CODEWITHSUNDEEP
c++
Thijmen
Thijmen

Reputation: 458

C++ index evaluation direction

Consider the following code: argv[1][2]

How does C++ handle the index evaluation? For example left to right: [1] is evaluated before [2], right to left: [2] is evaluated before [1] or does this depend on the compiler used?

Upvotes: 3

Views: 293

Answers (5)

aschepler
aschepler

Reputation: 72346

Strictly speaking, there are a number of evaluations going on in argv[1][2] (which is equivalent to (argv[1])[2])

  1. Evaluate argv
  2. Evaluate 1
  3. Evaluate 2
  4. Evaluate argv[1]
  5. Evaluate argv[1][2]

An operator expression can't really be evaluated without knowing what its operands' values are, so #1 and #2 must happen before #4, and #3 and #4 must happen before #5.

Of course, "evaluate 1" doesn't have much meaning since it's just a literal known value. But if the expression were something like argv[f1()][f2()] instead, then the order of subexpression evaluations can matter.

In versions of C++ up to C++14, it is unspecified in argv[f1()][f2()] whether f1() or f2() is called first. C++17 introduced a lot of additional guarantees on the order of subexpressions, including a rule for array subscripting: in A[B], all evaluations and side effects for subexpression A now happen before all evaluations and side effects of subexpression B. So C++17 guarantees in this case that f1() will be called before f2().

Upvotes: 10

Bathsheba
Bathsheba

Reputation: 234715

argv[1][2] is grouped strictly as ((argv[1])[2]).

So the expression is equivalent to *(*(argv + 1) + 2), which is a char type.

Upvotes: 1

Some programmer dude
Some programmer dude

Reputation: 409176

For any array or pointer a and index i, the expression a[i] is exactly equal to *(a + i). In short, all array indexing is "simple" pointer arithmetic.

For an array of arrays (or array of pointers) like argv normally is, then argv[i][j] is equal to (argv[i])[j] which is equal to (*(argv + i))[j] which is equal to (*(argv + i)) + j).

Upvotes: 2

bobah
bobah

Reputation: 18864

It's (argv[x])[y]. If argv is and array of pointers to C-strings (an argument to the main() for instance). Then the first index picks the string and the second argument — a character in the string.

Upvotes: 1

Fureeish
Fureeish

Reputation: 13424

It is converted exactly to: *(*(argv + 1) + 2). You might want to learn about pointer arithmetic

Upvotes: 1

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