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C Debugging - Part 8
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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81) main(int argc, char **argv)
{
printf("enter the character");
getchar();
sum(argv[1], argv[2]);
}
sum(num1,num2)
int num1,num2;
{
return num1+num2;
}
Answer:
Compiler error.
Explanation:
argv[1] & argv[2] are strings. They are passed to the function sum without converting it to integer values.
82) # include
int one_d[]={1,2, 3};
main()
{
int *ptr;
ptr=one_d;
ptr+=3;
printf("%d", *ptr);
}
Answer:
garbage value
Explanation:
ptr pointer is pointing to out of the array range of one_d.
83) # include
aaa() {
printf("hi") ;
}
bbb(){
printf("hello" );
}
ccc(){
printf("bye" );
}
main()
{
int (*ptr[3])();
ptr[0]=aaa;
ptr[1]=bbb;
ptr[2]=ccc;
ptr[2]();
}
Answer:
bye
Explanation:
ptr is array of pointers to functions of return type int.ptr[0] is assigned to address of the function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc respectively. ptr[2]() is in effect of writing ccc(), since ptr[2] points to ccc.
85) #include
main()
{
FILE *ptr;
char i;
ptr=fopen("zzz. c","r");
while((i=fgetch( ptr))!=EOF)
printf("%c", i);
}
Answer:
contents of zzz.c followed by an infinite loop
Explanation:
The condition is checked against EOF, it should be checked against NULL.
86) main()
{
int i =0;j=0;
if(i && j++)
printf("%d.. %d",i++,j) ;
printf("%d.. %d,i,j);
}
Answer:
0..0
Explanation:
The value of i is 0. Since this information is enough to determine the truth value of the boolean expression. So the statement following the if statement is not executed. The values of i and j remain unchanged and get printed.
87) main()
{
int i;
i = abc();
printf("%d", i);
}
abc()
{
_AX = 1000;
}
Answer:
1000
Explanation:
Normally the return value from the function is through the information from the accumulator. Here _AH is the pseudo global variable denoting the accumulator. Hence, the value of the accumulator is set 1000 so the function returns value 1000.
88) int i;
main(){
int t;
for ( t=4;scanf("% d",&i)-t;printf( "%d\n",i) )
printf("%d-- ",t--);
}
// If the inputs are 0,1,2,3 find the o/p
Answer:
4--0
3--1
2--2
Explanation:
Let us assume some x= scanf("%d",&i)-t the values during execution
will be,
t i x
4 0 -4
3 1 -2
2 2 0
89) main(){
int a= 0;int b = 20;char x =1;char y =10;
if(a,b,x,y)
printf("hello" );
}
Answer:
hello
Explanation:
The comma operator has associativity from left to right. Only the rightmost value is returned and the other values are evaluated and ignored. Thus the value of last variable y is returned to check in if. Since it is a non zero value if becomes true so, "hello" will be printed..
90) main(){
unsigned int i;
for(i=1;i>-2;i--)
printf("c aptitude");
}
Explanation:
i is an unsigned integer. It is compared with a signed value. Since the both types doesn't match, signed is promoted to unsigned value. The unsigned equivalent of -2 is a huge value so condition becomes false and control comes out of the loop.
Regards,
Ram...
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