带有const char的puts（）问题

Question

I'm having trouble understanding how c uses puts() to display parts of a message. Two ways which I would consider equivalent do not work the same way with the function. For example

 void skippie(char *msg)
 {
    puts(msg + 6);
 }

 char *msg = "Don't call me!";
 skippie(msg);

This compiles fine, however this does not

void skippie(char *msg)
{
    puts(msg[6]);
}

char *msg = "Don't call me!";
skippie(msg);

how does puts() distinguish between the two and only compile for one? The compiler complains that it wants a "const" char but even if I try and use that syntax it fails. Can anyone explain this?

Answer 1

The index operator also dereferences the pointer, so

msg[6] is equivalent to *(msg + 6) , not msg + 6 .

Furthermore, you cannot pass a const char* to a function, while it expects a char* . ie, you also have to update the function signature.

Answer 2

msg + 6 is not the same as msg[6].

As per your code, msg+6 is a char * , while msg[6] represents a char .

Quoting from the man page of puts() , the syntax is

int puts(const char *s);

so, the argument of puts() needs to be a const char * , not a char .