CODEWITHSUNDEEP
cprintf
user3126632
user3126632

Reputation: 467

Unable to understand the output of the program using printf weirdly

I was working with some of the interview questions when I found this code.

#include<stdio.h>
int main()
{
 short int a=5;

 printf("%d"+1,a);    //does not give compiler error 

 return 0;
}

It prints the following:

d

I am unable to understand how the printf function works here.

Upvotes: 8

Views: 579

Answers (4)

Bill Lynch
Bill Lynch

Reputation: 81986

Let's look at the first argument to the printf() call.

"%d" + 1

This points to the same thing as ptr does in the following code.

char *ptr = "%d";
ptr = ptr + 1;

So, what does it mean to increment a pointer? Well, we advance the pointer sizeof(*ptr) * 1 bytes forward.

So, in memory we have:

%d\0
^^
||
|This is where ("%d" + 1) points to.
This is where ("%d") points to.

So, your code is more or less functionally equivalent to doing:

short int a = 5;
printf("d", a);

Which will evaluate and then ignore the extra function argument and print d.

One more thing: You're very close to causing undefined behavior in that code. printf("%d", a) is using the wrong format string. The correct format string for a short int is "%hd".

You can find a full table of format strings here.

Upvotes: 15

Micah Smith
Micah Smith

Reputation: 4471

This is my answer based off @DCoder 's comment. "%d" is a pointer to the first character of the two character array %d. Incrementing this by one gives a pointer to d. The fact that there is a second argument now does not matter as the result of the expression "%d"+1 is simply the character d. If you try adding 0 your output is 5 as expected. If you try adding 2, there is no output as there is only "" is being passed to printf.

Upvotes: 1

ani627
ani627

Reputation: 6057

In this case printf("%d"+1,a); = printf("d",a).

You are specifying printf to print from +1 position which is "d", so printf will simply print d on screen.

Upvotes: 1

Yu Hao
Yu Hao

Reputation: 122493

"%d" + 1 does pointer arithmetic, the compiler sees it as "d", so

printf("%d"+1,a);

becomes:

printf("d", a);

You can see why it outputs d in your compiler.

As @sharth points out in the comment, the extra argument a here is evaluated and discarded.

Upvotes: 6

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