C 程序设计语言第二版——第四章练习题

1. Write the function strindex(s,t) which returns the position of the rightmost occurrence of t in s, or -1 if there is none.
2. Extend atof to handle scientific notation of the formwhere a floating-point number may be followed by e or E and an optionally signed exponent.
3. Given the basic framework, it's straightforward to extend the calculator. Add the modulus (%) operator and provisions for negative numbers.
4. Add the commands to print the top elements of the stack without popping, to duplicate it, and to swap the top two elements. Add a command to clear the stack.
5. Add access to library functions like sin, exp, and pow.
6. Add commands for handling variables. (It's easy to provide twenty-six variables with single-letter names.) Add a variable for the most recently printed value.
7. Write a routine ungets(s) that will push back an entire string onto the input. Should ungets know about buf and bufp, or should it just use ungetch?
8. Suppose that there will never be more than one character of pushback. Modify getch and ungetch accordingly.
？9. Our getch and ungetch do not handle a pushed-back EOF correctly. Decide what their properties ought to be if an EOF is pushed back, then implement your design.
10. An alternate organization uses getline to read an entire input line; this makes getch and ungetch unnecessary. Revise the calculator to use this approach.
11. Modify getop so that it doesn't need to use ungetch. Hint: use an internal static variable.
12. Adapt the ideas of printd to write a recursive version of itoa; that is, convert an integer into a string by calling a recursive routine.
13. Write a recursive version of the function reverse(s), which reverses the string s in place.
14. Define a macro swap(t,x,y) that interchanges two arguments of type t. (Block structure will help.)

1. Write the function strindex(s,t) which returns the position of the rightmost occurrence of t in s, or -1 if there is none.

#include<stdio.h>
#include<string.h>
#define LEN 100
char *s_gets(char *s, int n);
int strindex(char *s, char *t);
int main(void)
{char pattern[LEN];char s[LEN];int  index;printf("Please input the search string:\n");s_gets(pattern, LEN);printf("Please enter strings to be tested (empty line to quit):\n");while(s_gets(s, LEN) && s[0] != '\0')if ((index = strindex(s, pattern)) != -1)printf("%s\n\n", s+index);return 0;
}
char *s_gets(char *s, int n)
{char *ret, *find;ret = fgets(s, n, stdin);if(ret){find = strchr(s, '\n');if(find)*find = '\0';elsewhile(getchar() != '\n')continue;}return ret;
}
int strindex(char *s, char *t)
{int i, j, k;int len1 = strlen(s);int len2 = strlen(t);for(i = len1-1; i >= 0; i--){for(j = i, k = len2-1; k >= 0 && j >= 0 && s[j] == t[k]; j--, k--);if (k < 0)return (j+1);}return -1;
}

2. Extend atof to handle scientific notation of the formwhere a floating-point number may be followed by e or E and an optionally signed exponent.

