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diff: failed to confirm "static bug" in gcc - reinstating statics.

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microscopic code improvements.
This commit is contained in:
Denis Vlasenko 2007-03-09 10:08:53 +00:00
parent f5a157615d
commit 02f0c4c2bf
3 changed files with 81 additions and 78 deletions
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5
coreutils/chgrp.c
View File

@ -7,15 +7,12 @@
* Licensed under GPLv2 or later, see file LICENSE in this tarball for details.
*/
/* BB_AUDIT SUSv3 defects - unsupported options -H, -L, and -P. */
/* BB_AUDIT SUSv3 defects - none? */
/* BB_AUDIT GNU defects - unsupported long options. */
/* http://www.opengroup.org/onlinepubs/007904975/utilities/chgrp.html */
#include "busybox.h"
/* FIXME - move to .h */
extern int chown_main(int argc, char **argv);
int chgrp_main(int argc, char **argv);
int chgrp_main(int argc, char **argv)
{

153
coreutils/diff.c
View File

@ -65,18 +65,25 @@
#define FLAG_U (1<<12)
#define FLAG_w (1<<13)
/* XXX: FIXME: the following variables should be static, but gcc currently
/* The following variables should be static, but gcc currently
* creates a much bigger object if we do this. [which version of gcc? --vda] */
/* 4.x, IIRC also 3.x --bernhard */
/* Works for gcc 3.4.3. Sizes without and with "static":
# size busybox.t[34]/coreutils/diff.o
text data bss dec hex filename
6969 8 305 7282 1c72 busybox.t3/coreutils/diff.o
6969 8 305 7282 1c72 busybox.t4/coreutils/diff.o
--vda
*/
/* This is the default number of lines of context. */
int context = 3;
int status;
char *start;
const char *label1;
const char *label2;
struct stat stb1, stb2;
char **dl;
USE_FEATURE_DIFF_DIR(int dl_count;)
static int context = 3;
static int status;
static char *start;
static const char *label1;
static const char *label2;
static struct stat stb1, stb2;
static char **dl;
USE_FEATURE_DIFF_DIR(static int dl_count;)
struct cand {
int x;
@ -84,7 +91,7 @@ struct cand {
int pred;
};
struct line {
static struct line {
int serial;
int value;
} *file[2];
@ -188,7 +195,7 @@ static int readhash(FILE * f)
sum = 1;
space = 0;
if (!(option_mask32 & FLAG_b) && !(option_mask32 & FLAG_w)) {
if (!(option_mask32 & (FLAG_b | FLAG_w))) {
for (i = 0; (t = getc(f)) != '\n'; i++) {
if (t == EOF) {
if (i == 0)
@ -241,19 +248,18 @@ static int files_differ(FILE * f1, FILE * f2, int flags)
{
size_t i, j;
if ((flags & (D_EMPTY1 | D_EMPTY2)) || stb1.st_size != stb2.st_size ||
(stb1.st_mode & S_IFMT) != (stb2.st_mode & S_IFMT))
if ((flags & (D_EMPTY1 | D_EMPTY2)) || stb1.st_size != stb2.st_size
|| (stb1.st_mode & S_IFMT) != (stb2.st_mode & S_IFMT)
) {
return 1;
}
while (1) {
i = fread(bb_common_bufsiz1, 1, BUFSIZ/2, f1);
j = fread(bb_common_bufsiz1 + BUFSIZ/2, 1, BUFSIZ/2, f2);
if (i != j)
return 1;
if (i == 0 && j == 0) {
if (ferror(f1) || ferror(f2))
return 1;
return 0;
}
if (i == 0)
return (ferror(f1) || ferror(f2));
if (memcmp(bb_common_bufsiz1,
bb_common_bufsiz1 + BUFSIZ/2, i) != 0)
return 1;
@ -337,11 +343,11 @@ static void equiv(struct line *a, int n, struct line *b, int m, int *c)
static int isqrt(int n)
{
int y, x = 1;
int y, x;
if (n == 0)
return 0;
x = 1;
do {
y = x;
x = n / x;
@ -647,7 +653,6 @@ static void fetch(long *f, int a, int b, FILE * lb, int ch)
}
}
}
return;
}
@ -828,66 +833,66 @@ static void output(char *file1, FILE * f1, char *file2, FILE * f2)
}
/*
* The following code uses an algorithm due to Harold Stone,
* which finds a pair of longest identical subsequences in
* the two files.
* The following code uses an algorithm due to Harold Stone,
* which finds a pair of longest identical subsequences in
* the two files.
*
* The major goal is to generate the match vector J.
* J[i] is the index of the line in file1 corresponding
* to line i file0. J[i] = 0 if there is no
* such line in file1.
* The major goal is to generate the match vector J.
* J[i] is the index of the line in file1 corresponding
* to line i file0. J[i] = 0 if there is no
* such line in file1.
*
* Lines are hashed so as to work in core. All potential
* matches are located by sorting the lines of each file
* on the hash (called ``value''). In particular, this
* collects the equivalence classes in file1 together.
* Subroutine equiv replaces the value of each line in
* file0 by the index of the first element of its
* matching equivalence in (the reordered) file1.
* To save space equiv squeezes file1 into a single
* array member in which the equivalence classes
* are simply concatenated, except that their first
* members are flagged by changing sign.
