Write a program to print a histogram of the lengths of words in its input. It is easy to draw the histogram with the bars horizontal; a vertical orientation is more challenging.
统计输入中单词的长度，并且绘制相应的直方图。水平的直方图比较容易绘制，垂直的直方图较困难一些。
/* This program was the subject of a thread in comp.lang.c, because of the way it handled EOF. * The complaint was that, in the event of a text file's last line not ending with a newline, * this program would not count the last word. I objected somewhat to this complaint, on the * grounds that "if it hasn't got a newline at the end of each line, it isn't a text file". * * These grounds turned out to be incorrect. Whether such a file is a text file turns out to * be implementation-defined. I'd had a go at checking my facts, and had - as it turns out - * checked the wrong facts! (sigh) * * It cost me an extra variable. It turned out that the least disturbing way to modify the * program (I always look for the least disturbing way) was to replace the traditional * while((c = getchar()) != EOF) with an EOF test actually inside the loop body. This meant * adding an extra variable, but is undoubtedly worth the cost, because it means the program * can now handle other people's text files as well as my own. As Ben Pfaff said at the * time, "Be liberal in what you accept, strict in what you produce". Sound advice. * * The new version has, of course, been tested, and does now accept text files not ending in * newlines. * * I have, of course, regenerated the sample output from this program. Actually, there's no * "of course" about it - I nearly forgot. */ #include <stdio.h> #define MAXWORDLEN 10 int main(void) { int c; int inspace = 0; long lengtharr[MAXWORDLEN + 1]; int wordlen = 0; int firstletter = 1; long thisval = 0; //这个变量对于绘制垂直直方图，很有作用。 long maxval = 0; int thisidx = 0; int done = 0;
for(thisidx = 0; thisidx <= MAXWORDLEN; thisidx++) { lengtharr[thisidx] = 0; }
//为什么不利用 while((c = getchar()) != EOF) 进行接收判断一起进行？这可以处理那些不是以新行结尾的文件。 while(done == 0) { c = getchar();
//空格、换行、制表、EOF代表单词的结束，可以根据上一个单词的wordlen做统计了，并存放到lengtharr[wordlen - 1] if(c == ' ' || c == ' ' || c == ' ' || c == EOF) { if(inspace == 0) { firstletter = 0; inspace = 1; if(wordlen <= MAXWORDLEN) { if(wordlen > 0) { thisval = ++lengtharr[wordlen - 1]; if(thisval > maxval) { maxval = thisval; } } } else { thisval = ++lengtharr[MAXWORDLEN]; if(thisval > maxval) { maxval = thisval; } } } if(c == EOF) { done = 1; } } else //统计当前单词的长度 { if(inspace == 1 || firstletter == 1) { wordlen = 0; firstletter = 0; inspace = 0; } ++wordlen; } }
//绘制垂直直方图，这个算法是可以借鉴的。不利用数组存储，而是根据指标变量thisval，也就是当前最大的直方图峰值来绘图。 for(thisval = maxval; thisval > 0; thisval--) { printf("%4d | ", thisval); for(thisidx = 0; thisidx <= MAXWORDLEN; thisidx++) { if(lengtharr[thisidx] >= thisval) { printf("* "); } else { printf(" "); } } printf(" "); }
printf(" +"); for(thisidx = 0; thisidx <= MAXWORDLEN; thisidx++) { printf("---"); } printf(" "); for(thisidx = 0; thisidx < MAXWORDLEN; thisidx++) { printf("%2d ", thisidx + 1); } printf(">%d ", MAXWORDLEN); return 0; }
样本输出
Here's the output of the program when given its own
source as input:
113 | *
112 | *
111 | *
110 | *
109 | *
108 | *
107 | *
106 | *
105 | *
104 | *
103 | *
102 | *
101 | *
100 | *
99 | *
98 | *
97 | *
96 | *
95 | *
94 | * *
93 | * *
92 | * *
91 | * *
90 | * *
89 | * *
88 | * *
87 | * *
86 | * *
85 | * *
84 | * *
83 | * *
82 | * *
81 | * *
80 | * *
79 | * *
78 | * *
77 | * *
76 | * *
75 | * *
74 | * *
73 | * *
72 | * *
71 | * *
70 | * *
69 | * *
68 | * *
67 | * *
66 | * *
65 | * *
64 | * *
63 | * * *
62 | * * *
61 | * * *
60 | * * *
59 | * * *
58 | * * *
57 | * * *
56 | * * *
55 | * * *
54 | * * *
53 | * * *
52 | * * * *
51 | * * * *
50 | * * * *
49 | * * * *
48 | * * * *
47 | * * * *
46 | * * * *
45 | * * * *
44 | * * * *
43 | * * * * *
42 | * * * * *
41 | * * * * *
40 | * * * * *
39 | * * * * *
38 | * * * * *
37 | * * * * *
36 | * * * * *
35 | * * * * *
34 | * * * * *
33 | * * * * *
32 | * * * * *
31 | * * * * *
30 | * * * * * *
29 | * * * * * *
28 | * * * * * * *
27 | * * * * * * *
26 | * * * * * * *
25 | * * * * * * * *
24 | * * * * * * * *
23 | * * * * * * * *
22 | * * * * * * * * *
21 | * * * * * * * * *
20 | * * * * * * * * *
19 | * * * * * * * * *
18 | * * * * * * * * *
17 | * * * * * * * * *
16 | * * * * * * * * *
15 | * * * * * * * * *
14 | * * * * * * * * * *
13 | * * * * * * * * * *
12 | * * * * * * * * * *
11 | * * * * * * * * * *
10 | * * * * * * * * * *
9 | * * * * * * * * * * *
8 | * * * * * * * * * * *
7 | * * * * * * * * * * *
6 | * * * * * * * * * * *
5 | * * * * * * * * * * *
4 | * * * * * * * * * * *
3 | * * * * * * * * * * *
2 | * * * * * * * * * * *
1 | * * * * * * * * * * *
+---------------------------------
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