由于第3章第一题网上有很多种非常优秀的解法，我就不贴出来了，大家不妨自己探索。
练习3-2 编写一个函数escape（s，t），将字符串t复制到字符串s中，并在复制过程中将换行符、制表符等不可见字符分别转换成’\n’、’\t’等相应的可见的转义字符序列。要求使用switch语句。再编写一个具有相反功能的函数，在复制过程将转义字符序列转换成实际字符。
先解释一下两个run函数和两个escape分别实现上述两个功能：
1 #include <stdio.h> 2 3 void escape(char s[], char t[]); 4 void escape2(char s[], char t[]); 5 6 void runEscape() { 7 char testT[100] = "Hello,\ni am a\nGOOD\tBOY\t!!!!\n"; 8 char testS[100] = "Bye,\t littal Max!\n"; 9 10 printf("T:%sS:%s", testT, testS);11 escape(testS, testT);12 printf("Now,");13 printf("T:%sS:%s", testT, testS);14 }15 void runEscape2() {16 char testS[100] = "Hello,\ni am a\nGOOD\tBOY\t!!!!\n";17 char testT[100] = "Bye,\t littal Max!\n";18 19 printf("T:%sS:%s", testT, testS);20 escape2(testS, testT);21 printf("Now,");22 printf("T:%sS:%s", testT, testS);23 }24 void escape(char s[], char t[]) {25 int i = 0;26 int j = 0;27 28 while (t[i] != '\0') {29 switch (t[i])30 {31 case '\t':32 s[j++] = '\\';s[j++] = 't';33 break;34 case '\n':35 s[j++] = '\\'; s[j++] = 'n';36 break;37 default:38 s[j++] = t[i];39 break;40 }41 42 i++;43 }44 s[j] = '\0';45 }46 void escape2(char s[], char t[]) {47 int i = 0;48 int j = 0;49 50 while (t[i] != '\0') {51 if (t[i]=='\\')52 {53 switch (t[i+1])54 {55 case 't':56 s[j++] = '\t'; i++;57 break;58 case 'n':59 s[j++] = '\n'; i++;60 break;61 default:62 s[j++] = t[i];63 break;64 }65 }66 else 67 s[j++] = t[i];68 i++;69 }70 s[j] = '\0';71 }
练习3-3 编写函数expand(s1,s2)，将字符串s1中类似于a-z一类的速记符号在字符串s2中扩展为等价多的完整列表abc…xyz。该函数可以处理大小写字母和数字，并可以处理a-b-c、a-z0-9与-a-z等类似情况。作为前导和尾随的-字符原样排印。
1 #include <stdio.h> 2 3 void expand(char s1[], char s2[]); 4 5 void runExpand() { 6 char testS[100] = "0-9\na-z\nA-Z\na-b-c\na-a\na-Z\n"; 7 char testT[100] = "Bye, littal Tester!\n"; 8 9 printf("T:%sS:%s", testT, testS);10 expand(testS, testT);11 printf("\nNow,\n");12 printf("T:%sS:%s", testT, testS);13 }14 15 16 void expand(char s1[], char s2[]) {17 int i = 0;18 int j = 0;19 20 while (s1[i] != '\0') {21 if (s1[i] == '-'&&s1[i + 1] != '\0' && i > 0) {22 char head = s1[i - 1];23 char tail = s1[i + 1];24 25 if ('a' <= head&&head <= 'z'&&head < tail26 &&'a' <= tail&&tail <= 'z') {27 for (; head + 1 <= tail;) {28 s2[j++] = head + 1;29 head++;30 }31 }32 else if ('A' <= head&&head <= 'Z'&&head < tail33 &&'A' <= tail&&tail <= 'Z') {34 for (; head + 1 < tail;) {35 s2[j++] = head + 1;36 head++;37 }38 }39 else if ('0' <= head&&head <= '9'&&head < tail40 &&'0' <= tail&&tail <= '9') {41 for (; head + 1 < tail;) {42 s2[j++] = head + 1;43 head++;44 }45 }46 else47 s2[j++] = '-';48 }49 else50 s2[j++] = s1[i];51 52 i++;53 }54 }
练习3-4 在数的对二补码表示中，我们编写的itoa函数不能处理最大的负数，即n等于-2字长-1的情况。请解释原因。修改该函数，使它在任何机器上运行时都能打印出正确的值。
解释的原因我写在注释代码里面了~~
#include <stdio.h>void reverse(char s[]);void itoa(int n, char s[]);void runItoa() { int testn = -1987654321; char test[100]; itoa(testn, test); printf("test:%s", test);}/*不能正确处理最大负数的原因在于其二进制数形式为10000...如果对其进行n=-n运算，n将变成0，这显然是不符合我们期望的。因此我们可以进行检测，如果它是这个特殊的最大负数则将其变为无符号数。*/ void itoa(int n, char s[]) { int i, sign; sign = n; if (((unsigned)sign << 1) == 0) n = (unsigned)n; else if (sign < 0) n = -n; i = 0; do { s[i++] = n % 10 + '0'; } while ((n /= 10) > 0); if (sign < 0) s[i++] = '-'; s[i] = '\0'; reverse(s);}void reverse(char s[]) { int i = 0; int j = 0; char c[1000]; while (s[i] != '\0') c[i] = s[i++]; while (i > 0) s[j++] = c[--i]; s[j] = '\0';}
练习3-5 编写函数itob（n，s，b）将整数n转换为以b为底的数。 并将转换结果以字符的形式保存到字符串s中。
#include <stdio.h>void reverse(char s[]);void itob(int n, char s[], int);void runItob() { int testn = 1987654321; char test[100]; itob(testn, test, 10); printf("test:%s", test);}void itob(int n, char s[], int b) { int i = 0; do { s[i++] = n % b + '0'; } while ((n /= b) > 0); s[i] = '\0'; reverse(s);}
练习3-6 修改itoa函数，使得该函数可以接收三个参数。第三个参数为最小字段宽度。为了保证转换后结果至少具有第三个参数指定的最小宽度，必要时在结果左边填充一定的空格。
1 #include <stdio.h> 2 3 void reverse(char s[]); 4 void itoa2(int n, char s[], int width); 5 6 void runItoa2() { 7 int testn = -1987654321; 8 char test[100]; 9 10 itoa2(testn, test,20);11 printf("test:%send\n", test);12 }13 14 void itoa2(int n, char s[], int width) {15 int i, sign;16 sign = n;17 18 if (((unsigned)sign << 1) == 0)19 n = (unsigned)n;20 else if (sign < 0)21 n = -n;22 i = 0;23 do {24 s[i++] = n % 10 + '0';25 } while ((n /= 10) > 0);26 if (sign < 0)27 s[i++] = '-';28 while (width - i > 0) {29 s[i++] = ' ';30 }31 s[i] = '\0';32 reverse(s);33 }
都在努力为你驱逐烦恼焦躁，
