Question

在此声明中：

char *a = "string1"

字符串字面是什么？是string1吗？因为这个帖子What is the type of string literals in C and C++?说的不同。

据我所知

int main()
{
    char *a = "string1"; //is a string- literals allocated memory in read-only section.
    char b[] = "string2"; //is a array char where memory will be allocated in stack.

    a[0] = 'X'; //Not allowed. It is an undefined Behaviour. For me, it Seg Faults. 
    b[0] = 'Y'; //Valid. 

    return 0;
}

请添加除上述要点以外的一些细节。感谢。

调试输出显示错误 a[0] = 'Y';

Reading symbols from /home/jay/Desktop/MI/chararr/a.out...done.
(gdb) b main
Breakpoint 1 at 0x40056c: file ddd.c, line 4.
(gdb) r
Starting program: /home/jay/Desktop/MI/chararr/a.out 

Breakpoint 1, main () at ddd.c:4
4   {
(gdb) n
6   char *a = "string1";
(gdb) n
7   char b[] = "string2";
(gdb) 
9   a[0] = 'Y';
(gdb) 

Program received signal SIGSEGV, Segmentation fault.
0x0000000000400595 in main () at ddd.c:9

Answer 1

您可以将字符串文字视为“由双引号括起来的字符序列”。此字符串存储在只读内存，尝试修改此内存会导致未定义的行为。

那你怎么得到分段错误？
- 重点是char *ptr = "string literal"使ptr指向存储字符串文字的只读内存。因此，当您尝试访问此内存时：ptr[0] = 'X'（相当于*(ptr + 0) = 'X'），这是内存访问冲突。

另一方面：char b[] = "string2";分配内存并将字符串"string2"复制到其中，因此修改它是有效的。当b超出范围时，将释放此内存。

查看Literal string initializer for a character array

字符串文字：指针与字符数组

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