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Predict the output or error(s) for the following:
C aptitude 28.1
main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!=' ') ++*p++;
printf("%s %s",p,p1);
}
Answer: ibj!gsjfoet
Explanation: ++*p++ will be parse in the given order
*p that is value at the location currently pointed by p will be taken
++*p the retrieved value will be incremented
when ; is encountered the location will be incremented that is p++ will be executed
Hence, in the while loop initial value pointed by p is ‘h’, which is changed to ‘i’ by executing ++*p and pointer moves to point, ‘a’ which is similarly changed to ‘b’ and so on. Similarly blank space is converted to ‘!’. Thus, we obtain value in p becomes “ibj!gsjfoet” and since p reaches ‘ ’ and p1 points to p thus p1doesnot print anything.
C aptitude 28.2
#include <stdio.h>
#define a 10
main()
{
#define a 50
printf("%d",a);
}
Answer: 50
Explanation:The preprocessor directives can be redefined anywhere in the program. So the most recently assigned value will be taken.
C aptitude 28.3
main()
{
printf("%p",main);
}
Answer:Some address will be printed.
Explanation: Function names are just addresses (just like array names are addresses). main() is also a function. So the address of function main will be printed. %p in printf specifies that the argument is an address. They are printed as hexadecimal numbers.
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