我有一个c代码，它给我的错误是分段错误，错误意味着我没有得到它。这是代码：
#include <stdio.h>
#include <assert.h>
#include <stdlib.h>
#include <string.h>
#define STREQUAL(a,b) (strcmp(a,b) == 0)
/* State of the 54-card deck. This keeps a spare deck for copying
into. It also has three spare slots *behind* the start of the
deck: two so the deck can be moved backward if a joker is moved
from the bottom to the top in the first step, and one so that the
reference to the card before the first joker always points
somewhere even when there's a joker on the top of the pack. */
typedef struct SolState_t {
int a, b;
int *deck, *spare;
int deck1[57], deck2[57];
} SolState_t ;
SolState_t state;
int verbose = 0;
int lastout, cocount;
#define JOKER_STEP(var,ovar) \
(((var != 53) ? \
(source[var] = source[var +1], var++) : \
(source--, ovar++, source[0] = source[1], var = 1)), \
((var == ovar)?(ovar--):0))
/* Cycle the state for "rounds" outputs, skipping jokers
as usual. "lastout" is the last output, which is never a joker.
If "rounds" is zero though, cycle the state just once, even
if the output card is a joker. "lastout" may or may not be set.
This is only useful for key setup.
Note that for performance reasons, this updates the coincidence
statistics under all circumstances, so they need to be set to zero
immediately before the large batch run. */
static void cycle_deck(
int rounds
)
{
int *source, *s, *sb, *d;
int lo, hi;
int nlo, nhi, nccut;
int output;
do {
assert(state.a != state.b);
assert(state.deck[state.a] == 53);
assert(state.deck[state.b] == 53);
source = state.deck;
JOKER_STEP(state.a,state.b);
JOKER_STEP(state.b,state.a);
JOKER_STEP(state.b,state.a);
source[state.a] = 53;
source[state.b] = 53;
if (state.a < state.b) {
lo = state.a;
hi = state.b + 1;
} else {
lo = state.b;
hi = state.a + 1;
}
nlo = 54 - hi;
nhi = 54 - lo;
/* We do both the triple cut and the count cut as one
copying step; this means handling four separate cases. */
nccut = source[lo -1];
s = source;
if (lo == 0) {
/* There's a joker on the top of the pack. This can
only happen in one exact circumstance, but when it
does nccount is wrong. So we handle it specially. */
assert(state.a == 0);
assert(state.b == 2);
d = &state.spare[51];
sb = &source[3];
while(s < sb) {*d++ = *s++;}
d = &state.spare[0];
sb = &source[54];
while(s < sb) {*d++ = *s++;}
state.a = 51;
state.b = 53;
} else if (nccut <= nlo) {
/* The second cut is before the first joker. */
d = &state.spare[nhi - nccut];
sb = &source[lo -1];
while(s < sb) {*d++ = *s++;}
state.spare[53] = *s++;
d = &state.spare[nlo - nccut];
sb = &source[hi];
while(s < sb) {*d++ = *s++;}
d = &state.spare[53 - nccut];
sb = &source[nccut + hi]; /* ccut */
while(s < sb) {*d++ = *s++;}
d = &state.spare[0];
sb = &source[54];
while(s < sb) {*d++ = *s++;}
state.a += nlo - nccut - lo;
state.b += nlo - nccut - lo;
} else if (nccut < nhi) {
/* The second cut is between the two jokers */
d = &state.spare[nhi - nccut];
sb = &source[lo -1];
while(s < sb) {*d++ = *s++;}
state.spare[53] = *s++;
d = &state.spare[53 - nccut + nlo];
sb = &source[nccut - nlo + lo]; /* ccut */
while(s < sb) {*d++ = *s++;}
d = &state.spare[0];
sb = &source[hi];
while(s < sb) {*d++ = *s++;}
d = &state.spare[53 - nccut];
sb = &source[54];
while(s < sb) {*d++ = *s++;}
if (state.a < state.b) {
state.a = 53 - nccut + nlo;
state.b = nhi - nccut -1;
} else {
state.b = 53 - nccut + nlo;
state.a = nhi - nccut -1;
}
} else {
/* The second cut is after the last joker. */
