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diff: failed to confirm "static bug" in gcc - reinstating statics.
microscopic code improvements.
This commit is contained in:
Denis Vlasenko
parent
f5a157615d
commit
02f0c4c2bf
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@ -7,15 +7,12 @@
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* Licensed under GPLv2 or later, see file LICENSE in this tarball for details.
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*/
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/* BB_AUDIT SUSv3 defects - unsupported options -H, -L, and -P. */
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/* BB_AUDIT SUSv3 defects - none? */
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/* BB_AUDIT GNU defects - unsupported long options. */
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/* http://www.opengroup.org/onlinepubs/007904975/utilities/chgrp.html */
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#include "busybox.h"
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/* FIXME - move to .h */
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extern int chown_main(int argc, char **argv);
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int chgrp_main(int argc, char **argv);
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int chgrp_main(int argc, char **argv)
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{
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153
coreutils/diff.c
153
coreutils/diff.c
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@ -65,18 +65,25 @@
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#define FLAG_U (1<<12)
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#define FLAG_w (1<<13)
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/* XXX: FIXME: the following variables should be static, but gcc currently
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/* The following variables should be static, but gcc currently
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* creates a much bigger object if we do this. [which version of gcc? --vda] */
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/* 4.x, IIRC also 3.x --bernhard */
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/* Works for gcc 3.4.3. Sizes without and with "static":
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# size busybox.t[34]/coreutils/diff.o
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text data bss dec hex filename
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6969 8 305 7282 1c72 busybox.t3/coreutils/diff.o
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6969 8 305 7282 1c72 busybox.t4/coreutils/diff.o
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--vda
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*/
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/* This is the default number of lines of context. */
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int context = 3;
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int status;
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char *start;
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const char *label1;
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const char *label2;
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struct stat stb1, stb2;
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char **dl;
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USE_FEATURE_DIFF_DIR(int dl_count;)
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static int context = 3;
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static int status;
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static char *start;
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static const char *label1;
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static const char *label2;
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static struct stat stb1, stb2;
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static char **dl;
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USE_FEATURE_DIFF_DIR(static int dl_count;)
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struct cand {
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int x;
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@ -84,7 +91,7 @@ struct cand {
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int pred;
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};
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struct line {
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static struct line {
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int serial;
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int value;
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} *file[2];
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@ -188,7 +195,7 @@ static int readhash(FILE * f)
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sum = 1;
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space = 0;
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if (!(option_mask32 & FLAG_b) && !(option_mask32 & FLAG_w)) {
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if (!(option_mask32 & (FLAG_b | FLAG_w))) {
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for (i = 0; (t = getc(f)) != '\n'; i++) {
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if (t == EOF) {
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if (i == 0)
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@ -241,19 +248,18 @@ static int files_differ(FILE * f1, FILE * f2, int flags)
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{
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size_t i, j;
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if ((flags & (D_EMPTY1 | D_EMPTY2)) || stb1.st_size != stb2.st_size ||
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(stb1.st_mode & S_IFMT) != (stb2.st_mode & S_IFMT))
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if ((flags & (D_EMPTY1 | D_EMPTY2)) || stb1.st_size != stb2.st_size
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|| (stb1.st_mode & S_IFMT) != (stb2.st_mode & S_IFMT)
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) {
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return 1;
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}
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while (1) {
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i = fread(bb_common_bufsiz1, 1, BUFSIZ/2, f1);
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j = fread(bb_common_bufsiz1 + BUFSIZ/2, 1, BUFSIZ/2, f2);
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if (i != j)
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return 1;
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if (i == 0 && j == 0) {
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if (ferror(f1) || ferror(f2))
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return 1;
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return 0;
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}
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if (i == 0)
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return (ferror(f1) || ferror(f2));
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if (memcmp(bb_common_bufsiz1,
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bb_common_bufsiz1 + BUFSIZ/2, i) != 0)
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return 1;
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@ -337,11 +343,11 @@ static void equiv(struct line *a, int n, struct line *b, int m, int *c)
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static int isqrt(int n)
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{
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int y, x = 1;
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int y, x;
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if (n == 0)
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return 0;
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x = 1;
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do {
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y = x;
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x = n / x;
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@ -647,7 +653,6 @@ static void fetch(long *f, int a, int b, FILE * lb, int ch)
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}
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}
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}
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return;
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}
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@ -828,66 +833,66 @@ static void output(char *file1, FILE * f1, char *file2, FILE * f2)
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}
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/*
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* The following code uses an algorithm due to Harold Stone,
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* which finds a pair of longest identical subsequences in
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* the two files.
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* The following code uses an algorithm due to Harold Stone,
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* which finds a pair of longest identical subsequences in
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* the two files.
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*
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* The major goal is to generate the match vector J.
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* J[i] is the index of the line in file1 corresponding
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* to line i file0. J[i] = 0 if there is no
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* such line in file1.
