Question

Few days ago I’ve a weird idea came into my mind that manipulate if(); statement in a weird way.Let’s go on to a simple code.

The Code :

if(printf("blahblah\n");{  
}

My Idea :

1.)To me I think this code will always evaluated to be true(my assumption) since the test condition is substituted with a function call.

So today I’m doing an exercise provided by a book(just to help me refresh what i learn few days ago).This is the code.

The Code :

#include <stdio.h>

int main(void) // This program would add up the value enter by user , for e.g with the 
{              //input of 20 , it will print out the sum of 1+2+3+4+5.....20.

 int count , sum , size;

 count = 0;

 sum = 0;

    printf("Enter a value to find the sum of it from 1 : ");
    scanf("%d" , &size);

    while (count++ < size)

    sum = sum + count;

    printf("sum = %d\n" , sum);

 return 0;
}

By using my idea on the first code , I modified the second code into this.

#include <stdio.h>

int main(void)
{
    int count , sum , size;

    count = 0;

    sum = 0;

    printf("Enter a value to find the sum of it from 1 : ");

    while (scanf("%d" , &size) && count++ < size )

            sum = sum + count;

        printf("sum = %d\n" , sum);

    return 0;
}

The Question :

1.)Based on the assumption made by me in the first code , the scanf() function suppose to be always evaluated to true.That’s why the second test condition count++ < size is the one that determine whether the statement in while statement will be executed or not.

2.)But when I run the program , I input 30 but it doesn’t work , the program just stop there without doing anything after i hit enter.

3.)I try to switch the to test condition with the `count++ < size as left operand while the input function as right operand.

4.)After doing so , the result i get is different.When i try to run the program , the program execute the second printf() function statement , and print out sum = 0.

Your help is much appreciated , do correct me for mistakes.I’m willing to learn from it.

Answer 1

To me I think this code will always evaluated to be true(my assumption) since the test condition is substituted with a function call.

This is incorrect. The function (in this case, printf) returns a value (in this case, an int). When you use it as the condition in the if statement, the function is called and the value it returns becomes the condition: if it returns zero it evaluates to false; if it returns nonzero it evaluates to true.

There is no difference between

if (printf("Hello, World!")) { }

and

int i;
i = printf("Hello, World!");
if (i) { }

(aside, of course, from the additional variable in the second example.)

---

In your modified second example, scanf is called each time the loop condition is checked. You could rewrite the loop like this:

while (1)
{
    int result_of_scanf;

    result_of_scanf = scanf("%d", &size);
    if (result_of_scanf == 0)
        break;

    if (count++ >= size)
        break;

    sum += count;
}

scanf doesn’t just get called once; it gets called for each iteration of the loop. scanf returns the number of elements that it read successfully, so in this case it will return either 1 (if you input a valid integer within the range of int) or 0 (if you give any other input).