92a34e118f
I had changed this from a for loop to a memset during an earlier cleanup. This change was incorrect so revert it. While here, clean up Reported by: flo
406 lines
8.9 KiB
C
406 lines
8.9 KiB
C
/*
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* This program may be freely redistributed,
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* but this entire comment MUST remain intact.
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*
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* Copyright (c) 1984, 1989, William LeFebvre, Rice University
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* Copyright (c) 1989, 1990, 1992, William LeFebvre, Northwestern University
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*
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* $FreeBSD$
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*/
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/*
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* This file contains various handy utilities used by top.
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*/
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#include "top.h"
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#include "utils.h"
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#include <sys/param.h>
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#include <sys/sysctl.h>
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#include <sys/user.h>
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#include <stdlib.h>
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#include <stdio.h>
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#include <string.h>
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#include <fcntl.h>
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#include <paths.h>
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#include <kvm.h>
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int
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atoiwi(const char *str)
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{
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size_t len;
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len = strlen(str);
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if (len != 0)
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{
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if (strncmp(str, "infinity", len) == 0 ||
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strncmp(str, "all", len) == 0 ||
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strncmp(str, "maximum", len) == 0)
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{
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return(Infinity);
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}
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else if (str[0] == '-')
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{
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return(Invalid);
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}
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else
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{
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return(atoi(str));
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}
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}
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return(0);
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}
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/*
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* itoa - convert integer (decimal) to ascii string for positive numbers
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* only (we don't bother with negative numbers since we know we
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* don't use them).
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*/
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/*
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* How do we know that 16 will suffice?
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* Because the biggest number that we will
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* ever convert will be 2^32-1, which is 10
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* digits.
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*/
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_Static_assert(sizeof(int) <= 4, "buffer too small for this sized int");
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char *
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itoa(unsigned int val)
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{
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char *ptr;
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static char buffer[16]; /* result is built here */
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/* 16 is sufficient since the largest number
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we will ever convert will be 2^32-1,
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which is 10 digits. */
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ptr = buffer + sizeof(buffer);
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*--ptr = '\0';
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if (val == 0)
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{
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*--ptr = '0';
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}
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else while (val != 0)
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{
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*--ptr = (val % 10) + '0';
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val /= 10;
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}
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return(ptr);
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}
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/*
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* itoa7(val) - like itoa, except the number is right justified in a 7
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* character field. This code is a duplication of itoa instead of
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* a front end to a more general routine for efficiency.
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*/
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char *
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itoa7(int val)
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{
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char *ptr;
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static char buffer[16]; /* result is built here */
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/* 16 is sufficient since the largest number
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we will ever convert will be 2^32-1,
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which is 10 digits. */
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ptr = buffer + sizeof(buffer);
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*--ptr = '\0';
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if (val == 0)
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{
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*--ptr = '0';
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}
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else while (val != 0)
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{
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*--ptr = (val % 10) + '0';
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val /= 10;
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}
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while (ptr > buffer + sizeof(buffer) - 7)
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{
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*--ptr = ' ';
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}
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return(ptr);
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}
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/*
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* digits(val) - return number of decimal digits in val. Only works for
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* positive numbers. If val <= 0 then digits(val) == 0.
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*/
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int
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digits(int val)
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{
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int cnt = 0;
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while (val > 0)
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{
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cnt++;
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val /= 10;
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}
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return(cnt);
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}
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/*
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* string_index(string, array) - find string in array and return index
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*/
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int
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string_index(const char *string, const char * const *array)
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{
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size_t i = 0;
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while (*array != NULL)
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{
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if (strcmp(string, *array) == 0)
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{
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return(i);
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}
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array++;
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i++;
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}
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return(-1);
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}
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/*
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* argparse(line, cntp) - parse arguments in string "line", separating them
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* out into an argv-like array, and setting *cntp to the number of
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* arguments encountered. This is a simple parser that doesn't understand
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* squat about quotes.
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*/
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const char * const *
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argparse(char *line, int *cntp)
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{
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const char **ap;
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static const char *argv[1024] = {0};
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*cntp = 1;
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ap = &argv[1];
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while ((*ap = strsep(&line, " ")) != NULL) {
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if (**ap != '\0') {
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(*cntp)++;
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if (*cntp >= (int)nitems(argv)) {
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break;
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}
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ap++;
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}
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}
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return (argv);
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}
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/*
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* percentages(cnt, out, new, old, diffs) - calculate percentage change
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* between array "old" and "new", putting the percentages i "out".
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* "cnt" is size of each array and "diffs" is used for scratch space.
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* The array "old" is updated on each call.
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* The routine assumes modulo arithmetic. This function is especially
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* useful on for calculating cpu state percentages.
