# Subex Placement: Sample Questions 8 - 9 of 19

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## Question 8

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### Write in Short

What will be the output of the following code?

1. `main()`
2. `{`
3. `    inti=3;`
4. `    while(i--)`
5. `    {`
6. `        inti=100;`
7. `        i--;`
8. `        printf(“n%d”,i);`
9. `    }`
10. `}`

### Explanation

• In the main () function one integer variable i initialized with 3. The while () loop gets executed till the condition is true. Here, i- is passed in the while loop. i- is the post decremet operator. So, first while (3) will be executed as 3 > 0.
• Here, again i is initialized with new value 100. i-= 99. So, when control reaches to printf () function 99 will be printed on new line of the output screen. Now value of i is decremented by 1 and becomes 2. So, while (2) gets executed as 2 > 0.
• Here, again i is initialized with new value 100. i-= 99. So, when control reaches to printf () function 99 will be printed on new line of the output screen. Now value of i is decremented by 1 and becomes 1. Still 1 > 0 so while (1) gets executed as 2 > 0.
• Here, again i is initialized with new value 100. i-= 99. So, when control reaches to printf () function 99 will be printed on new line of the output screen.

## Question 9

Edit

### Write in Short

What will be the output of the following code?

1. `main()`
2. `{`
3. `    int x;`
4. `    printf("n%d",x=0,x=20,x=40);`
5. `}`

### Explanation

Here, in main () function integer variable x is declared. In the printf () function it is initialized with three different values like 0,20, 40. Here, multiple times varialbe x is initialized. But output will be 0 because in the printf () function 0 is assigned first to the variable x. It does not consider the further initialization of variable x.

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