers in functions #include <stdio.h> void hiroko(int *height){ if (*height < 180) *height = 180; int sato = 178; int sanaka = 175; int masaki = 179; hiroko(&masaki); printf("sato printf("sanaka printf("masaki = %d\n", sato); = %d\n", sanaka); = %d\n", masaki); The formal arguments int *height in the definition hiroko means it takes an address of int-type object. In the function call, we have to give an address for the argument of hiroko. return 0; Takako Nemoto (JAIST) 17 December 11 / 17

14 Some modification #include <stdio.h> /*10-5.c*/ void sum_diff(int n1, int n2, int *sum, int *diff){ *sum = n1 + n2; *diff = (n1 > n2)? n1 - n2 : n2 -n1; int na, nb; int wa = 0, sa = 0; puts("input two integers: "); printf("integer 1: "); scanf("%d", &na); printf("integer 2: "); scanf("%d", &nb); sum_diff(na, nb, &wa, &sa); printf("the sum is %d and the difference is %d.\n", wa, sa); return 0; Takako Nemoto (JAIST) 17 December 12 / 17

15 Pointers void sum_diff(int n1, int n2, int *sum, int *diff){ *sum = n1 + n2; *diff = (n1 > n2)? n1 - n2 : n2 -n1; The formal arguments int *sum and int *diff in void sum_diff(int n1, int n2, int *sum, int *diff) means that the function takes the address of the variable sum and diff. sum_diff(na, nb, &wa, &sa); Since the 3rd and 4th argument of sum_diff are addresses of int type objects, we have to give the address of wa and sa. The above function call gives the values of na and nb, and the addresses of wa and sa as arguments. In the function body, *sum and *diff act as aliases (alternative names) for wa and sa, respectively. Takako Nemoto (JAIST) 17 December 13 / 17

16 #include <stdio.h>/*10-6.c*/ void swap(int *px, int *py){ int temp = *px; *px = *py; *py = temp; int na, nb; puts("input two integers."); printf("integer A: "); scanf("%d", &na); printf("integer B: "); scanf("%d", &nb); swap(&na, &nb); puts("swap A and B!"); printf("integer A: %d\n", na); printf("integer B: %d\n", nb); return 0; Takako Nemoto (JAIST) 17 December 14 / 17

17 #include <stdio.h> void swap(int *px, int *py){ int temp = *px; *px = *py; *py = temp; void sort2(int *n1, int *n2){ if (*n1 > *n2) swap(n1, n2); In swap(n1, n2);, we don t need & in front of n1 and n2, since they declared as pointers and so has the value of some addresses. int na, nb; puts("input two integers."); printf("integer A: "); scanf("%d", &na); printf("integer B: "); scanf("%d", &nb); sort2(&na, &nb); puts("result of sorting in ascending order:"); printf("integer A: %d\n", na); printf("integer B: %d\n", nb); return 0; Takako Nemoto (JAIST) 17 December 15 / 17

18 #include <stdio.h> /*10-8.c*/ void swap(int *px, int *py){ int temp = *px; *px = *py; *py = temp; double da, db; puts("input two real numbers."); printf("a: "); scanf("%lf", &da); printf("b: "); scanf("%lf", &db); swap(&da, &db); puts("result of sorting in ascending order:"); printf("a: %f\n", da); printf("b: %f\n", db); return 0; Takako Nemoto (JAIST) 17 December 16 / 17

19 Today s homework 1 Do Exercise 10-1 ( 10-1): Make a function void adjust_value(int *n) s.t. if the value pointed by n is smaller than 0, then it updates the value as 0 and if the value is greater than 100, then it updates the value as 100. Make also main function so that the user can input the value and can see the value updated by the above function. 2 Do Exercise 10-3 ( 10-3): Make a function void sort3(int *n1, int *n2, int *n3) s.t. it sorts the 3 values pointed by n1, n2 and n3 in ascending order. Make also the main function so that the user can input 3 values and can see the result of sorting. 3 Read sections 10 and 11. As usual, send the programs and the text file as attached files to me via , with the title Homework Lecture 10 by the next lecture. Please make sure to include your name and student ID number. Takako Nemoto (JAIST) 17 December 17 / 17