Solutions to questions in Programming In C, completed in middle 2015

Author: zero4drift

Chapter10Exercises/1/1.1.c

@@ -0,0 +1,14 @@
+#include <stdio.h>
+
+int main(void)
+{
+	int count = 10, x;
+	int *int_pointer;
+	
+	int_pointer = &count;
+	x = *int_pointer;
+	
+	printf("count = %i, x = %i\n", count, x);
+	
+	return 0;
+}

Chapter10Exercises/1/1.10.c

@@ -0,0 +1,49 @@
+#include <stdio.h>
+
+struct entry
+	{
+		int          value;
+		struct entry *next;
+	};
+	
+struct entry *findEntry(struct entry *listPtr, int match)
+{
+	while(listPtr != (struct entry *) 0)
+		if(listPtr->value == match)
+			return listPtr;
+		else
+			listPtr = listPtr->next;
+	
+	return (struct entry *) 0;
+}
+
+
+int main(void)
+{
+	struct entry *findEntry(struct entry *listPtr, int match);
+	struct entry n1, n2, n3;
+	struct entry *listPtr, *listStart = &n1;
+	
+	int search;
+	
+	n1.value = 100;
+	n1.next = &n2;
+	
+	n2.value = 200;
+	n2.next = &n3;
+	
+	n3.value = 300;
+	n3.next = (struct entry *) 0;
+	
+	printf("Enter value to locate: ");
+	scanf("%i", &search);
+	
+	listPtr = findEntry(listStart, search);
+	
+	if(listPtr != (struct entry *) 0)
+		printf("Found %i.\n", listPtr->value);
+	else
+		printf("Not found.\n");
+	
+	return 0;
+}

Chapter10Exercises/1/1.11.c

@@ -0,0 +1,24 @@
+//此函数对一个整数数组中的各元素进行求和
+
+#include <stdio.h>
+
+int arraySum(int array[], const int n)
+{
+	int sum = 0, *ptr;
+	int *const arrayEnd = array + n;
+	
+	for(ptr = array; ptr < arrayEnd; ++ptr)
+		sum += *ptr;
+	
+	return sum;
+}
+
+int main(void)
+{
+	int arraySum(int array[], const int n);
+	int values[10] = {3, 7, -9, 3, 6, -1, 7, 9, 1, -5};
+	
+	printf("The sum is %i\n", arraySum(values, 10));
+	
+	return 0;
+}

Chapter10Exercises/1/1.12.c

@@ -0,0 +1,24 @@
+//此函数对一个整数数组中的各元素进行求和（第二版）
+
+#include <stdio.h>
+
+int arraySum(int array[], const int n)
+{
+	int sum = 0;
+	int *const arrayEnd = array + n;
+	
+	for(; array < arrayEnd; ++array)
+		sum += *array;
+	
+	return sum;
+}
+
+int main(void)
+{
+	int arraySum(int array[], const int n);
+	int values[10] = {3, 7, -9, 3, 6, -1, 7, 9, 1, -5};
+	
+	printf("The sum is %i\n", arraySum(values, 10));
+	
+	return 0;
+}

Chapter10Exercises/1/1.13.c

@@ -0,0 +1,24 @@
+#include <stdio.h>
+
+void copyString(char *to, char *from)
+{
+	for(; *from != '\0'; ++from, ++to)
+		*to = *from;
+	
+	*to = '\0';
+}
+
+int main(void)
+{
+	void copyString(char *to, char *from);
+	char string1[] = "A string to be copied.";
+	char string[50];
+	
+	copyString(string2, string1);
+	printf("%s\n", string2);
+	
+	copyString(string2, "So is this.");
+	printf("%s\n", string2);
+	
+	return 0;
+}

Chapter10Exercises/1/1.14.c

@@ -0,0 +1,24 @@
+#include <stdio.h>
+
+void copyString(char *to, char *from)
+{
+	while(*from)
+		*to++ = *from++;
+	
+	*to = '\0';
+}
+
+int main(void)
+{
+	void copyString(char *to, char *from);
+	char string1[] = "A string to be copied.";
+	char string2[50];
+	
+	copyString(string2, string1);
+	printf("%s\n", string2);
+	
+	copyString(string2, "So is this.");
+	printf("%s\n", string2);
+	
+	return 0;
+}

Chapter10Exercises/1/1.15.c

@@ -0,0 +1,24 @@
+//此函数求字符串中的字符数，指针版本
+
+#include <stdio.h>
+
+int stringLength(const char *string)
+{
+	const char *cptr = string;
+	
+	while(*cptr)
+		++cptr;
+	
+	return cptr - string;
+}
+
+int main(void)
+{
+	int stringLength(const char *string);
+	
+	printf("%i ", stringLength("StringLength test"));
+	printf("%i ", stringLength(""));
+	printf("%i\n", stringLength("complete"));
+	
+	return 0;
+}

