Predict the output or
error(s) for the following:
1. void main()
{
int
const * p=5;
printf("%d",++(*p));
}
Answer:
Compiler error: Cannot modify a constant value.
Explanation:
p is a pointer to a "constant integer". But we tried to change the value of the
"constant integer".
2. main()
{
char s[
]="man";
int i;
for(i=0;s[ i
];i++)
printf("\n%c%c%c%c",s[
i ],*(s+i),*(i+s),i[s]);
}
Answer:
mmmm
aaaa
nnnn
Explanation:
s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea.
Generally array name is the base address for that array. Here s is the
base address. i is the index number/displacement from the base address. So,
indirecting it with * is same as s[i]. i[s] may be surprising. But in the
case of C it is same as s[i].
3. main()
{
float me =
1.1;
double you =
1.1;
if(me==you)
printf("I love U");
else
printf("I hate U");
}
Answer:
I hate U
Explanation:
For floating point numbers (float, double, long double) the values cannot be
predicted exactly. Depending on the number of bytes, the precession with of the
value represented varies. Float takes 4 bytes and long double takes 10
bytes. So float stores 0.9 with less precision than long double.
Rule of Thumb:
Never compare or at-least be cautious when using floating point numbers with
relational operators (== , >, <, <=, >=,!= ) .
4. main()
{
static int
var = 5;
printf("%d
",var--);
if(var)
main();
}
Answer:
5 4 3 2 1
Explanation:
When static storage class is given, it is initialized once. The change in the
value of a static variable is retained even between the function calls. Main is
also treated like any other ordinary function, which can be called recursively.
5. main()
{
int c[
]={2.8,3.4,4,6.7,5};
int
j,*p=c,*q=c;
for(j=0;j<5;j++) {
printf(" %d ",*c);
++q; }
for(j=0;j<5;j++){
printf(" %d ",*p);
++p; }
}
Answer:
2 2 2 2 2 2 3 4 6 5
Explanation:
Initially pointer c is assigned to both p and q. In the first loop, since only q
is incremented and not c , the value 2 will be printed 5 times. In second loop p
itself is incremented. So the values 2 3 4 6 5 will be printed.
6. main()
{
extern int i;
i=20;
printf("%d",i);
}
Answer:
Linker Error : Undefined symbol '_i'
Explanation:
extern storage class in the following declaration,
extern int i;
specifies to the compiler that the memory for i is allocated in some other
program and that address will be given to the current program at the time of
linking. But linker finds that no other variable of name i is available in any
other program with memory space allocated for it. Hence a linker error has
occurred .
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