c程序段错误对我没有意义（c初学者）

Question

这个程序可以编译，但是当运行时输出 find() 方法中的一些第一个打印语句，然后在 for 循环命中时发出 Segmentation fault:11。

所以我尝试使用 gdb 进行调试，但无法找出问题所在。根据我从 gdb 收集的信息，由于 strstr() 方法之一发生了段错误。另外，我注意到程序在调用 find 方法中的 for 循环之前关闭。 为什么这个程序在运行时编译但崩溃了？

这是 gdb 的输出：

Program received signal EXC_BAD_ACCESS, Could not access memory.
Reason: KERN_INVALID_ADDRESS at address: 0x0000000000000001
0x00007fff8d6469c4 in strstr ()
(gdb) 

感谢所有帮助。我实际上是 C 的初学者，但我正在努力学习更多。

#include <stdio.h>
#include <string.h>


int acdc_or_metallica(char *s);
int surfing_or_TV(char *s);
int exo_mustard(char *s);
int arts_theater_or_dining(char *s);

void find(int(*match)(char*));


int NUM_STUFF = 7;
char *STUFF[] = {

"Butthead likes ACDC and TV",
"Beavis likes Metallica and TV",
"Cartman likes to eat",
"Spiderman likes theater and working-out",
"Silver Surfer likes surfing and space",
"GunSword likes mustard and exoskeletons"
"Meatwad likes TV"

};


int main() 
{

find(acdc_or_metallica);
find(surfing_or_TV);
find(exo_mustard);
find(arts_theater_or_dining);

return 0;

}

int acdc_or_metallica(char *s)
{
return strstr(s, "acdc") || strstr(s, "metallica");
}

int surfing_or_TV(char *s)
{
return strstr(s, "surfing") || strstr(s, "TV");
}

int exo_mustard(char *s)
{
return strstr(s, "exoskeleton") && strstr(s, "mustard");
}

int arts_theater_or_dining(char *s)
{
return strstr(s, "arts") || strstr(s, "theater") || strstr(s, "dining");
}

void find(int(*match)(char*))
{
int i;
puts("Search results:");
puts("error below this line...");

puts("-----------------------------------------------");
for (i = 0;i < NUM_STUFF; i++) {
     if (match(STUFF[i])) {
        printf("%s\n", STUFF[i]);       
    }
}
puts("-----------------------------------------------");
}

Answer 1

您正在使用NULL 指针调用match。

您将 NUM_STUFF 声明为 7（在 C 中应该是 #define NUM_STUFF 7），并且您的 STUFF 数组有 5 个字符串，因为缺少两个逗号

"Silver Surfer likes surfing and space", /* you forgot the comma */
"GunSword likes mustard and exoskeletons", /* you forgot the comma */

（因为你忘记了逗号，而下一个标记是一个字符串，两行都连接在一个字符串中）

^{我在 Linux 上通过indent-ing 你的代码发现了该错误，然后将其编译为gcc -Wall -g u1.c -o u1 并运行gdb u1 进行调试。}

你应该声明

#define NUM_STUFF 7
char *STUFF[NUM_STUFF] = {

---

实际上，您应该添加NULL 终止数组的约定。 那么你会有：

char *STUFF[] = {
  "Butthead likes ACDC and TV",
  "Beavis likes Metallica and TV",
  "Cartman likes to eat",
  "Spiderman likes theater and working-out",
  "Silver Surfer likes surfing and space",
  "GunSword likes mustard and exoskeletons", 
  "Meatwad likes TV",
  NULL
};

^{使用终止符NULL 编码可以更轻松地添加新句子。在这种情况下无需定义或更改NUM_STUFF。}

那么你的主循环就是

for (i = 0; STUFF[i] != NULL; i++)

---

风格提示：我发现

void find(int(*match)(char*));

很难阅读。我个人更喜欢声明 typedef 来作为潜在指向函数的签名，即我更喜欢编码

typedef int matchfun_t (char*);
void find (matchfun_t*);

Answer 2

 char *STUFF[] = {

"Butthead likes ACDC and TV",
"Beavis likes Metallica and TV",
"Cartman likes to eat",
"Spiderman likes theater and working-out",
"Silver Surfer likes surfing and space",
"GunSword likes mustard and exoskeletons" //<---- , 
"Meatwad likes TV"

};

你忘记了“，”，因为数据不够