char a[10] = "pqrstuvwxyz";
printf("%s", a[3]);
printf("%s", 3[a]);
Both the printfs give the same output. Why?
It must be something to do with the way the string is represented in the memory by the C compiler, but what is it?
I would have thought that the offset is multiplied by the size of char, in which case they should give different outputs.
Because those are equivalent expressions. Indexing is equivalent to pointer arithmetic.
int x[10];
int y = x[1];
// same as (and is converted by the compiler to)...
int y = *(x + 1);
Since addition is commutative, it is also the same as...
int y = *(1 + x);
And...
int y = 1[x];
On a side note, you are invoking undefined behavior by using the %s format specifier and passing in a char. %s expects a null terminated string, i.e., a pointer to char terminated by a null character. Heed your warnings.
If you want to print a substring in that array, use:
printf("%s", a + 3);
// or
printf("%s", &a[3]);
You need to use the %c format specifier for a[3] as it is a character, not a string.%s is used for strings, not individual characters.
As for your question why both have same output, well Both are Same!! Both are interpreted as *(a+3).In C you can do it both ways.
And make some corrections to your code:
printf("%s", a[3]);
printf("%s", 3[a]);
would be
printf("%c", a[3]);
printf("%c", 3[a]);
To print the whole string , you can use either :
printf("%s", a);
OR
printf("%s", &a[0]);
In C a[i] is converted to *(a+i) internally. i[a] is converted to *(i+a). Array names also act as pointers in c. so (a+i) or (i+a) give an address (using pointer arithmetic) that is dereferenced using *
