Predict the output or
error(s) for the following:
100. main()
{
int a=10,*j;
void *k;
j=k=&a;
j++;
k++;
printf("\n %u
%u ",j,k);
}
Answer:
Compiler error: Cannot increment a void pointer
Explanation:
Void pointers are generic pointers and they can be used only when the type is
not known and as an intermediate address storage type. No pointer arithmetic can
be done on it and you cannot apply indirection operator (*) on void pointers.
101. Printf can be implemented by
using __________ list.
Answer:
Variable length argument lists
102. char *someFun()
{
char *temp =
“string constant";
return temp;
}
int main()
{
puts(someFun());
}
Answer:
string constant
Explanation:
The program
suffers no problem and gives the output correctly because the character
constants are stored in code/data area and not allocated in stack, so this
doesn’t lead to dangling pointers.
103. char *someFun1()
{
char temp[ ]
= “string";
return temp;
}
char
*someFun2()
{
char temp[ ]
= {‘s’, ‘t’,’r’,’i’,’n’,’g’};
return temp;
}
int main()
{
puts(someFun1());
puts(someFun2());
}
Answer:
Garbage values.
Explanation:
Both the
functions suffer from the problem of dangling pointers. In someFun1() temp is a
character array and so the space for it is allocated in heap and is initialized
with character string “string”. This is created dynamically as the function is
called, so is also deleted dynamically on exiting the function so the string
data is not available in the calling function main() leading to print some
garbage values. The function someFun2() also suffers from the same problem but
the problem can be easily identified in this case.
104. There were 10 records stored in
“somefile.dat” but the following program printed 11 names. What went wrong?
void main()
{
struct student
{
char name[30], rollno[6];
}stud;
FILE *fp = fopen(“somefile.dat”,”r”);
while(!feof(fp))
{
fread(&stud, sizeof(stud), 1 , fp);
puts(stud.name);
}
}
Explanation:
fread reads 10 records and prints the names successfully. It will return EOF
only when fread tries to read another record and fails reading EOF (and
returning EOF). So it prints the last record again. After this only the
condition feof(fp) becomes false, hence comes out of the while loop.
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