xref: /freebsd/usr.bin/top/utils.c (revision c5fda9bac0325eb8c5b447717862d279006f318f)
1 /*
2  *  This program may be freely redistributed,
3  *  but this entire comment MUST remain intact.
4  *
5  *  Copyright (c) 1984, 1989, William LeFebvre, Rice University
6  *  Copyright (c) 1989, 1990, 1992, William LeFebvre, Northwestern University
7  *
8  * $FreeBSD$
9  */
10 
11 /*
12  *  This file contains various handy utilities used by top.
13  */
14 
15 #include "top.h"
16 #include "utils.h"
17 
18 #include <sys/param.h>
19 #include <sys/sysctl.h>
20 #include <sys/user.h>
21 
22 #include <stdlib.h>
23 #include <stdio.h>
24 #include <string.h>
25 #include <fcntl.h>
26 #include <paths.h>
27 #include <kvm.h>
28 
29 int
30 atoiwi(const char *str)
31 {
32     size_t len;
33 
34     len = strlen(str);
35     if (len != 0)
36     {
37 	if (strncmp(str, "infinity", len) == 0 ||
38 	    strncmp(str, "all",      len) == 0 ||
39 	    strncmp(str, "maximum",  len) == 0)
40 	{
41 	    return(Infinity);
42 	}
43 	else if (str[0] == '-')
44 	{
45 	    return(Invalid);
46 	}
47 	else
48 	{
49 	    return(atoi(str));
50 	}
51     }
52     return(0);
53 }
54 
55 /*
56  *  itoa - convert integer (decimal) to ascii string for positive numbers
57  *  	   only (we don't bother with negative numbers since we know we
58  *	   don't use them).
59  */
60 
61 				/*
62 				 * How do we know that 16 will suffice?
63 				 * Because the biggest number that we will
64 				 * ever convert will be 2^32-1, which is 10
65 				 * digits.
66 				 */
67 _Static_assert(sizeof(int) <= 4, "buffer too small for this sized int");
68 
69 char *
70 itoa(unsigned int val)
71 {
72     char *ptr;
73     static char buffer[16];	/* result is built here */
74     				/* 16 is sufficient since the largest number
75 				   we will ever convert will be 2^32-1,
76 				   which is 10 digits. */
77 
78     ptr = buffer + sizeof(buffer);
79     *--ptr = '\0';
80     if (val == 0)
81     {
82 	*--ptr = '0';
83     }
84     else while (val != 0)
85     {
86 	*--ptr = (val % 10) + '0';
87 	val /= 10;
88     }
89     return(ptr);
90 }
91 
92 /*
93  *  itoa7(val) - like itoa, except the number is right justified in a 7
94  *	character field.  This code is a duplication of itoa instead of
95  *	a front end to a more general routine for efficiency.
96  */
97 
98 char *
99 itoa7(int val)
100 {
101     char *ptr;
102     static char buffer[16];	/* result is built here */
103     				/* 16 is sufficient since the largest number
104 				   we will ever convert will be 2^32-1,
105 				   which is 10 digits. */
106 
107     ptr = buffer + sizeof(buffer);
108     *--ptr = '\0';
109     if (val == 0)
110     {
111 	*--ptr = '0';
112     }
113     else while (val != 0)
114     {
115 	*--ptr = (val % 10) + '0';
116 	val /= 10;
117     }
118     while (ptr > buffer + sizeof(buffer) - 7)
119     {
120 	*--ptr = ' ';
121     }
122     return(ptr);
123 }
124 
125 /*
126  *  digits(val) - return number of decimal digits in val.  Only works for
127  *	positive numbers.  If val <= 0 then digits(val) == 0.