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*/
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long
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percentages(int cnt, int *out, long *new, long *old, long *diffs)
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{
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int i;
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long change;
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long total_change;
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long *dp;
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long half_total;
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/* initialization */
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total_change = 0;
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dp = diffs;
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/* calculate changes for each state and the overall change */
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for (i = 0; i < cnt; i++)
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{
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if ((change = *new - *old) < 0)
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{
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/* this only happens when the counter wraps */
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change = (int)
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((unsigned long)*new-(unsigned long)*old);
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}
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total_change += (*dp++ = change);
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*old++ = *new++;
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}
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/* avoid divide by zero potential */
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if (total_change == 0)
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{
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total_change = 1;
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}
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/* calculate percentages based on overall change, rounding up */
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half_total = total_change / 2l;
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/* Do not divide by 0. Causes Floating point exception */
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if(total_change) {
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for (i = 0; i < cnt; i++)
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{
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*out++ = (int)((*diffs++ * 1000 + half_total) / total_change);
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}
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}
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/* return the total in case the caller wants to use it */
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return(total_change);
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}
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/* format_time(seconds) - format number of seconds into a suitable
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* display that will fit within 6 characters. Note that this
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* routine builds its string in a static area. If it needs
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* to be called more than once without overwriting previous data,
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* then we will need to adopt a technique similar to the
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* one used for format_k.
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*/
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/* Explanation:
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We want to keep the output within 6 characters. For low values we use
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the format mm:ss. For values that exceed 999:59, we switch to a format
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that displays hours and fractions: hhh.tH. For values that exceed
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999.9, we use hhhh.t and drop the "H" designator. For values that
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exceed 9999.9, we use "???".
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*/
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char *
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format_time(long seconds)
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{
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static char result[10];
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/* sanity protection */
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if (seconds < 0 || seconds > (99999l * 360l))
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{
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strcpy(result, " ???");
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}
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else if (seconds >= (1000l * 60l))
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{
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/* alternate (slow) method displaying hours and tenths */
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sprintf(result, "%5.1fH", (double)seconds / (double)(60l * 60l));
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/* It is possible that the sprintf took more than 6 characters.
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If so, then the "H" appears as result[6]. If not, then there
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is a \0 in result[6]. Either way, it is safe to step on.
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*/
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result[6] = '\0';
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}
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else
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{
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/* standard method produces MMM:SS */
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/* we avoid printf as must as possible to make this quick */
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sprintf(result, "%3ld:%02ld",
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(long)(seconds / 60), (long)(seconds % 60));
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}
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return(result);
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}
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/*
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* format_k(amt) - format a kilobyte memory value, returning a string
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* suitable for display. Returns a pointer to a static
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* area that changes each call. "amt" is converted to a
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* string with a trailing "K". If "amt" is 10000 or greater,
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* then it is formatted as megabytes (rounded) with a
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* trailing "M".
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*/
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/*
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* Compromise time. We need to return a string, but we don't want the
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* caller to have to worry about freeing a dynamically allocated string.
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* Unfortunately, we can't just return a pointer to a static area as one
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* of the common uses of this function is in a large call to sprintf where
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* it might get invoked several times. Our compromise is to maintain an
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* array of strings and cycle thru them with each invocation. We make the
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* array large enough to handle the above mentioned case. The constant
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* NUM_STRINGS defines the number of strings in this array: we can tolerate
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* up to NUM_STRINGS calls before we start overwriting old information.
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* Keeping NUM_STRINGS a power of two will allow an intelligent optimizer
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* to convert the modulo operation into something quicker. What a hack!
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*/
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#define NUM_STRINGS 8
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char *
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format_k(int amt)
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{
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static char retarray[NUM_STRINGS][16];
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static int index = 0;
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char *p;
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char *ret;
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char tag = 'K';
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p = ret = retarray[index];
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index = (index + 1) % NUM_STRINGS;
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if (amt >= 10000)
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{
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amt = (amt + 512) / 1024;
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tag = 'M';
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if (amt >= 10000)
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{
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amt = (amt + 512) / 1024;
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tag = 'G';
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}
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}
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p = stpcpy(p, itoa(amt));
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*p++ = tag;
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*p = '\0';
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return(ret);
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}
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char *
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format_k2(unsigned long long amt)
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{
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static char retarray[NUM_STRINGS][16];
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static int index = 0;
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char *p;
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char *ret;
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char tag = 'K';
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p = ret = retarray[index];
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index = (index + 1) % NUM_STRINGS;
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if (amt >= 100000)
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{
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amt = (amt + 512) / 1024;
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tag = 'M';
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if (amt >= 100000)
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{
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amt = (amt + 512) / 1024;
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tag = 'G';
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}
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}
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p = stpcpy(p, itoa((int)amt));
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*p++ = tag;
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*p = '\0';
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return(ret);
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}
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int
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find_pid(pid_t pid)
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{
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kvm_t *kd = NULL;
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struct kinfo_proc *pbase = NULL;
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int nproc;
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int ret = 0;
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kd = kvm_open(NULL, _PATH_DEVNULL, NULL, O_RDONLY, NULL);
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if (kd == NULL) {
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fprintf(stderr, "top: kvm_open() failed.\n");
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quit(TOP_EX_SYS_ERROR);
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}
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pbase = kvm_getprocs(kd, KERN_PROC_PID, pid, &nproc);
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if (pbase == NULL) {
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goto done;
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}
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if ((nproc == 1) && (pbase->ki_pid == pid)) {
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ret = 1;
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}
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done:
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kvm_close(kd);
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return ret;
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}
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