见c 语言实现 atof() 函数

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<ctype.h>
#define LEN 40
char *s_gets(char *s, int n);
double myatof(const char *s);
int main(void)
{char s[LEN];double num;puts("Enter strings, enter a newline to quit:");while(s_gets(s, LEN) && s[0] != '\0'){num = myatof(s);printf("String: %s, corrsponding number: %f.\n",s, num);}puts("Bye!");return 0;
}
char *s_gets(char *s, int n)
{char *ret, *find;ret = fgets(s, n, stdin);if(ret){find = strchr(s, '\n');if(find)*find = '\0';elsewhile(getchar() != '\n')continue;}return ret;
}
double myatof(const char *s)
{int sign, sign_e;//数值的符号，指数部分的符号int hasdot = 0;int hase = 0;double intpart = 0.0;//小数的整数部分double decpart = 0.0; //小数的小数部分int decdigit = 1; //小数的小数位数int exp = 0; //指数部分double ret;int i;//跳过开头的空格for (i = 0; isspace(s[i]); i++);//判断符号，如有，跳过sign = (s[i] == '-') ? -1 : 1;if (s[i] == '-' || s[i] == '+')i++;//判断浮点数//第一部分：for (; s[i] != '\0'; i++){if (isdigit(s[i])) //数字intpart = 10 * intpart + s[i] - '0';//计算小数的整数部分else if (s[i] == '.') //小数点{hasdot = 1;i++;break;}else if (s[i] == 'e' || s[i] == 'E') //科学计数符{hase = 1;i++;break;}else //非法字符return sign * intpart;}/*第一部分结束，有如下情况：1. 扫描数字知道非法字符或字符串结尾，2d 2342. 数字加小数点 2.3. 小数点 .4. 数字加科学计数符 3e5. 科学计数符 e   这种情况是非法表示，但最终计算的结果为0,因此可当作正常计算，不单独列出6. 非法字符, 直接退出*///第二部分，接着上述情况扫描//能进入下面循环，排除遇到字符串结尾，非法字符//因此只能遇到点号或科学计数符for (; s[i] != '\0'; i++){//第一种：.3 或 3.4，均为合法，计算小数的小数部分if (hasdot && isdigit(s[i]))decpart += (s[i] - '0') / pow(10, decdigit++);//第二种：.e 或 2.e 或 .2e 或 3.3e 第一种非法，但计算结果为0else if (hasdot && (s[i] == 'e' || s[i] == 'E')){hase = 1;i++;break;}//第三种：第一部分以e结束，3e eelse if (hase)break;//第四种：第一部分以点号结束，现在扫描非数字，非科学计数符的其他非法字符elsereturn sign * (intpart + decpart);}/*第三部分从第二部分退出后继续后面的程序，有如下情况：以科学计算符 e 结束第二部分，前面有小数点或者没有小数部分计算完，下面讨论指数部分*///判断指数部分符号sign_e = (s[i] == '-') ? -1 : 1;if (s[i] == '+' || s[i] == '-')i++;for(; s[i] != '\0'; i++){if(isdigit(s[i]))exp = exp * 10 + s[i] - '0';elsebreak;}ret = sign * ((intpart + decpart) * pow(10, sign_e * exp));return ret;
}

3. Given the basic framework, it’s straightforward to extend the calculator. Add the modulus (%) operator and provisions for negative numbers.

分析：与书中的例子相比，加上一个求余计算符，识别负数。区别负号和减号的方法是负号后直接跟数字或小数点，减号与操作数之间有空格。

参考：
Reverse Polish Calculator （逆波兰计算器）方案的分析——如何解决问题，从需要到实现

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<ctype.h>
#define MAXOP 100
#define NUMBER '0'
int getop(char *s);
void push(double);
double pop(void);
int getch(void);
void ungetch(int c);
int main(void)
{int type;double op2;char s[MAXOP];while ((type = getop(s)) != EOF){switch (type){case NUMBER:push(atof(s));break;case '+':push(pop() + pop());break;case '*':push(pop() * pop());break;case '-':op2 = pop();push(pop() - op2);break;case '/':op2 = pop();if (op2 != 0.0)push(pop() / op2);elseprintf("error: zero divisor\n");break;case '%':op2 = pop();if (op2 != 0.0)push(fmod(pop(), op2));elseprintf("error: zero divisor\n");break;case '\n':printf("\t%.8g\n",pop());break;default:printf("error: unknow command %s\n",s);break;}}return 0;
}
//push() pop()
#define MAXVAL 100
int sp = 0;
double val[MAXVAL];
void push(double f)
{if (sp < MAXVAL)val[sp++] = f;elseprintf("error: stack full, can't push %g\n",f);
}
double pop(void)
{if (sp > 0)return val[--sp];elseprintf("error: stack empty\n");return 0.0;
}
// getch() ungetch()
#define BUFSIZE 100
char buf[BUFSIZE];
int bufp = 0;
int getch(void)
{return (bufp > 0) ? buf[--bufp] : getchar();
}
void ungetch(int c)
{if (bufp >= BUFSIZE)printf("ungetch: too many characters\n");elsebuf[bufp++] = c;
}
//getop()
int getop(char *s)
{int i, c;while ((s[0] = c = getch()) == ' ' || c == '\t');s[1] = '\0';if (!isdigit(c) && c != '.' && c != '-')return c;i = 0;if (c == '-')if (isdigit(c = getch()) || c == '.')s[++i] = c;else{if (c != EOF)ungetch(c);return '-';}if (isdigit(c))while (isdigit(s[++i] = c = getch()));if (c == '.')while (isdigit(s[++i] = c = getch()));s[i] = '\0';if (c != EOF)ungetch(c);return NUMBER;
}

