自编题：任意给定两个字符串，比如"The UNIX.... " 和 "The only ...." 如何如下显示？ 用纯C实现。
1 #include <stdio.h> 2 #include <string.h> 3 #define L 23 4 #define C 80 5 typedef struct{ 6 int l, c; // lines, cols 7 } pos_t; 8 9 typedef struct { 10 char *str; 11 pos_t pos; 12 int w, h; // the width and height of a recttangle 13 void(*set)(void *); // the way of display 14 } str_obj; 15 16 char buf[L][C]; 17 18 void init(void) 19 { 20 memset(buf, ' ', L*C); 21 } 22 void display(void) 23 { 24 buf[L][C] = 0; 25 printf("%s", buf); 26 } 27 void set_pixel(int x, int y, char c) 28 { 29 if(y>0&&y<=L && x>0&&x<=C) 30 buf[y-1][x-1] = c; 31 } 32 33 static void hor_line(int c, int l, int length) 34 { 35 set_pixel(c, l, '+'); 36 int i; 37 for(i=c+1; i<length+c-1; i++) 38 set_pixel(i, l, '-'); 39 set_pixel(i,l, '+'); 40 } 41 42 static void ver_line(int c, int l, int length) 43 { 44 set_pixel(c, l, '+'); 45 int i; 46 for(i=l+1; i<length+l-1; i++) 47 set_pixel(c, i, '|'); 48 set_pixel(c, i, '+'); 49 } 50 51 void print(void *obj) 52 { 53 str_obj *p = obj; 54 char *str = p->str; 55 int x = p->pos.c, y = p->pos.l; 56 int w = p->w, h = p->h; 57 hor_line(x-2, y-1, w+4); 58 ver_line(x-2, y-1, h+2); 59 hor_line(x-2, y+h, w+4); 60 ver_line(x+w+1, y-1, h+2); 61 int i,j; 62 int len = strlen(str), count = 0; 63 for(i=0; i<h; i++){ 64 for(j=0; j<w; j++){ 65 if(count < len ){ 66 set_pixel(j+x,y+i, str[count++]); 67 } 68 } 69 } 70 } 71 72 int main(int argc, char **argv) 73 { 74 75 init(); 76 77 str_obj xstr = { 78 "The UNIX operating system provides its services through " 79 "a set of system calls, which are in effect functions wit" 80 "hin the operating system that may be called by user prog" 81 "rams. This chapter describes how to use some of the most" 82 " important system calls from C programs.", 83 {2,5}, 15,20, print 84 }; 85 86 xstr.set(&xstr); 87 xstr.pos.c += xstr.w+3; 88 xstr.set(&xstr); 89 90 str_obj ystr = { 91 "The only way to learn a new programming language is by " 92 "writing programs in it.", 93 {5, 50}, 20, 5, print 94 }; 95 ystr.set(&ystr); 96 display(); 97 return 0; 98 }