而scanf用布尔运算符

Question

I saw this piece of code today:

while(scanf("%d %d",&x,&y),x||y)
{
    ....

From what I've understand, it enters the loop if some of the values (x or y) is true.

Since the scanf docs says:

On success, the function returns the number of items of the argument list successfully filled. This count can match the expected number of items or be less (even zero) due to a matching failure, a reading error, or the reach of the end-of-file.

I have rewritten the code to:

while(scanf("%d %d",&x,&y) >= 1)
{
   ....

But on an online programming challenge site the first while works, the second fails.

Am I right on my assumptions ? What are the differences between this two pieces of code?

(I am tagging as C++, because I have tested in C++ 4.8.2 - GNU C++ Compiler)

Answer 1

scanf returns the number of arguments it matches, but the first code fragment throws out that result and just checks to see if either x or y is true. The second fragment returns that you matched at least one integer, regardless of value.

Consider the input "0 0" . In the first case, scanf() returns 2, but x || y x || y returns false . In the second case, your conditional is true .

Answer 2

The first code gives while the result of x||y , this is correct

However, the second code compares the return value of scanf and 1 , and then gives while the comparing result.

run this code and you'll be clear.

#include<stdlib.h>
#include<iostream>
using namespace std;

int main()
{
    int x, y;
    cout<<(scanf("%d %d", &x, &y), x||y)<<endl;
    cout<<(scanf("%d %d", &x, &y))<<endl;
    return 0;
}