结果：

3 3 +6
s
error: unknow command s
error: stack empty //输入非法字符 s 后还有换行符，因此出现该行提示0

4. Add the commands to print the top elements of the stack without popping, to duplicate it, and to swap the top two elements. Add a command to clear the stack.

5. Add access to library functions like sin, exp, and pow.

6. Add commands for handling variables. (It’s easy to provide twenty-six variables with single-letter names.) Add a variable for the most recently printed value.

三题功能放一个程序中：
加入几个功能仿真计算器：

输出栈顶的值但不pop，用 ? 表示
复制栈顶元素，用 d 表示
交换栈顶的两个元素，用 s 表示
清空栈内元素，用 c
添加数学函数 sin(), cos(), tan(), exp(), pow()

#include<stdio.h>
#include<string.h>
#include<ctype.h>
#include<math.h>
#include<stdlib.h>
#define MAXOP 100
#define NUMBER '0'
#define NAME 'N'
int getop(char *s);
void push(double f);
double pop(void);
int getch(void);
void ungetch(int c);
void mathfunc(char *s);
double top(void);
void duplicate(void);
void swap(void);
void clear(void);
int main(void)
{int type;double op2;char s[MAXOP];while ((type = getop(s)) != EOF) {switch (type) {case NUMBER:push(atof(s));break;case '+':push(pop() + pop());break;case '*':push(pop() * pop());break;case '-':op2 = pop();push(pop() - op2);break;case '/':op2 = pop();if (op2 != 0.0)push(pop() / op2);elseprintf("error: zero divisor\n");break;case '%':op2 = pop();if (op2 != 0.0)push(fmod(pop(), op2));elseprintf("error: zero divisor\n");break;case '?':printf("\t%.8g\n",top());break;case 'd':duplicate();break;case 'c':clear();break;case 's':swap();break;case NAME:mathfunc(s);break;case '\n':printf("\t%.8g\n", pop());break;default:printf("error: unknown command %s\n", s);break;}}return 0;
}
//pop() push()
#define MAXVAL 100
int sp = 0;
double val[MAXVAL];
void push(double f)
{if (sp < MAXVAL)val[sp++] = f;elseprintf("error: stack full, can't push %g\n",f);
}
double pop(void)
{if (sp > 0)return val[--sp];elseprintf("error: stack empty\n");return 0.0;
}
//top() duplicate() swap() clear()
double top(void)
{if (sp > 0)return val[sp-1];elseprintf("error: stack empty\n");return 0.0;
}
void duplicate(void)
{push(top());
}
void swap(void)
{double elem1 = pop();double elem2 = pop();push(elem1);push(elem2);
}
void clear(void)
{sp = 0;
}
//getch() ungetch()
#define BUFSIZE 100
char buf[BUFSIZE];
int bufp = 0;
int getch(void)
{return (bufp > 0) ? buf[--bufp] : getchar();
}
void ungetch(int c)
{if (bufp >= BUFSIZE)printf("ungetch: too many characters\n");elsebuf[bufp++] = c;
}
//void mathfunc()
void mathfunc(char *s)
{double op2;if (strcmp(s, "sin") == 0)push(sin(pop()));else if (strcmp(s, "cos") == 0)push(cos(pop()));else if (strcmp(s, "tan") == 0)push(tan(pop()));else if (strcmp(s, "exp") == 0)push(exp(pop()));else if (strcmp(s, "pow") == 0) {op2 = pop();push(pow(pop(), op2));}elseprintf("error: unknow command %s\n", s);
}
//getop()
int getop(char *s)
{int i, c, c1;while ((s[0] = c = getch()) == ' ' || c == '\t');s[1] = '\0';i = 0;if (islower(c)) {while (islower(s[++i] = c1 = getch()));s[i] = '\0';if (c1 != EOF)ungetch(c1);if (strlen(s) > 1)return NAME;return c;}if (!isdigit(c) && c != '.' && c != '-')return c;if (c == '-')if (isdigit(c = getch()) || c == '.')s[++i] = c;else {if (c != EOF)ungetch(c);return '-';}if (isdigit(c))while (isdigit(s[++i] = c = getch()));if (c == '.')while (isdigit(s[++i] = c = getch()));s[i] = '\0';if (c != EOF)ungetch(c);return NUMBER;
}