* Lines are hashed so as to work in core. All potential
* matches are located by sorting the lines of each file
* on the hash (called ``value''). In particular, this
* collects the equivalence classes in file1 together.
* Subroutine equiv replaces the value of each line in
* file0 by the index of the first element of its
* matching equivalence in (the reordered) file1.
* To save space equiv squeezes file1 into a single
* array member in which the equivalence classes
* are simply concatenated, except that their first
* members are flagged by changing sign.
*
* Next the indices that point into member are unsorted into
* array class according to the original order of file0.
* Next the indices that point into member are unsorted into
* array class according to the original order of file0.
*
* The cleverness lies in routine stone. This marches
* through the lines of file0, developing a vector klist
* of "k-candidates". At step i a k-candidate is a matched
* pair of lines x,y (x in file0 y in file1) such that
* there is a common subsequence of length k
* between the first i lines of file0 and the first y
* lines of file1, but there is no such subsequence for
* any smaller y. x is the earliest possible mate to y
* that occurs in such a subsequence.
* The cleverness lies in routine stone. This marches
* through the lines of file0, developing a vector klist
* of "k-candidates". At step i a k-candidate is a matched
* pair of lines x,y (x in file0 y in file1) such that
* there is a common subsequence of length k
* between the first i lines of file0 and the first y
* lines of file1, but there is no such subsequence for
* any smaller y. x is the earliest possible mate to y
* that occurs in such a subsequence.
*
* Whenever any of the members of the equivalence class of
* lines in file1 matable to a line in file0 has serial number
* less than the y of some k-candidate, that k-candidate
* with the smallest such y is replaced. The new
* k-candidate is chained (via pred) to the current
* k-1 candidate so that the actual subsequence can
* be recovered. When a member has serial number greater
* that the y of all k-candidates, the klist is extended.
* At the end, the longest subsequence is pulled out
* and placed in the array J by unravel
* Whenever any of the members of the equivalence class of
* lines in file1 matable to a line in file0 has serial number
* less than the y of some k-candidate, that k-candidate
* with the smallest such y is replaced. The new
* k-candidate is chained (via pred) to the current
* k-1 candidate so that the actual subsequence can
* be recovered. When a member has serial number greater
* that the y of all k-candidates, the klist is extended.
* At the end, the longest subsequence is pulled out
* and placed in the array J by unravel
*
* With J in hand, the matches there recorded are
* checked against reality to assure that no spurious
* matches have crept in due to hashing. If they have,
* they are broken, and "jackpot" is recorded--a harmless
* matter except that a true match for a spuriously
* mated line may now be unnecessarily reported as a change.
* With J in hand, the matches there recorded are
* checked against reality to assure that no spurious
* matches have crept in due to hashing. If they have,
* they are broken, and "jackpot" is recorded--a harmless
* matter except that a true match for a spuriously
* mated line may now be unnecessarily reported as a change.
*
* Much of the complexity of the program comes simply
* from trying to minimize core utilization and
* maximize the range of doable problems by dynamically
* allocating what is needed and reusing what is not.
* The core requirements for problems larger than somewhat
* are (in words) 2*length(file0) + length(file1) +
* 3*(number of k-candidates installed), typically about
* 6n words for files of length n.
* Much of the complexity of the program comes simply
* from trying to minimize core utilization and
* maximize the range of doable problems by dynamically
* allocating what is needed and reusing what is not.
* The core requirements for problems larger than somewhat
* are (in words) 2*length(file0) + length(file1) +
* 3*(number of k-candidates installed), typically about
* 6n words for files of length n.
*/
static unsigned diffreg(char * ofile1, char * ofile2, int flags)
{

1
include/libbb.h
View File

@ -514,6 +514,7 @@ int bb_test(int argc, char** argv);
#if ENABLE_ROUTE
void bb_displayroutes(int noresolve, int netstatfmt);
#endif
int chown_main(int argc, char **argv);
#if ENABLE_GUNZIP
int gunzip_main(int argc, char **argv);
#endif