Output question

Question

What is the output for this code?
char str[]="%d %c", arr[]="Sololearn";
printf(str, 0[arr], 2[arr+3]);
How is the output 83 e?

Answer 1

Same answer as Pąľľąvī , but I've explained a bit about what 0[arr] and 2[arr + e] means.

arr[0] literally means *(arr + 0)
because in C, variables containing array values are actually pointers to the first element of that array. So in the same way,
0[arr] literally means *(0 + arr), which is the same as *(arr + 0), which in turn, is the same as arr[0].

So you are passing arr[0], which is 'S', to the %d format specifier, which prints the ordinal value of 'S'

In the same way as explained above
2[arr + 3] is the same as
*(arr + 3 +2)
= *(arr + 5)
= arr[5]

So you are passing arr[5], which is 'e', to the %c format specifier. The %c specifier is used to print values of char type to the screen. So it simply prints 'e'.

Answer 2

Thank you