结果：

1 4 5 + tan t
error: unknown command t-0.45231566
2 2 1 + s pow9

7. Write a routine ungets(s) that will push back an entire string onto the input. Should ungets know about buf and bufp, or should it just use ungetch?

分析：
ungets() 只需调用 ungetch() 函数就行了，而是否超出缓冲区大小由 ungetch() 函数判断。

#include<stdio.h>
#include<string.h>
#define LEN 100
#define BUFSIZE 100
char buf[BUFSIZE];
int bufp = 0;
char *s_gets(char *s, int n);
int getch(void);
void ungetch(int c);
void ungets(char *s);
int main(void)
{char s[LEN];int c;printf("Please enter a string:\n");s_gets(s, LEN);printf("Push back the string:\n");ungets(s);while (( c = getch()) != EOF)putchar(c);return 0;
}
char *s_gets(char *s, int n)
{char *ret, *find;ret = fgets(s, n, stdin);if (ret) {find = strchr (s, '\n');if (find)*find = '\0';elsewhile (getchar() != '\0')continue;}return ret;
}
//getch() ungetch()
int getch(void)
{return (bufp > 0) ? buf[--bufp] : getchar();
}
void ungetch(int c)
{if (bufp >= BUFSIZE)printf("ungetch: too many characters\n");elsebuf[bufp++] = c;
}
void ungets(char *s)
{int len = strlen(s);int i = len - 1;while (i >= 0) ungetch(s[i--]);
}

8. Suppose that there will never be more than one character of pushback. Modify getch and ungetch accordingly.

参考：
c 程序设计语言笔记
重温
题目意思： BUFSIZE 1，这样只能存一个字符。因此不用声明数组，只有一个元素。

int buf = EOF; //表示 buf 为空
int getch(void)
{int temp; if (buf != EOF) {temp = buf;buf = EOF;}elsetemp = getchar();return temp;
}
void ungetch(int c)
{if (buf != EOF)printf("too many characters\n");elsebuf = c;
}

？9. Our getch and ungetch do not handle a pushed-back EOF correctly. Decide what their properties ought to be if an EOF is pushed back, then implement your design.

这个问题暂时不懂，为什么 getop() 中用

if (c != EOF)ungetch();

将 if （c != EOF) 去掉后：
用gdb调试程序，在 getop 函数设置断点单步运行，输入数字3再按[ctrl] 和d键，程序并没有等按换行键再跳到下一步，直接下一步，未存储 EOF，因此后面本应该扫描到EOF时又没有值，即EOF不在输入缓冲区内，提示输入新的值。

这时输入3和EOF时，**按键输入EOF后没反应，重新按一次EOF后程序进行下一步，**此时成功识别EOF，且将其ungetch 放回buf内，查看buf[0] 的值为 -1，此时getop 返回 NUMBER，将数字字符转换为数字，接着继续循环到getop，此时 c 为 EOF，值为 -1，程序将其判断为不属于合法字符的条件中，因此返回 -1，回到主函数，type 为 -1，跳出循环，程序结束。

目前没明白，先记录，以后补充。

10. An alternate organization uses getline to read an entire input line; this makes getch and ungetch unnecessary. Revise the calculator to use this approach.

