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C Debugging - Part 3
Posted Date: 05 May 2008 Resource Type: Articles/Knowledge Sharing Category: Placement Papers
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Posted By: Ramkumar Member Level: Diamond
Rating:  Points: 3
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Note : All the programs are tested under Turbo C/C++ compilers.
It is assumed that,
Ø Programs run under DOS environment,
Ø The underlying machine is an x86 system,
Ø Program is compiled using Turbo C/C++ compiler.
The program output may depend on the information based on this assumptions (for example sizeof(int) == 2 may be assumed).
21. #define square(x) x*x
main()
{
int i;
i = 64/square(4) ;
printf("%d", i);
}
Answer:
64
Explanation:
the macro call square(4) will substituted by 4*4 so the expression becomes i = 64/4*4 . Since / and * has equal priority the expression will be evaluated as (64/4)*4 i.e. 16*4 = 64
22. main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!='\ 0') ++*p++;
printf("%s %s",p,p1);
}
Answer:
ibj!gsjfoet
Explanation:
++*p++ will be parse in the given order
Ø *p that is value at the location currently pointed by p will be taken
Ø ++*p the retrieved value will be incremented
Ø 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 ‘\0’ and p1 points to p thus p1doesnot print anything.
23. #include
#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.
24. #define clrscr() 100
main()
{
clrscr();
printf("%d\n" ,clrscr() );
}
Answer:
100
Explanation:
Preprocessor executes as a seperate pass before the execution of the compiler. So textual replacement of clrscr() to 100 occurs.The input program to compiler looks like this :
main()
{
100;
printf("%d\n" ,100);
}
Note:
100; is an executable statement but with no action. So it doesn't give any problem
25. 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.
27) main()
{
clrscr();
}
clrscr();
Answer:
No output/error
Explanation:
The first clrscr() occurs inside a function. So it becomes a function call. In the second clrscr(); is a function declaration (because it is not inside any function).
28) enum colors {BLACK,BLUE, GREEN}
main()
{
printf("%d.. %d..%d",BLACK, BLUE,GREEN) ;
return(1);
}
Answer:
0..1..2
Explanation:
enum assigns numbers starting from 0, if not explicitly defined.
29) void main()
{
char far *farther,*farthest;
printf("%d.. %d",sizeof( farther), sizeof(farthest) );
}
Answer:
4..2
Explanation:
the second pointer is of char type and not a far pointer
30) main()
{
int i=400,j=300;
printf("%d.. %d");
}
Answer:
400..300
Explanation:
printf takes the values of the first two assignments of the program. Any number of printf's may be given. All of them take only the first two values. If more number of assignments given in the program,then printf will take garbage values.
Regards,
Ram...
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