How printf statement works in the below code?

Question

void main()
{
printf("Adi%d"+2,3);
}

output= i3

This printf statement worked, but how the statement worked ?

Now why does this work? printf(&2??("Adi%d" :> ,010-05); Same kind of pointless obfuscation question. — Lundin
– Lundin, Commented Sep 8, 2016 at 6:46

gmoniava · Accepted Answer · 2016-09-08 06:39:04Z

8

printf("Adi%d"+2,3);

"Adi%d" - is interpreted as start of the address of the memory where the string literal "Adi%d" is stored. When you add 2 to it, it became address of memory where string "i%d" is stored. So basically you passed to printf string: "i%d". Then %d and printf came into play replacing %d with 3, hence the output i3.

answered Sep 8, 2016 at 6:39

gmoniava

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Comments

Ashish Ranjan · Accepted Answer · 2016-09-08 06:56:42Z

1

Its part of pointer to character, nothing to do with printf, "Adi" + 2 will make it read from position 0 + 2 = 2 that will be i

int main()
{
    char* a = "Adi" + 2;
    printf(a); // output i

}

answered Sep 8, 2016 at 6:56

Ashish Ranjan

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