128  */
129 
130 int
131 digits(int val)
132 {
133     int cnt = 0;
134 
135     while (val > 0)
136     {
137 	cnt++;
138 	val /= 10;
139     }
140     return(cnt);
141 }
142 
143 /*
144  * string_index(string, array) - find string in array and return index
145  */
146 
147 int
148 string_index(const char *string, const char * const *array)
149 {
150     size_t i = 0;
151 
152     while (*array != NULL)
153     {
154 	if (strcmp(string, *array) == 0)
155 	{
156 	    return(i);
157 	}
158 	array++;
159 	i++;
160     }
161     return(-1);
162 }
163 
164 /*
165  * argparse(line, cntp) - parse arguments in string "line", separating them
166  *	out into an argv-like array, and setting *cntp to the number of
167  *	arguments encountered.  This is a simple parser that doesn't understand
168  *	squat about quotes.
169  */
170 
171 const char * const *
172 argparse(char *line, int *cntp)
173 {
174     const char **ap;
175     static const char *argv[1024] = {0};
176 
177     *cntp = 1;
178     ap = &argv[1];
179     while ((*ap = strsep(&line, " ")) != NULL) {
180         if (**ap != '\0') {
181             (*cntp)++;
182             if (*cntp >= (int)nitems(argv)) {
183                 break;
184             }
185 	    ap++;
186         }
187     }
188     return (argv);
189 }
190 
191 /*
192  *  percentages(cnt, out, new, old, diffs) - calculate percentage change
193  *	between array "old" and "new", putting the percentages i "out".
194  *	"cnt" is size of each array and "diffs" is used for scratch space.
195  *	The array "old" is updated on each call.
196  *	The routine assumes modulo arithmetic.  This function is especially
197  *	useful on for calculating cpu state percentages.
198  */
199 
200 long
201 percentages(int cnt, int *out, long *new, long *old, long *diffs)
202 {
203     int i;
204     long change;
205     long total_change;
206     long *dp;
207     long half_total;
208 
209     /* initialization */
210     total_change = 0;
211     dp = diffs;
212 
213     /* calculate changes for each state and the overall change */
214     for (i = 0; i < cnt; i++)
215     {
216 	if ((change = *new - *old) < 0)
217 	{
218 	    /* this only happens when the counter wraps */
219 	    change = (int)
220 		((unsigned long)*new-(unsigned long)*old);
221 	}
222 	total_change += (*dp++ = change);
223 	*old++ = *new++;
224     }
225 
226     /* avoid divide by zero potential */
227     if (total_change == 0)
228     {
229 	total_change = 1;
230     }
231 
232     /* calculate percentages based on overall change, rounding up */
233     half_total = total_change / 2l;
234 
235     /* Do not divide by 0. Causes Floating point exception */
236     if(total_change) {
237         for (i = 0; i < cnt; i++)
238         {
239           *out++ = (int)((*diffs++ * 1000 + half_total) / total_change);
240         }
241     }
242 
243     /* return the total in case the caller wants to use it */
244     return(total_change);
245 }
246 
247 /* format_time(seconds) - format number of seconds into a suitable
248  *		display that will fit within 6 characters.  Note that this
249  *		routine builds its string in a static area.  If it needs
250  *		to be called more than once without overwriting previous data,
251  *		then we will need to adopt a technique similar to the
252  *		one used for format_k.
253  */
254 
255 /* Explanation:
256    We want to keep the output within 6 characters.  For low values we use
257    the format mm:ss.  For values that exceed 999:59, we switch to a format
258    that displays hours and fractions:  hhh.tH.  For values that exceed
259    999.9, we use hhhh.t and drop the "H" designator.  For values that
260    exceed 9999.9, we use "???".
261  */
262 
263 char *
264 format_time(long seconds)
265 {
266     static char result[10];
267 
268     /* sanity protection */
269     if (seconds < 0 || seconds > (99999l * 360l))
270     {
271 	strcpy(result, "   ???");
272     }
273     else if (seconds >= (1000l * 60l))
274     {
275 	/* alternate (slow) method displaying hours and tenths */
276 	sprintf(result, "%5.1fH", (double)seconds / (double)(60l * 60l));
277 
278 	/* It is possible that the sprintf took more than 6 characters.
