Consider the following:
char abc[14] = "C Programming"; printf("%s", abc + abc[3] - abc[4]);
The output of the above printf statement is "rogramming". I can't seem to figure how this output is obtained.
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2 Answers
Because chars are a form of integers.
abc + abc[3] - abc[4]
==> abc + 'r' - 'o'
==> abc + 3
And thus you print the string abc starting at index 3.
7 Comments
fjardon
@GRC Why don't you share your thoughts :)
fjardon
@GRC He wrote exactly the same thing. I quote:
'r' - 'o' = 3
haccks
He (@GRC) is talking about pointer arithmetic which is addressed in my answer.
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abc is an array. When used in expression, in most cases, it converted to pointer to its first element. abc[3] is a char which is 'r'. abc[4] is 'o'. abc[3] - abc[4] = 'r' - 'o' = 3. abc + 3 = &abc[3].
So, the expression abc + abc[3] - abc[4] is equivalent to pointer to 3rd charater of string "C Programming".