| Q #1: Which one of the following items are vaild C identifiers? Give reason(s) for your answer.
|
-
i) ComputerScience&Engineering
- NOT VALID
- because the & cannot be used as an identifier.
- ii) Vitamin-A
- NOT VALID
- because "-" cannot be used as an identifier.
- iii) 805_North
- NOT VALID
- because identifiers must start with a letter or underscore.
|
Q #2: The following declarations are given.
int x = 5, y = 9, z = 20, w = 11, v = -2, a;
What is the value of each of the follwing C language expressions when they are executed in isolation? List the sequence of computations you made to arrive at answers for each item.
|
- i) a = x != y && z
- Answer : 1
- since != has higher precedence than any other operators in this statement, x != y is the first thing to be executed. and then && and = follows. Thus:
a = ( ( x != y ) && z )
a = ( ( 5 != 9 ) && z)
a = ( 1 && z )
a = ( 1 && 20 )
a = 1 Thus, the answer is 1.
- ii) a = y & x + v
- Answer : 1
- since the % has higher precedence than other operators in this statemetn, y % x is the first operator to be evauluated. And then the "+" operator, then "=" operator follows. Thus:
a = ( ( y % x ) + v )
a = ( ( 9 % 5 ) + v )
a = ( 4 + v )
a = ( 4 + -3 )
a = 1 Thus, the answer is 1.
- iii) a = ( z - y ) == w
- Answer : 1
- since "( )" have the higest, then "==" next, then "=" next, the stateuemtn will look like a = ( ( z - v ) == w ). Thus:
a = ( ( 20 - 9 ) == w )
a = ( 11 == w )
a = ( 11 == 11 )
a = 1 Thus, the answer is 1
|
| Q #3: If x, y, z, and w are declared as integer variables, which of the following expressions are valid in C?
|
- i) x + 5 = y + z
- Answer : NO
- The left side of a "=" operator cannot be a statement. It must be an identifier.
- ii) x *= 1 + a + b
- Answer : NO
- because identifier a and b are not declared, they can't be used.
- iii) x = y = z + w
- Answer : YES
|
| Q #4: Write a sample C program using the "while" loop statement to compute the avarage of first 100 non-negative integers.
|
- First non-negative integer means nubmer from 0 to 99. The avarage of the first non-negative inters means dividing the sum of number 0 to 99 and divide it by 100. The follwing is one of the possiblities
#include < stdio.h >
int main() {
// initialization
int sum = 0, counter = 0, ave = 0;
// find the sum
while ( counter < 100 ) {
sum += counter;
counter++;
}
// calculate the average
// print the result
ave = sum / 100;
printf (" The average is %d\n", ave);
}
There are many other ways of doing this. This is just a possibility
|
| Q: Rewrite the following program fragment using a switch statement. The resulting fragment should be semantically equivalent to the original. Assume a, b, c, d are integer variables.
|
if ( a == 5 ) {
b = c + d;
}
if ( a == -5 ) {
b = c % d;
}
if ( a == 0 ) {
b = c - d;
} else b = c * d;
A lot of people missed this question. We tried to give as much partial credits as possible. The answer is as follows:
switch(a) {
case 5:
b = c + d;
break;
case -5:
b = c % d;
break;
case 0:
b = c - d;
break;
default:
b = c * d;
}
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