use printf("%s",..) to print a struct, the struct's first variable type is 'char *', why can get a right string stored in 'char *'?

Question

In C language,define a struct like this： 　　　　

typedef struct str
{
    char *s;
    int len;
}str;


int main()
{
    str a;
    a.s = "abc"
    printf("%s", a);
    return 0;
}

the output is: "abc", I want to konw why can get this?
I guess the complier think printf("%s", a) as printf("%s", *&a) because &a is equal to &a.s, so *&a is equal to *&a.s, right?

but if so, if I put the int len at the first in the struct body like this:

typedef struct str
{
    int len;
    char *s;
}str;
int main()
{
    str a;
    a.len = 10;
    printf("%d", a);
}

this time the output is: 10, why?
Maybe compiler read %d so it knows that should print a integer value, only print 4 bytes? I want to get some explain about it.

Answer 1

You must specify which value you want to print, printf and C in general cannot guess :)

typedef struct str
{
    int len;
    char *s;
}str;
int main()
{
    str a;
    a.len = 10; // As you did there, you specify which value from your struct !
    printf("%d", a.len);
}

Otherwise it's undefined behaviour and Segfault is coming !

Answer 2

You're invoking undefined behavior. Enable warnings and errors in your compiler (e.g. gcc -Wall -Wextra -Werror) and your ability to do these shenanigans will disappear.

If you really want to know, it's probably because when you pass a by value to printf(), its first member is the char* and that's what printf() sees when it looks for the string to print. In other words, you got lucky.