Strange C String Output

I was trying to learn strin开发者_JAVA百科gs in C when I came across this code.

#include <stdio.h>
int main(){
        char s[] = "Hello world";
        printf("%s" , s);
        printf("%s" , &s);
        return 0;
}

Both gave Hello World as output. According to my understanding, this output is Ok for First case. How is it working for the second one? Please clarify.

Taking the address of an array is the same as taking the address of it's first element. When the array's name is used, then it also decays to the address of it's first element- so the expressions s and &s yield the same result.

s returns the address of the first item in the array and &s returns the address of the array itself -- these happen to be the same.

In general, if you wish to be more explicit, the expression &s[0] can also be used to return the address of the first item.

s and &s return the same address and hence. This address is the location where "H" from "Hello world" is stored.

Because,
The name of the array decays to the address of the first element in an array &
The address of first element is same as address of the array.

Just for what it may be worth, if you want to get technical, your second version:

printf("%s" , &s);

has undefined behavior, and only works by accident. By explicitly taking the address, you're getting the address of the array (which is fine) but the result has the type "pointer to array of 12 characters", rather than the type "pointer to char", as required for printf's %s conversion. Since the types don't match, the result is undefined behavior.

In reality, however, that's purely a technicality -- the code will work just fine on every implementation of C of which I'm aware.

If you wanted to demonstrate that the difference exists, you could do so pretty easily though. For example:

char string[] = "hello world";

printf("without &: %p, %p\n", (void *)string, (void *)(string+1));
printf("with &:    %p, %p\n", (void *)&string, (void *)(&string+1));

In the first case, string decays to a pointer to char, so on the first line, the second pointer will be exactly one greater than the first. On the second line, we're adding one to a pointer to an array of characters, so when we add one, it'll actually add the size of the array. Running this on my machine, I get results like this:

without &: 0038F96C, 0038F96D
with &:    0038F96C, 0038F978

char s[] is similar to char *s that is a charecter pointer that points to the first element of the array (it contains the address of the first element stored). we can also store strings by storing the address of the first character. during the time of execution, computer start taking characters from that address one by one and make a string until it reaches a null character('\0'). in the above example 's' and '%s' represents the same value (the address of the starting character) hope you will get it. if you use char s[10] (fixed length array) you will understand everything.

s is equivalent to &s[0] so we are passing address of s[0] not the address of pointer that is pointing to s[0] so it will print Hellow world in second case.

s is name of array not a pointer.