279 	   If so, then the "H" appears as result[6].  If not, then there
280 	   is a \0 in result[6].  Either way, it is safe to step on.
281 	 */
282 	result[6] = '\0';
283     }
284     else
285     {
286 	/* standard method produces MMM:SS */
287 	/* we avoid printf as must as possible to make this quick */
288 	sprintf(result, "%3ld:%02ld",
289 	    (long)(seconds / 60), (long)(seconds % 60));
290     }
291     return(result);
292 }
293 
294 /*
295  * format_k(amt) - format a kilobyte memory value, returning a string
296  *		suitable for display.  Returns a pointer to a static
297  *		area that changes each call.  "amt" is converted to a
298  *		string with a trailing "K".  If "amt" is 10000 or greater,
299  *		then it is formatted as megabytes (rounded) with a
300  *		trailing "M".
301  */
302 
303 /*
304  * Compromise time.  We need to return a string, but we don't want the
305  * caller to have to worry about freeing a dynamically allocated string.
306  * Unfortunately, we can't just return a pointer to a static area as one
307  * of the common uses of this function is in a large call to sprintf where
308  * it might get invoked several times.  Our compromise is to maintain an
309  * array of strings and cycle thru them with each invocation.  We make the
310  * array large enough to handle the above mentioned case.  The constant
311  * NUM_STRINGS defines the number of strings in this array:  we can tolerate
312  * up to NUM_STRINGS calls before we start overwriting old information.
313  * Keeping NUM_STRINGS a power of two will allow an intelligent optimizer
314  * to convert the modulo operation into something quicker.  What a hack!
315  */
316 
317 #define NUM_STRINGS 8
318 
319 char *
320 format_k(int amt)
321 {
322     static char retarray[NUM_STRINGS][16];
323     static int index = 0;
324     char *p;
325     char *ret;
326     char tag = 'K';
327 
328     p = ret = retarray[index];
329     index = (index + 1) % NUM_STRINGS;
330 
331     if (amt >= 10000)
332     {
333 	amt = (amt + 512) / 1024;
334 	tag = 'M';
335 	if (amt >= 10000)
336 	{
337 	    amt = (amt + 512) / 1024;
338 	    tag = 'G';
339 	}
340     }
341 
342     p = stpcpy(p, itoa(amt));
343     *p++ = tag;
344     *p = '\0';
345 
346     return(ret);
347 }
348 
349 char *
350 format_k2(unsigned long long amt)
351 {
352     static char retarray[NUM_STRINGS][16];
353     static int index = 0;
354     char *p;
355     char *ret;
356     char tag = 'K';
357 
358     p = ret = retarray[index];
359     index = (index + 1) % NUM_STRINGS;
360 
361     if (amt >= 100000)
362     {
363 	amt = (amt + 512) / 1024;
364 	tag = 'M';
365 	if (amt >= 100000)
366 	{
367 	    amt = (amt + 512) / 1024;
368 	    tag = 'G';
369 	}
370     }
371 
372     p = stpcpy(p, itoa((int)amt));
373     *p++ = tag;
374     *p = '\0';
375 
376     return(ret);
377 }
378 
379 int
380 find_pid(pid_t pid)
381 {
382 	kvm_t *kd = NULL;
383 	struct kinfo_proc *pbase = NULL;
384 	int nproc;
385 	int ret = 0;
386 
387 	kd = kvm_open(NULL, _PATH_DEVNULL, NULL, O_RDONLY, NULL);
388 	if (kd == NULL) {
389 		fprintf(stderr, "top: kvm_open() failed.\n");
390 		quit(TOP_EX_SYS_ERROR);
391 	}
392 
393 	pbase = kvm_getprocs(kd, KERN_PROC_PID, pid, &nproc);
394 	if (pbase == NULL) {
395 		goto done;
396 	}
397 
398 	if ((nproc == 1) && (pbase->ki_pid == pid)) {
399 		ret = 1;
400 	}
401 
402 done:
403 	kvm_close(kd);
404 	return ret;
405 }
406