Question

我正在关注K＆amp; R.在第1章末尾，提议的练习之一是编写文本包装程序。这是我写的代码：

#include <stdio.h>
#define MAXLINE 1000
#define WRAP 10

int getl(char s[]);
void wrap(char s[], int l);

/* wrap.c: wraps text */
main ()
{
    char line[MAXLINE];
    int c;

    while ((c = getl(line)) >= 0)
        wrap(line, c);
    return 0;
}

int getl(char s[])
{
    int i, c;

    for (i = 0; (c = getchar()) != '\n' && c != EOF && i < MAXLINE; ++i)
        s[i] = c;
    s[i] = '\n';
    if (c == EOF)
        i = -1;
    return i;
}

void wrap(char s[], int l)
{
    int i = 1,ii, c;

    if (l == 0)
        putchar('\n');
    for(c = 0; s[i-1] != '\n'; c = c + i)
    {
        if (l - c <= WRAP)
            for (i = c; i <= l; ++i)
                putchar(s[i]);
        else
        {
            for (i = c + WRAP - 1; i >= c && s[i] != ' ' && s[i] != '\t'; --i)
                ;
            if (i == (c-1))
            {
                for (i = c + WRAP; s[i] != ' ' && s[i] != '\t' && s[i] != '\n'; ++i)
                    ;
                for (ii = c; ii < i; ++ii)
                    putchar(s[ii]);
                putchar('\n');
                if (s[i] != '\n')
                    ++i;
            }
            else if (i == c)
                ++i;
            else
            {
                for (ii = c; ii < i; ++ii)
                    putchar(s[ii]);
                putchar('\n');
                ++i;
            }
        }
    }
}

我写了一个名为wraptest的文件，其中包括以下内容：

asd

asdf
asd fg

asdflkjasdad sasdf
adsfa asdfasfd adfsdf
asd asdfadfasdfad
this line

运行时

./wrap < wraptest

我得到以下

asd

asdf
asd fg

asdflkjasdad
sasdf
adsfa
asdfasfd

asd
asdfadfasdfad
Segmentation fault (core dumped)

Valgrind给了我以下内容：

==6729== Conditional jump or move depends on uninitialised value(s)
==6729==    at 0x80486EE: wrap (wrap.c:37)
==6729==    by 0x80484CC: main (wrap.c:15)
==6729==  Uninitialised value was created by a stack allocation  
==6729==    at 0x80484A2: main (wrap.c:10)
==6729== 
==6729== Invalid read of size 1 
==6729==    at 0x80486E9: wrap (wrap.c:37)
==6729==    by 0x80484CC: main (wrap.c:15)
==6729==  Address 0xbe8a3623 is not stack'd, malloc'd or (recently) free'd
==6729== 
==6729== 
==6729== Process terminating with default action of signal 11 (SIGSEGV)
==6729==  Access not within mapped region at address 0xBE8A3623
==6729==    at 0x80486E9: wrap (wrap.c:37)
==6729==    by 0x80484CC: main (wrap.c:15)
==6729==  If you believe this happened as a result of a stack
==6729==  overflow in your program's main thread (unlikely but
==6729==  possible), you can try to increase the size of the
==6729==  main thread stack using the --main-stacksize= flag.
==6729==  The main thread stack size used in this run was 8388608.
==6729== 
==6729== HEAP SUMMARY:
==6729==     in use at exit: 0 bytes in 0 blocks
==6729==   total heap usage: 0 allocs, 0 frees, 0 bytes allocated
==6729== 
==6729== All heap blocks were freed -- no leaks are possible
==6729== 
==6729== For counts of detected and suppressed errors, rerun with: -v
==6729== ERROR SUMMARY: 6 errors from 2 contexts (suppressed: 0 from 0)
Segmentation fault (core dumped)

那么错误是什么？我一直在寻找大约一个小时，但我还没有找到它。

Answer 1

您永远不会检查i的值，但它会在wrap函数中索引您的字符串数组。这是i值在崩溃之前在我的机器上增长到1000的位置。我建议将它放在调试器上，简化代码，或者至少检查索引到字符串缓冲区的变量，以确保它不会超过缓冲区的末尾。 HTH！

修改

既然你问过，我会在Ubuntu 12.04 LTS上向你展示我是如何调试它的。

从shell中调用名为sotest.c
的文件
# use the -g for debug info gcc -g sotest.c -o sotest # invoke the gnu debugger gdb sotest

你会看到：


GNU gdb（Ubuntu / Linaro 7.4-2012.04-0ubuntu2.1）7.4-2012.04版权所有   （C）2012 Free Software Foundation，Inc。许可证GPLv3 +：GNU GPL   版本3或更高版本http://gnu.org/licenses/gpl.html这是免费的   软件：您可以自由更改并重新分发它。没有   保证，在法律允许的范围内。输入“显示复制”和   详情请参阅“保修”。此GDB配置为   “i686的Linux的GNU的”。有关错误报告说明，请参阅：   http://bugs.launchpad.net/gdb-linaro/ ...从中读取符号   /home/macduff/sotest...done。   （GDB）

然后，使用输入文件调用程序：

run < wraptest

然后你会得到一些带有堆栈跟踪的输出文本


编程接收信号SIGSEGV，分段故障。 0x08048666 in   包裹（       s = 0xbfffebd4“asd \ n \ 030 \ 313 \ 375 \ 267 \ 254 | \ 376 \ 267 \ f \ 356 \ 377 \ 277”，l = 3）       在sotest.c：58 58 for（i = c + WRAP - 1; i＆gt; = c＆amp;＆amp; s [i]！=''＆amp;＆amp; s [i]！='\ t'; - -i）

c中的文本换行程序无法正常工作

1 个答案: