Placement C Progrmming Language Question with Solution

1)

1. const char *a;

2. char* const a; 

3. char const *a;

-Differentiate the above declarations.

Answer:

1. 'const' applies to char * rather than 'a' ( pointer to a constant char )

*a='F'      : illegal

a="Hi"      : legal

2. 'const' applies to 'a'  rather than to the value of a (constant pointer to char )

*a='F'       : legal

a="Hi"       : illegal

3. Same as 1.

2)

main()

{

int i=5,j=10;

i=i&=j&&10;

printf("%d %d",i,j);

}

Answer:-

1 10

Explanation:

The expression can be written as i=(i&=(j&&10)); The inner expression (j&&10) evaluates to 1 because j==10. i is 5. i = 5&1 is 1. Hence the result. 

3)

main()

{

int i=4,j=7;

j = j || i++ && printf("YOU CAN");

printf("%d %d", i, j);

}

Answer:

4 1 

Explanation:

The boolean expression needs to be evaluated only till the truth value of the expression is not known. j is not equal to zero itself means that the expression’s truth value is 1. Because it is followed by || and true || (anything) => true where (anything) will not be evaluated. So the remaining expression is not evaluated and so the value of i remains the same.

Similarly when && operator is involved in an expression, when any of the operands become false, the whole expression’s truth value becomes false and hence the remaining expression will not be evaluated.     

false && (anything) => false where (anything) will not be evaluated.

4)

main()

{

register int a=2;

printf("Address of a = %d",&a);

printf("Value of a   = %d",a);

}

Answer:

Compier Error: '&' on register variable

Rule to Remember:

& (address of ) operator cannot be applied on register variables.

5)

main()

{

float i=1.5;

switch(i)

{

case 1: printf("1");

case 2: printf("2");

default : printf("0");

}

}

Answer:

Compiler Error: switch expression not integral

Explanation:

Switch statements can be applied only to integral types.

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