CODEWITHSUNDEEP
cprintf
Shabnam Sheikhha
Shabnam Sheikhha

Reputation: 49

why won't this program give me the output i want?

I've written this simple code so I can see what's wrong with a more complex program that I have written.

#include<stdio.h>
int main()
{
    int n = 0, i = 1, a = 0;
    scanf("%d", &n);
    while (i <= n)
    {
        scanf(" %d", &a);
        printf("%d", &a);
        i++;
    }
}

but when I run the program it goes like this: 4 1 6487620 what's wrong with it?

Upvotes: 1

Views: 297

Answers (3)

chqrlie
chqrlie

Reputation: 144740

You pass the address of a instead of its value to printf. You should also output a linefeed to separate the numbers:

printf("%d\n", a);

Upvotes: 0

Nutan
Nutan

Reputation: 786

When you use printf("%d", &a);

this means that it will print the address of a

and to print the value of a you have to wright

printf("%d", a);

and after making the changes compile the program and try to rerun :)

Upvotes: 1

Sourav Ghosh
Sourav Ghosh

Reputation: 134336

In your code

 printf("%d", &a);

should be

printf("%d", a); // don;t print address....

FWIW, passing an address (a pointer type) as an argument to %d is a mismatch and invokes undefined behavior.

Upvotes: 1

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