CODEWITHSUNDEEP
c
vijay_shekhar
vijay_shekhar

Reputation: 11

Can anyone explain me the output of this code?

char c[] = "hello"; 

printf("%*d", c);

the output is :

infinite loop of spaces

enter image description here

Upvotes: 1

Views: 255

Answers (2)

Noman
Noman

Reputation: 163

c is an array of char which holds hello but what is an array. it is a pointer with a chunk of memory mean above we create an array means we create a pointer whose point blocks of memory which hold hello string like c variable. it means c contain Adresse of the first block of hello.

now we look into printf("%*d")

printf("%*d") * represent the width and second parameter d represent a integer

u can see that how * symbol work in below example

printf("output is:%*c",3,c): its output:' h'

here is width three and print a char which hold c and that is h

now see why screen has infinite loop let's see printf("%'*'d",c)

we know the first argument is the width and second output is data here our first argument is c which is an array and u know array hold address of the first block Code executed it give an address to width and u know the address is too long and it sometimes negative then why your screen has too much space not infinite because u give width as a long as a address further doubt comment me

Upvotes: 0

bruno
bruno

Reputation: 32596

Can anyone explain me the output of this code?

infinite loop of spaces

char c[] = "hello";

printf("%*d", c);

the %*d say the first arg after the format indicates the width, here it is the address of c interpreted as a huge number, and the default added character to respect the width is a space.

note there is a missing arg normaly giving the value to print

if I use a valid code like that :

#include <stdio.h>

int main()
{
  printf("%0*d\n", 3, 1);
  return 0;
}

the result is 001 because I ask for to write '1' with a width of 3 and the added character is '0'

Upvotes: 11

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