Chapter10Exercises/1/1.2.c

@@ -0,0 +1,17 @@
+#include <stdio.h>
+
+int main(void)
+{
+	char c = 'Q';
+	char *char_pointer = &c;
+	
+	printf("%c %c\n", c, *char_pointer);
+	
+	c = '/';
+	printf("%c %c\n", c, *char_pointer);
+	
+	*char_pointer = '(';
+	printf("%c %c\n", c, *char_pointer);
+	
+	return 0;
+}

Chapter10Exercises/1/1.3.c

@@ -0,0 +1,18 @@
+//关于指针的更多知识
+
+#include <stdio.h>
+
+int main(void)
+{
+	int i1, i2;
+	int *p1, *p2;
+	
+	i1 = 5;
+	p1 = &i1;
+	i2 = *p1 / 2 + 10;
+	p2 = p1;
+	
+	printf("i1 = %i, i2 = %i, *p1 = %i, *p2 = %i\n", i1, i2, *p1, *p2);
+	
+	return 0;
+}

Chapter10Exercises/1/1.4.c

@@ -0,0 +1,26 @@
+//´Ë³ÌÐòÑÝʾ½á¹¹ÌåÖ¸Õë
+
+#include <stdio.h>
+
+int main(void)
+{
+	struct date
+	{
+		int month;
+		int day;
+		int year;
+	};
+	
+	struct date today, *datePtr;
+	
+	datePtr = &today;
+	
+	datePtr->month = 9;
+	datePtr->day = 25;
+	datePtr->year = 2015;
+	
+	printf("Today's date is %i/%i/%.2i.\n",
+	       datePtr->month, datePtr->day, datePtr->year % 100);
+	
+	return 0;
+}

Chapter10Exercises/1/1.5.c

@@ -0,0 +1,24 @@
+//此函数使用包含指针的结构体
+
+#include <stdio.h>
+
+int main(void)
+{
+	struct intPtrs
+	{
+		int *p1;
+		int *p2;
+	};
+	
+	struct intPtrs pointers;
+	int i1 = 100, i2;
+	
+	pointers.p1 = &i1;
+	pointers.p2 = &i2;
+	*pointers.p2 = -97;
+	
+	printf("i1 = %i, *pointers.p1 = %i\n", i1, *pointers.p1);
+	printf("i2 = %i, *pointers.p2 = %i\n", i2, *pointers.p2);
+	
+	return 0;
+}

Chapter10Exercises/1/1.6.c

@@ -0,0 +1,29 @@
+//此函数使用链表
+
+#include <stdio.h>
+
+int main(void)
+{
+	struct entry
+	{
+		int          value;
+		struct entry *next;
+	};
+	
+	struct entry n1, n2, n3;
+	int i;
+	
+	n1.value = 100;
+	n2.value = 200;
+	n3.value = 300;
+	
+	n1.next = &n2;
+	n2.next = &n3;
+	
+	i = n1.next->value;
+	printf("%i ", i);
+	
+	printf("%i\n", n2.next->value);
+	
+	return 0;
+}

Chapter10Exercises/1/1.7.c

@@ -0,0 +1,31 @@
+//此程序遍历一个链表
+
+#include <stdio.h>
+
+int main(void)
+{
+	struct entry
+	{
+		int          value;
+		struct entry *next;
+	};
+	
+	struct entry n1, n2, n3;
+	struct entry *list_pointer = &n1;
+	
+	n1.value = 100;
+	n1.next = &n2;
+	
+	n2.value = 200;
+	n2.next = &n3;
+	
+	n3.value = 300;
+	n3.next = (struct entry *) 0;
+	
+	while(list_pointer != (struct entry *) 0){
+		printf("%i\n", list_pointer->value);
+		list_pointer = list_pointer->next;
+	}
+	
+	return 0;
+}

Chapter10Exercises/1/1.8.c

@@ -0,0 +1,21 @@
+//此程序演示指针与函数的使用
+
+#include <stdio.h>
+
+void test(int *int_pointer)
+{
+	*int_pointer = 100;
+}
+
+int main(void)
+{
+	void test(int *int_pointer);
+	int i = 50, *p = &i;
+	
+	printf("Before the call to test i = %i\n", i);
+	
+	test(p);
+	printf("After the call to test i = %i\n", i);
+	
+	return 0;
+}