The following code seems to segfault and I cannot figure out why.
以下代码似乎是段错误，我无法弄清楚原因。
#include <string.h>
static char src[] = "aaa";
int main()
{
char* target[2] = {"cccc","bbbbbbbbbb"};
strcpy(target[1],src);
return 0;
}
1 个解决方案
#1
11
Because target[1] is a pointer to "bbbbbbbbbb" and you are not allowed to modify string constants. It's undefined behaviour.
因为target [1]是指向“bbbbbbbbbb”的指针，所以不允许修改字符串常量。这是未定义的行为。
It's no different to:
它与以下内容没有什么不同：
char *x = "bbb";
x[0] = 'a';
I think you may be confusing it with:
我想你可能会混淆它：
char x[] = "bbb";
x[0] = 'a';
which is valid since it creates an array that you are allowed to modify. But what yours gives you:
这是有效的，因为它创建了一个允许您修改的数组。但是你的东西给了你：
char* target[2] = {"cccc","bbbbbbbbbb"};
is an array of pointers, all of which point to non-modifiable memory.
是一个指针数组，所有这些都指向不可修改的内存。
If you were to try:
如果您尝试：
char t0[] = "cccc";
char t1[] = "bbbbbbbbbb";
char* target[2] = {t0, t1};
strcpy(target[1],src);
you would find that it works since target[1] now points to t1, which is modifiable.
你会发现它有效，因为目标[1]现在指向t1，这是可修改的。
#1
11
Because target[1] is a pointer to "bbbbbbbbbb" and you are not allowed to modify string constants. It's undefined behaviour.
因为target [1]是指向“bbbbbbbbbb”的指针，所以不允许修改字符串常量。这是未定义的行为。
It's no different to:
它与以下内容没有什么不同：
char *x = "bbb";
x[0] = 'a';
I think you may be confusing it with:
我想你可能会混淆它：
char x[] = "bbb";
x[0] = 'a';
which is valid since it creates an array that you are allowed to modify. But what yours gives you:
这是有效的，因为它创建了一个允许您修改的数组。但是你的东西给了你：
char* target[2] = {"cccc","bbbbbbbbbb"};
is an array of pointers, all of which point to non-modifiable memory.
是一个指针数组，所有这些都指向不可修改的内存。
If you were to try:
如果您尝试：
char t0[] = "cccc";
char t1[] = "bbbbbbbbbb";
char* target[2] = {t0, t1};
strcpy(target[1],src);
you would find that it works since target[1] now points to t1, which is modifiable.
你会发现它有效，因为目标[1]现在指向t1，这是可修改的。