未用头文件 stdio.h 中的getline() 函数，自己写的版本

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<ctype.h>
#include<math.h>
#define MAXOP 100
#define NUMBER '0'
#define NAME 'N'
#define BUFSIZE 100
//mygetline() 参数
int bufp = 0;//一次扫描存入字符串个数
int getp = 0;//取字符的位置
double buf[BUFSIZE];
//getline() 参数结束
int getop(char *s);
void push(double f);
double pop(void);
int mygetline(void);
void mathfunc(char *s);
double top(void);
void duplicate(void);
void swap(void);
void clear(void);
int main(void)
{int type;double op2;char s[MAXOP];while ((type = getop(s)) != EOF) {switch (type) {case NUMBER:push(atof(s));break;case '+':push(pop() + pop());break;case '*':push(pop() * pop());break;case '-':op2 = pop();push(pop() - op2);break;case '/':op2 = pop();if (op2 != 0.0)push(pop() / op2);elseprintf("Error: zero divisor\n");break;case '%':op2 = pop();if (op2 != 0.0)push(fmod(pop(), op2));elseprintf("Error: zero divisor\n");break;case '?':printf("\t%.8g\n",top());break;case 'd':duplicate();break;case 'c':clear();break;case 's':swap();break;case NAME:mathfunc(s);break;case '\n':printf("\t%.8g\n",pop());break;default:printf("Error: unknown command %s\n",s);break;}}return 0;
}
// pop() push()
#define MAXVAL 100
int sp = 0;
double val[MAXVAL];
void push(double f)
{if (sp < MAXVAL)val[sp++] = f;elseprintf("Error: stack full, can't push %g\n",f);
}
double pop(void)
{if (sp > 0)return val[--sp];elseprintf("Error: stack empty\n");return 0.0;
}
//top() duplicate() clear() swap()
double top(void)
{if (sp > 0)return val[sp-1];elseprintf("Error: stack empty\n");return 0.0;
}
void duplicate(void)
{push(pop());
}
void swap(void)
{double elem1 = pop();double elem2 = pop();push(elem1);push(elem2);
}
void clear(void)
{sp = 0;
}
//mygetline() 头文件 stdio.h 中自带 getline() 函数，
//但该函数返回读取字符串个数，这里区分，自己写
int mygetline(void)
{int c;//buffer 内还有剩余字符且没有取完if (bufp > 0 && getp < bufp) return buf[getp++];//buffer 内为空 或者 buffer 内的字符全部取完else if ((bufp == 0) || (getp == bufp)) {bufp = 0;getp = 0;while (bufp < BUFSIZE && (c = getchar()) != EOF && c != '\n') buf[bufp++] = c;if (bufp != BUFSIZE)buf[bufp++] = c;//清除之前已经取过的多余字符for (int i = bufp; i < BUFSIZE && buf[i] != '\0'; i++)buf[i] = '\0';return buf[getp++];}
}
//mathfunc()
void mathfunc(char *s)
{double op2;if (strcmp(s, "sin") == 0)push(sin(pop()));else if (strcmp(s, "cos") == 0)push(cos(pop()));else if (strcmp(s, "tan") == 0)push(tan(pop()));else if (strcmp(s, "exp") == 0)push(exp(pop()));else if (strcmp(s, "pow") == 0) {op2 = pop();push(pow(pop(), op2));}elseprintf("Error: unknow command %s\n",s);
}
//getop()
int getop(char *s)
{int i, c, c1;while ((s[0] = c = mygetline()) == ' ' || c == '\t');s[1] = '\0';i = 0;if (islower(c)) {while (islower(s[++i] = c1 = mygetline()));s[i] = '\0';getp--;if (strlen(s) > 1)return NAME;return c;}if (!isdigit(c) && c != '.' && c != '-')return c;if (c == '-')if (isdigit(c = mygetline()) || c == '.')s[++i] = c;else {getp--;return '-';}if (isdigit(c))while (isdigit(s[++i] = c = mygetline()));if (c == '.')while (isdigit(s[++i] = c = mygetline()));s[i] = '\0';getp--;//排除情况 .t 或 -. ，这种返回的只有点号或者负号加点号//如果按照正常返回 NUMBER 则 atof 转化为0//现将这两种情况判为异常，返回 #if (strlen(s) < 3 && !isdigit(s[0]) && !isdigit(s[1]))return '#';return NUMBER;
}

