我目前正在尝试解决K + R书籍的练习2.4,并遇到了一个奇怪的错误,我无法在其他地方重现。我正在使用:
Apple LLVM version 8.0.0 (clang-800.0.42.1)
Target: x86_64-apple-darwin16.4.0
Thread model: posix
InstalledDir: /Applications/Xcode.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/bin
代码是:
#include <stdio.h>
#include <string.h>
/*
* Write an alternate version of `squeeze(s1, s2)' that deletes each
character
* in s1 that matches any character in the string s2.
*/
void squeeze(char *s1, const char *s2);
int main(int argc, char **argv) {
char *tests[] = {"hello", "world", "these", "are", "some", "tests"};
for (unsigned int i = 0; i < sizeof(tests) / sizeof(tests[0]); i++) {
printf("'%s' = ", tests[i]);
squeeze(tests[i], "aeiou");
printf("'%s'\n", tests[i]);
}
return 0;
}
void squeeze(char *s1, const char *s2) {
const size_t s2len = strlen(s2);
s1[0] = s1[0];
unsigned int j = 0;
for (unsigned int i = 0; s1[i] != '\0'; i++) {
unsigned int k;
for (k = 0; k < s2len; k++)
if (s1[i] == s2[k]) break;
if (k == s2len) // we checked every character once, didn't find a bad char
s1[j++] = s1[i];
}
s1[j] = '\0';
}
GDB说:
Thread 2 received signal SIGBUS, Bus error.
0x0000000100000e57 in squeeze (s1=0x100000f78 "hello", s2=0x100000fa1
"aeiou")
at exercise2-4.c:23
23 s1[0] = s1[0];
错误最初发生在s1[j++] = s1[i]
,但是我插入s1[0] = s1[0]
来测试它而不依赖于变量,它也在那里发生。显然,我在这里遗漏了一些东西。
如果有任何相关性,我会使用clang -O0 -g -Weverything exercise2-4.c -o exercise2-4
进行编译。
非常感谢您的时间和对不起,如果之前已经回答过这个问题,我还没有发现任何问题,在这么奇怪的地方发生错误。
答案 0 :(得分:1)
您不能更改字符串文字。任何更改字符串文字的尝试都会导致未定义的行为。
来自C标准(6.4.5字符串文字)
7未指明这些阵列是否与它们不同 元素具有适当的值。 如果程序尝试 修改这样的数组,行为是未定义的。
而不是指向字符串文字的指针数组
char *tests[] = {"hello", "world", "these", "are", "some", "tests"};
你应该声明一个二维字符数组。例如
char tests[][6] = {"hello", "world", "these", "are", "some", "tests"};
如果要向其中添加新字符,还必须为数组的每个元素保留足够的空间。