对字符串指针数组的打印过程感到困惑

Question

#include<stdio.h>
int main(){
    char *names[2] = {"Lord", "Voldemort"};
    printf("%s %s\n",names[0], names[1]);
    return 0;
}

Why is the above code working? I mean, it should print the addresses of the strings it is containing in their respective indexes and we should be using *names[0] and *names[1]. But why it is working though?

Answer 1

If you had an array of integers like

int values[2] = { 1, 2 };

then what would you get if use used eg values[1] ? You would get the second element in the array. Printing this values array would be done like

printf("%d %d\n", values[0], values[1]);

You are with me this far?

Now back your array. If you use names[1] what do you get then? You still get the second element in the array. And what is the second element? It is a pointer to char (ie char * ). And what can a pointer to char be used as? A string .

And the "%s" format with printf expects a string, a char * more specifically. Which is just what you give as arguments.

---

If you use the dereference operator like *names[1] then thanks to operator precedence it is equal to *(names[1]) which is equal to names[1][0] . In other words it gives you the first character in the string pointed to by names[1] .

Answer 2

names[0] and names[1] are both const char* types.

When you use %s with a const char* argument, printf outputs the characters starting at the beginning of the argument, until \\0 is reached.

If you want the addresses then use

printf("%p %p\n", (const void*)names[0], (const void*)names[1]);