CODEWITHSUNDEEP
carraysstringprintf
Rafed Nole
Rafed Nole

Reputation: 122

Address Of String Literals in C

I have read that the type of string literal is char[n+1], where n is the length.The storage of string literals is an implementation issue.But still it must be unique at an instant.

printf
("%u\t %s\t  %d\t  %c\t  %f\t  %e\t  %x\t  %p\t",
&"XY",&"XY",&"XY",&"XY",&"XY",&"XY",&"XY",&"XY");

The output of this code is

4206628 XY 4206628 $ 0.000000 1.800980e-307 7ffde000 00000032

Why %f gives zero, %s gives XY (no effect of &?), and %p gives a totally different value?

Upvotes: 2

Views: 280

Answers (2)

Yu Hao
Yu Hao

Reputation: 122383

Why %f gives zero?

Because %f expects a double while it's not, this leads to undefined behavior.

%s gives XY (no effect of &?)

Possibly because for an array arr: arr and &arr have the same value. However, the type is different, which means you are passing the unexpected type to printf, again, undefined behavior.

and %p gives a totally different value?

That's the pointer value you are looking for.

Upvotes: 1

unwind
unwind

Reputation: 399793

You cannot pass values of the wrong type (a type that doesn't match what the formatting specifier says is expected) and not get undefined behavior.

For instance, it's quite possible that a double which is what %f expects is bigger than a pointer (which is what you're actually) passing, thus leading to a mis-match between the passed values and the values consumed by printf(), and more or less mayhem as a result.

Upvotes: 2

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