输出：

2 3 +5
3 2 + -.t3 2 ++
Error: unknown command -.
Error: unknown command t10

11. Modify getop so that it doesn’t need to use ungetch. Hint: use an internal static variable.

if (bufp < BUFSIZE)buf[bufp++] = c;
elseprintf("too many characters\n");

12. Adapt the ideas of printd to write a recursive version of itoa; that is, convert an integer into a string by calling a recursive routine.

#include<string.h>
#include<stdlib.h>
#include<limits.h>
#define LEN 34
void itoa(int n, char *s, int b);
int input(int min, int max);
char digits[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
int main(void)
{int n;int base;char s[LEN];printf("Please enter an integer (%d ~ %d): ",INT_MIN, INT_MAX);n = input(INT_MIN, INT_MAX);printf("Please enter the base (2 ~ 36): ");base = input(2, 36);itoa(n, s, base);printf("%d base %d is %s\n",n, base, s);return 0;
}
int input(int min, int max)
{int status;long n; //识别出超出int范围的数int ok = 1;while(ok) {status = scanf("%ld",&n);if (status != 1 || n < min || n > max || getchar() != '\n') {while (getchar() != '\n')continue;printf("Enter again (%d ~ %d): ",min, max);continue;}elsereturn (int)n;}
}
void itoa(int n, char *s, int b)
{static int i = 0;if (n / b)itoa(n/b, s, b);else {if (n < 0) s[i++] = '-';}s[i++] = digits[(unsigned)(abs(n)) % b];s[i] = '\0';
}

为了识别出超出输入超出int范围的数，在input() 函数中：

long n;
return (int)n;

通过对负数进行转换：

s[i++] = digits[(unsigned)(abs(n)) % b];

保证输入是最小负数时转化正确。
输出：

Please enter an integer (-2147483648 ~ 2147483647): -2147483649
Enter again (-2147483648 ~ 2147483647): -2147483648
Please enter the base (2 ~ 36): 1
Enter again (2 ~ 36): 10
-2147483648 base 10 is -2147483648

13. Write a recursive version of the function reverse(s), which reverses the string s in place.

#include<stdio.h>
#include<string.h>
#define LEN 100
char *s_gets(char *s, int n);
void reverse(char *s, int n);
int main(void)
{char s[LEN];int n;printf("Please enter a string:\n");s_gets(s, LEN);n = strlen(s);reverse(s, n);printf("Reversed string:\n%s\n",s);return 0;
}
char *s_gets(char *s, int n)
{char *ret, *find;ret = fgets(s, n, stdin);if (ret) {find = strchr(s, '\n');if (find)*find = '\0';elsewhile (getchar() != '\n')continue;}return ret;
}
void swap(char *s1, char *s2)
{char t;t = *s1;*s1 = *s2;*s2 = t;
}
void reverse(char *s, int n)
{static int i = 0;if (i < n-1) {swap(&s[i++], &s[n-1]);reverse(s, n-1);}
}

14. Define a macro swap(t,x,y) that interchanges two arguments of type t. (Block structure will help.)

#include<stdio.h>
#define swap(t, s, y) { t x; x = s; s = y; y = x;}
int main(void)
{int i, j;i = 1;j = 3;swap(int, i, j);printf("i = %d, j = %d\n",i, j);return 0;
}

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