d = &state.spare[53 - nccut + nhi];
sb = &source[nccut - nhi]; /* ccut */
while(s < sb) {*d++ = *s++;}
d = &state.spare[0];
sb = &source[lo -1];
while(s < sb) {*d++ = *s++;}
state.spare[53] = *s++;
d = &state.spare[53 - nccut + nlo];
sb = &source[hi];
while(s < sb) {*d++ = *s++;}
d = &state.spare[53 - nccut];
sb = &source[54];
while(s < sb) {*d++ = *s++;}
state.a += 53 - nccut + nlo - lo;
state.b += 53 - nccut + nlo - lo;
}
source = state.deck;
state.deck = state.spare;
state.spare = source;
output = state.deck[state.deck[0]];
if (output >= 26) {
if (output >= 52) {
if (output > 52)
continue;
output = 0;
} else {
output -= 26;
}
}
cocount += (lastout == output);
lastout = output;
rounds--;
} while (rounds > 0);
}
static void print_deck(
)
{
int i;
for (i = 0; i < 54; i++) {
if (state.deck[i] < 53) {
putchar(' ' + state.deck[i]);
} else if (i == state.a) {
putchar('U');
} else {
assert(i == state.b);
putchar('V');
}
}
}
/* Key the deck with a passphrase. */
static void key_deck(
char *key
)
{
int i, kval, *tmp;
state.deck = state.deck1 + 3;
state.spare = state.deck2 + 3;
for (i = 0; i < 52; i++) {
state.deck[i] = i+1;
}
state.deck[state.a = 52] = 53;
state.deck[state.b = 53] = 53;
for (; *key != '\0'; key++) {
if ( *key >= 'A' && *key <= 'Z' ) {
cycle_deck(0); /* Special value '0' is only useful here... */
/* And now perform a second count cut based on the key letter */
kval = *key - 'A' + 1;
for (i = 0; i < 53; i++)
state.spare[i] = state.deck[(i + kval) % 53];
state.spare[53] = state.deck[53];
if (state.a != 53)
state.a = (state.a + 53 - kval) % 53;
if (state.b != 53)
state.b = (state.b + 53 - kval) % 53;
tmp = state.deck;
state.deck = state.spare;
state.spare = tmp;
if (verbose) {
print_deck();
printf(" after %c\n", *key);
}
}
}
/* These are touched by the keying: fix them. */
lastout = 100; cocount = 0;
}
/* Encrypt a single character. */
static char encrypt_char(
char char_in
)
{
char char_out;
cycle_deck(1);
char_out = 'A' + (char_in - 'A' + lastout) % 26;
if (verbose) {
print_deck();
printf(" %c -> %c\n", char_in, char_out);
}
return char_out;
}
int main(
int argc,
char *argv[]
)
{
char **av = argv, *tmp;
int slow_mode = 0;
long rounds;
/* Skip the name of the program */
av++; argc--;
if (argc < 2) {
printf("Usage: [flags] key message|len\n");
}
while (argc > 2) {
if (STREQUAL(*av, "-v")) {
verbose = 1;
} else if (STREQUAL(*av, "-s")) {
slow_mode = 1;
} else {
printf ("Unrecognised flag: %s\n", *av);
exit(-1);
}
av++; argc--;
}
key_deck(av[0]);
rounds = strtol(av[1], &tmp, 0);
if (*tmp != '\0') {
/* It's not a number - so it's a string! */
char *text = av[1];
int i = 0;
for (; *text != '\0'; text++) {
if (*text >= 'A' && *text <= 'Z') {
if (i > 0 && (i % 5) == 0)
putchar(' ');
putchar(encrypt_char(*text));
i++;
}
}
while ((i % 5) != 0) {
putchar(encrypt_char('X'));
i++;
}
putchar('\n');
} else {
/* Treat it as a sequence of 'A's. */
int i;
if (rounds <= 0) {
printf("Rounds number must be greater than zero\n");
exit(-1);
}
if (verbose || slow_mode) {
for (i = 0; i < rounds; i++)
encrypt_char('A');
} else {
cycle_deck(rounds);
}
printf("Coincidences: %d / %ld\n", cocount, rounds -1);
}
return 0;
}
答案 0 :(得分：4)
编辑答案以响应您的编辑，尝试使用GDB： 跑： gdb myapp.exe 然后给出命令“start”并在“step”之后直到你遇到问题为止。它将帮助您将问题区域缩小到一个特定的功能。
我应该提到在调试时获取完整信息，使用“-g”开关进行编译（使用gcc）。有关GDB的完整教程可用here
答案 1 :(得分：4)
segmentation fault表示内存访问冲突。它通常是在C语言编程时取消引用dangling pointer或访问超出范围的数组索引的结果。
尝试使用gdb逐步完成程序并确定发生分段错误的位置。