为什么该程序无法打印所需的输出？

Question

I just know about the %i format specifier from this link

Difference between format specifiers %i and %d in printf

and I tried to implement it with this program.

#include <stdio.h>

int main(){

    long long a,b;

    printf("Input: ");

    scanf("%i %lld",&b,&a);

    printf("Output: %i %lld",b,a);
}

%i worked properly but %lld stores a garbage value in variable a.

This is the output of this program.

Input : 033 033

Output : 27 141733920846

Process returned 0 (0x0) execution time : 4.443 s Press any key to continue.

Can anyone explain, why I am getting the garbage value in variable a?

Answer 1

scanf %i takes an int * , but you're passing &b , which is a long long int * . This has undefined behavior.

You should be using %lli .

The same problem occurs in printf : Use %lli to print b , not %i .

You should also check scanf 's return value to make sure two values were successfully read.

Answer 2

First of all, using %i for a long long int is undefined behavior, so use %lli instead.

The same issue persists in the printf statement, too.

Fixed code:

#include <stdio.h>


int main(){

    long long a,b;
    int retval;


    printf("Input: \n");

    retval = scanf("%lli %lld",&b,&a);

    printf("Output: %lli %lld",b,a);
    printf("\nRetval: %d",retval);
    return 1;
}

Input:

033 033

Output:

Input: Output: 27 33 Retval: 2

Live Demo

Note: Always check the return value of scanf. It returns the number of scanned items, which you should test against your expectations.