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* The major goal is to generate the match vector J.
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* J[i] is the index of the line in file1 corresponding
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* to line i file0. J[i] = 0 if there is no
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* such line in file1.
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*
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* Lines are hashed so as to work in core. All potential
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* matches are located by sorting the lines of each file
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* on the hash (called ``value''). In particular, this
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* collects the equivalence classes in file1 together.
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* Subroutine equiv replaces the value of each line in
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* file0 by the index of the first element of its
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* matching equivalence in (the reordered) file1.
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* To save space equiv squeezes file1 into a single
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* array member in which the equivalence classes
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* are simply concatenated, except that their first
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* members are flagged by changing sign.
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* Lines are hashed so as to work in core. All potential
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* matches are located by sorting the lines of each file
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* on the hash (called ``value''). In particular, this
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* collects the equivalence classes in file1 together.
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* Subroutine equiv replaces the value of each line in
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* file0 by the index of the first element of its
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* matching equivalence in (the reordered) file1.
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* To save space equiv squeezes file1 into a single
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* array member in which the equivalence classes
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* are simply concatenated, except that their first
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* members are flagged by changing sign.
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*
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* Next the indices that point into member are unsorted into
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* array class according to the original order of file0.
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* Next the indices that point into member are unsorted into
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* array class according to the original order of file0.
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*
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* The cleverness lies in routine stone. This marches
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* through the lines of file0, developing a vector klist
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* of "k-candidates". At step i a k-candidate is a matched
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* pair of lines x,y (x in file0 y in file1) such that
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* there is a common subsequence of length k
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* between the first i lines of file0 and the first y
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* lines of file1, but there is no such subsequence for
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* any smaller y. x is the earliest possible mate to y
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* that occurs in such a subsequence.
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* The cleverness lies in routine stone. This marches
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* through the lines of file0, developing a vector klist
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* of "k-candidates". At step i a k-candidate is a matched
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* pair of lines x,y (x in file0 y in file1) such that
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* there is a common subsequence of length k
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* between the first i lines of file0 and the first y
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* lines of file1, but there is no such subsequence for
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* any smaller y. x is the earliest possible mate to y
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* that occurs in such a subsequence.
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*
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* Whenever any of the members of the equivalence class of
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* lines in file1 matable to a line in file0 has serial number
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* less than the y of some k-candidate, that k-candidate
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* with the smallest such y is replaced. The new
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* k-candidate is chained (via pred) to the current
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* k-1 candidate so that the actual subsequence can
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* be recovered. When a member has serial number greater
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* that the y of all k-candidates, the klist is extended.
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* At the end, the longest subsequence is pulled out
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* and placed in the array J by unravel
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* Whenever any of the members of the equivalence class of
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* lines in file1 matable to a line in file0 has serial number
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* less than the y of some k-candidate, that k-candidate
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* with the smallest such y is replaced. The new
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* k-candidate is chained (via pred) to the current
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* k-1 candidate so that the actual subsequence can
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* be recovered. When a member has serial number greater
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* that the y of all k-candidates, the klist is extended.
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* At the end, the longest subsequence is pulled out
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* and placed in the array J by unravel
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*
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* With J in hand, the matches there recorded are
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* checked against reality to assure that no spurious
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* matches have crept in due to hashing. If they have,
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* they are broken, and "jackpot" is recorded--a harmless
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* matter except that a true match for a spuriously
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* mated line may now be unnecessarily reported as a change.
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* With J in hand, the matches there recorded are
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* checked against reality to assure that no spurious
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* matches have crept in due to hashing. If they have,
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* they are broken, and "jackpot" is recorded--a harmless
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* matter except that a true match for a spuriously
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* mated line may now be unnecessarily reported as a change.
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*
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* Much of the complexity of the program comes simply
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* from trying to minimize core utilization and
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* maximize the range of doable problems by dynamically
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* allocating what is needed and reusing what is not.
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* The core requirements for problems larger than somewhat
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* are (in words) 2*length(file0) + length(file1) +
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* 3*(number of k-candidates installed), typically about
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* 6n words for files of length n.
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* Much of the complexity of the program comes simply
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* from trying to minimize core utilization and
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* maximize the range of doable problems by dynamically
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* allocating what is needed and reusing what is not.
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* The core requirements for problems larger than somewhat
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* are (in words) 2*length(file0) + length(file1) +
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* 3*(number of k-candidates installed), typically about
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* 6n words for files of length n.
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*/
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static unsigned diffreg(char * ofile1, char * ofile2, int flags)
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{
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@ -514,6 +514,7 @@ int bb_test(int argc, char** argv);
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#if ENABLE_ROUTE
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void bb_displayroutes(int noresolve, int netstatfmt);
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#endif
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int chown_main(int argc, char **argv);
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#if ENABLE_GUNZIP
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int gunzip_main(int argc, char **argv);
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#endif
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