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C Debugging - Part 14
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: 1
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141) main()
{
extern i;
printf("%d\n" ,i);
{
int i=20;
printf("%d\n" ,i);
}
}
Answer:
Linker Error : Unresolved external symbol i
Explanation:
The identifier i is available in the inner block and so using extern has no use in resolving it.
142) main()
{
int a=2,*f1,*f2;
f1=f2=&a;
*f2+=*f2+=a+ =2.5;
printf("\n%d %d %d",a,*f1,*f2) ;
}
Answer:
16 16 16
Explanation:
f1 and f2 both refer to the same memory location a. So changes through f1 and f2 ultimately affects only the value of a.
143) main()
{
char *p="GOOD";
char a[ ]="GOOD";
printf("\n sizeof(p) = %d, sizeof(*p) = %d, strlen(p) = %d", sizeof(p), sizeof(*p), strlen(p));
printf("\n sizeof(a) = %d, strlen(a) = %d", sizeof(a), strlen(a));
}
Answer:
sizeof(p) = 2, sizeof(*p) = 1, strlen(p) = 4
sizeof(a) = 5, strlen(a) = 4
Explanation:
sizeof(p) => sizeof(char* ) => 2
sizeof(*p) => sizeof(char) => 1
Similarly,
sizeof(a) => size of the character array => 5
When sizeof operator is applied to an array it returns the sizeof the array and it is not the same as the sizeof the pointer variable. Here the sizeof(a) where a is the character array and the size of the array is 5 because the space necessary for the terminating NULL character should also be taken into account.
144) #define DIM( array, type) sizeof(array) /sizeof(type)
main()
{
int arr[10];
printf(“The dimension of the array is %d”, DIM(arr, int));
}
Answer:
10
Explanation:
The size of integer array of 10 elements is 10 * sizeof(int). The macro expands to sizeof(arr)/ sizeof(int) => 10 * sizeof(int) / sizeof(int) => 10.
145) int DIM(int array[])
{
return sizeof(array) /sizeof(int );
}
main()
{
int arr[10];
printf(“The dimension of the array is %d”, DIM(arr));
}
Answer:
1
Explanation:
Arrays cannot be passed to functions as arguments and only the pointers can be passed. So the argument is equivalent to int * array (this is one of the very few places where [] and * usage are equivalent). The return statement becomes, sizeof(int *)/ sizeof(int) that happens to be equal in this case.
146) main()
{
static int a[3][3]={1,2, 3,4,5,6,7, 8,9};
int i,j;
static *p[]={a,a+1, a+2};
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
printf("%d\t% d\t%d\t%d\ n",*(*(p+ i)+j),
*(*(j+p)+i), *(*(i+p)+ j),*(*(p+ j)+i));
}
}
Answer:
1 1 1 1
2 4 2 4
3 7 3 7
4 2 4 2
5 5 5 5
6 8 6 8
7 3 7 3
8 6 8 6
9 9 9 9
Explanation:
*(*(p+i)+j) is equivalent to p[i][j].
147) main()
{
void swap();
int x=10,y=8;
swap(&x,&y);
printf("x=%d y=%d",x,y);
}
void swap(int *a, int *b)
{
*a ^= *b, *b ^= *a, *a ^= *b;
}
Answer:
x=10 y=8
Explanation:
Using ^ like this is a way to swap two variables without using a temporary variable and that too in a single statement.
Inside main(), void swap(); means that swap is a function that may take any number of arguments (not no arguments) and returns nothing. So this doesn’t issue a compiler error by the call swap(&x,&y); that has two arguments.
This convention is historically due to pre-ANSI style (referred to as Kernighan and Ritchie style) style of function declaration. In that style, the swap function will be defined as follows,
void swap()
int *a, int *b
{
*a ^= *b, *b ^= *a, *a ^= *b;
}
where the arguments follow the (). So naturally the declaration for swap will look like, void swap() which means the swap can take any number of arguments.
148) main()
{
int i = 257;
int *iPtr = &i;
printf("%d %d", *((char*)iPtr) , *((char*)iPtr+ 1) );
}
Answer:
1 1
Explanation:
The integer value 257 is stored in the memory as, 00000001 00000001, so the individual bytes are taken by casting it to char * and get printed.
149) main()
{
int i = 258;
int *iPtr = &i;
printf("%d %d", *((char*)iPtr) , *((char*)iPtr+ 1) );
}
Answer:
2 1
Explanation:
The integer value 257 can be represented in binary as, 00000001 00000001. Remember that the INTEL machines are ‘small-endian’ machines. Small-endian means that the lower order bytes are stored in the higher memory addresses and the higher order bytes are stored in lower addresses. The integer value 258 is stored in memory as: 00000001 00000010.
150) main()
{
int i=300;
char *ptr = &i;
*++ptr=2;
printf("%d", i);
}
Answer:
556
Explanation:
The integer value 300 in binary notation is: 00000001 00101100. It is stored in memory (small-endian) as: 00101100 00000001. Result of the expression *++ptr = 2 makes the memory representation as: 00101100 00000010. So the integer corresponding to it is 00000010 00101100 => 556.
Regards,
Ram...
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