c - 我是否误解了这个关于字符串文字范围的例子？

Question

我正在阅读常见的 C 陷阱并在 some famous Uni website 上找到这篇文章. (这是谷歌上出现的第二个链接)。

该页面上的最后一个示例是，

// Memory allocation on the stack
void b(char **p) {
    char * str="print this string";
    *p = str;
}

int main(void) {
    char * s;
    b(&s);
    s[0]='j'; //crash, since the memory for str is allocated on the stack, 
              //and the call to b has already returned, the memory pointed to by str 
              //is no longer valid.
    return 0;
}

评论中的解释让我想到，字符串文字的内存不是静态的吗？

那里的实际错误不是您不应该修改字符串文字，因为它是未定义的行为吗？或者那里的评论是否正确，而我对该示例的理解是错误的？

进一步搜索后，我看到了这个问题:referencing a char that went out of scope我从那个问题中了解到，以下是有效代码。

#include <malloc.h>
char* a = NULL;
{
    char* b = "stackoverflow";
    a = b;
}

int main() {
    puts(a);
}

还有这个question同意其他 stackoverflow 问题和我的想法，但反对该网站代码的评论。

为了测试它，我尝试了以下方法，

#include <stdio.h>
#include <malloc.h>

void b(char **p)
{
    char * str = "print this string";
    *p = str;
}

int main(void)
{
    char * s;
    b(&s);
    // s[0]='j'; //crash, since the memory for str is allocated on the stack,
                //and the call to b has already returned, the memory pointed to by str is no longer valid.
    printf("%s \n", s);
    return 0;
}

正如预期的那样，不会出现段错误。

Answer 1

标准说(强调是我的):

6.4.5 String literals

6. [...] The multibyte character sequence is then used to initialize an array of static storage duration and length just sufficient to contain the sequence. [...]

7. [...] If the program attempts to modify such an array, the behavior is undefined. [...]