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C Debugging - Part 4
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).
Predict the output or error(s) for the following:
31) main()
{
char *p;
p="Hello";
printf("%c\n" ,*&*p);
}
Answer:
H
Explanation:
* is a dereference operator & is a reference operator. They can be applied any number of times provided it is meaningful. Here p points to the first character in the string "Hello". *p dereferences it and so its value is H. Again & references it to an address and * dereferences it to the value H.
32) main()
{
int i=1;
while (i<=5)
{
printf("%d", i);
if (i>2)
goto here;
i++;
}
}
fun()
{
here:
printf("PP") ;
}
Answer:
Compiler error: Undefined label 'here' in function main
Explanation:
Labels have functions scope, in other words The scope of the labels is limited to functions . The label 'here' is available in function fun() Hence it is not visible in function main.
33) main()
{
static char names[5][20] ={"pascal" ,"ada","cobol" ,"fortran" ,"perl"};
int i;
char *t;
t=names[3];
names[3]=names[ 4];
names[4]=t;
for (i=0;i<=4;i++)
printf("%s", names[i]) ;
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
Array names are pointer constants. So it cannot be modified.
34) void main()
{
int i=5;
printf("%d", i++ + ++i);
}
Answer:
Output Cannot be predicted exactly.
Explanation:
Side effects are involved in the evaluation of i
35) void main()
{
int i=5;
printf("%d", i+++++i);
}
Answer:
Compiler Error
Explanation:
The expression i+++++i is parsed as i ++ ++ + i which is an illegal combination of operators.
36) #include
main()
{
int i=1,j=2;
switch(i)
{
case 1: printf("GOOD" );
break;
case j: printf("BAD" );
break;
}
}
Answer:
Compiler Error: Constant expression required in function main.
Explanation:
The case statement can have only constant expressions (this implies that we cannot use variable names directly so an error).
Note:
Enumerated types can be used in case statements.
37) main()
{
int i;
printf("%d", scanf("%d" ,&i)); // value 10 is given as input here
}
Answer:
1
Explanation:
Scanf returns number of items successfully read and not 1/0. Here 10 is given as input which should have been scanned successfully. So number of items read is 1.
38) #define f(g,g2) g##g2
main()
{
int var12=100;
printf("%d", f(var,12) );
}
Answer:
100
39) main()
{
int i=0;
for(;i++;printf( "%d",i)) ;
printf("%d", i);
}
Answer:
1
Explanation:
before entering into the for loop the checking condition is "evaluated". Here it evaluates to 0 (false) and comes out of the loop, and i is incremented (note the semicolon after the for loop).
40) #include
main()
{
char s[]={'a','b' ,'c','\n' ,'c','\0' };
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d", ++*p + ++*str1-32);
}
Answer:
M
Explanation:
p is pointing to character '\n'.str1 is pointing to character 'a' ++*p meAnswer:"p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10. then it is incremented to 11. the value of ++*p is 11. ++*str1 meAnswer:"str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98. both 11 and 98 is added and result is subtracted from 32.
i.e. (11+98-32)=77( "M");
Regards,
Ram...
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