Programmers can make mistakes. Basic pointers and arrays of C language

// a.c
char name[] = "tom";
// b.c
extern char *name;
void main()
{
    
	printf("%s", name);
}

Execution will fail , Why? ？

• Background knowledge 0
  The executable contains 4 Class segment content ：
  .text Store code snippets
  .data Store data segments . Initialized global or static variables
  .rodata Store read-only data segments . Like string constants
  .bss Store uninitialized global or static variables .

• Background knowledge 1
  char a[] = “abc”;
  char *b = “def”;
  a and b What's the difference? ？ In which paragraph ？
  difference 1：b It's a variable , Have your own address and stored data ;a It's just a name ,a Namely “abc” The address of .（ You can print a and &a, It's the same ）
  difference 2：“abc” Stored in data paragraph ,“def” Stored in rodata paragraph
  difference 3：“abc” The content can be modified ,“def” The content cannot be modified

• Background knowledge 2
  char a[] = “abc”;
  char *b = a;
  ask ：a[0] and b[0] What's the difference? ？
  a[0] Direct value ;b[0] First find b Stored values （a Address ）, And then according to a Address value
  
  Take the address casually a The value of is 0x3315,a[0] The address is also 0x3315

// a.c
char name[] = "tom";
// b.c
extern char *name;
void main()
{
    
	printf("%s", name);
}

So in b.c in ：extern char *name; Statement name It's a pointer , So when printing name When , Will find name Stored content ：“tom”, And then “tom” Take value as address .

resolvent 1：

// a.c
char name[] = "tom";
// b.c
extern char name[];
void main()
{
    
	printf("%s", name);
}

resolvent 2：

// a.c
char *name = "tom";
// b.c
extern char *name;
void main()
{
    
	printf("%s", name);
}