3

It is a known thing that array names decay into pointer such as a[i] is equivalent to *(a+i). I stumbled upon a question asking me predict the output of: 

int main(void)
{
   printf("%c", "abcdefgh"[4]);
   return 0;
}

The answer they say is e and the reason given by them is as a[4] gets converted into *(a+4) where a is base address of array a on similar lines we can find the output of above question.

But how I don't understand as with a[i] decaying into *(a+i) because compiler treats it that way but in the code snippet above how and why we will treat "abcdef"[4] this?

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3 Answers 3

Reset to default
4

You will wonder even more if I say that you can write even the following way :)

printf( "%c\n", 4["abcdefgh"] );

String literals in C have types of character arrays. Thus string literal "abcdefgh" has type char[9]. It includes also the terminating zero.

In expressions arrays as you correctly mentioned decay to pointers to their first elements. So the string literal decays to a pointer of type char *that points to the first character of the literal that is to 'a'

According to the C Standard (6.5.2.1 Array subscripting)

2 A postfix expression followed by an expression in square brackets [] is a subscripted designation of an element of an array object. The definition of the subscript operator [] is that E1[E2] is identical to (*((E1)+(E2))). Because of the conversion rules that apply to the binary + operator, if E1 is an array object (equivalently, a pointer to the initial element of an array object) and E2 is an integer, E1[E2] designates the E2th element of E1 (counting from zero).

Thus this expression (*((E1)+(E2))) does not depend on whether the initial record was E1[E2] or E2[E1]

Returning to your code shippet you have that in this expression

4["abcdefgh"]

the string literal is converted to the pointer to its first character. You can imagine this like

char *p = "abcdefgh";

and you will get

4[p] is equivalent to *( 4 + p ) and results in lvalue of character 'e'

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3

"abcdefgh"[4] is equivalent to *("abcdefgh" + 4). When used in an expression (except when it is an operand of unary & and sizeof operator) "abcdefgh" represents the base address of string.

In general, remember that one can use a string literal wherever char * pointer is allowed. In the snippet 

char *ptr;
p = "abcdefgh";

the assignment doesn't copy the characters to p instead it makes p points the first character of the string. C allows pointers to be subscripted, so we can subscript string literals: 

char ch;  
ch = "abcdefgh"[4];
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"When used in an expression" – no, not in any expression, only when it's not operand of the sizeof or & operators.
0

"abcdefgh" is an array. It is an array of characters, similar (but not identical) to as if you had declared a variable of type char[9] and static storage duration:

static char str[9] = { 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 0 };

Hence, you can index it just like any other array, using the [] subscript operator.

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