CODEWITHSUNDEEP
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saagarm
saagarm

Reputation: 23

Printf with strings as arguments

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.

Upvotes: 2

Views: 197

Answers (2)

haccks
haccks

Reputation: 106012

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".

Upvotes: 4

fjardon
fjardon

Reputation: 7996

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.

Upvotes: 5

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