Difference between revisions of "C Questions on Pointers"

Revision as of 19:52, 16 June 2014 (view source) Arjun Suresh (talk | contribs) ← Older edit		Latest revision as of 20:18, 16 June 2014 (view source) Arjun Suresh (talk | contribs)
(3 intermediate revisions by the same user not shown)
Line 88:		Line 88:
	{		{
	int *a = (int*) 1;		int *a = (int*) 1;
−	printf("%d", a);	+	printf("%d", a);
	}		}
	</syntaxhighlight>		</syntaxhighlight>
	|type="()"		|type="()"
	/}		/}
−	- 0	+	-Garbage value
	+1		+1
	-0		-0
Line 107:		Line 107:
	{		{
	int a = 1, *p, **pp;		int a = 1, *p, **pp;
−	p = &a;	+	p = &a;
−	pp = p;	+	pp = p;
−	printf("%d", **pp);	+	printf("%d", **pp);
	}		}
	</syntaxhighlight>		</syntaxhighlight>
	|type="()"		|type="()"
	/}		/}
		+	-Garbage value
	-1		-1
−	-0
	-Compile error		-Compile error
	+Segmentation fault		+Segmentation fault
Line 127:		Line 127:
	{		{
	int a = 1, *p, **pp;		int a = 1, *p, **pp;
−	p = &a;	+	p = &a;
−	pp = p;	+	pp = p;
−	printf("%d", *pp);	+	printf("%d", *pp);
	}		}
	</syntaxhighlight>		</syntaxhighlight>
Line 135:		Line 135:
	/}		/}
	+1		+1
−	-0	+	-Garbage value
	-Compile error		-Compile error
	-Segmentation fault		-Segmentation fault

---

Latest revision as of 20:18, 16 June 2014

<quiz display="simple"> {What will be the output of the following code? <syntaxhighlight lang="c">

1. include<stdio.h>

int main() {

 int a = 5;
 int* p = &a;
 printf("%d", ++*p);

} </syntaxhighlight> |type="{}" /} { 6 } ||p is pointing to the address of a. *p will return the content of a which is 5 and ++ will increment it to 6.

{What will be the output of the following code? <syntaxhighlight lang="c">

1. include<stdio.h>

int main() {

 char a[] = "Hello World";
 char* p = &a;
 printf("%s", p+2 );

} </syntaxhighlight> |type="()" /} -Compiler Error -World +llo World -Runtime Error

||Since p is a char pointer p+2 will add 2 to p (since sizeof(char) is 1). So, p+2 will be pointing to the string "llo World".

{What will be the output of the following code? <syntaxhighlight lang="c">

1. include<stdio.h>

int main() {

 int a;
 int* p = &a;
 printf("%zu", sizeof( *(char*)p  ));

} </syntaxhighlight> |type="()" /}

+1 -2 -4 -Compile Error

||p is typecasted to char pointer and then dereferenced. So, returned type will be char and sizeof(char) is 1.

{Is the following code legal? <syntaxhighlight lang="c">

1. include<stdio.h>

int main() {

 int a = 1;
 if((char*)  &a)
 {
   printf("My machine is little endian");
 }
 else
 {
    printf("My machine is big endian\n");
 }

} </syntaxhighlight> |type="()" /} +Yes -No ||On a little endian machine the lower address will contain the least significant byte. Suppose a is stored at address 1000 and contains 1, then character at 1000 will be 1, if the machine is little endian.

{What will be the output of the following code? <syntaxhighlight lang="c">

1. include<stdio.h>

int main() {

 int *a = (int*) 1;
 printf("%d", a);

} </syntaxhighlight> |type="()" /} -Garbage value +1 -0 -Compile error -Segmentation fault ||Assigning int values to pointer variable is possible. Only when we dereference the variable using *, we get a segmentation fault.

{What will be the output of the following code? <syntaxhighlight lang="c">

1. include<stdio.h>

int main() {

 int a = 1, *p, **pp;
 p = &a;
 pp = p;
 printf("%d", **pp);

} </syntaxhighlight> |type="()" /} -Garbage value -1 -Compile error +Segmentation fault ||Here, p is having address of a and the same is copied to pp. So, *pp will give 1 which is contained in a, but **p will use 1 as an address and tries to access the memory location 1, giving segmentation fault.

{What will be the output of the following code? <syntaxhighlight lang="c">

1. include<stdio.h>

int main() {

 int a = 1, *p, **pp;
 p = &a;
 pp = p;
 printf("%d", *pp);

} </syntaxhighlight> |type="()" /} +1 -Garbage value -Compile error -Segmentation fault ||Here, p is having address of a and the same is copied to pp. So, *pp will give 1 which is contained in a.

{Assuming a little endian machine, what will be the output of the following program? <syntaxhighlight lang="c">

1. include<stdio.h>

fun(int a) {

 char *arr[] = {"0000", "0001","0010","0011","0100","0101","0110","0111","1000","1001","1010","1011","1100","1101","1110","1111"};
 unsigned char* p = (unsigned char*) &a ;
 p+=3;
 int i;
 for(i = 0; i < sizeof a; i++)
 {
   int d = (*p)>>4; 
   printf("%s", arr[d]);
   d = (*p) & 0xf;
   printf("%s ", arr[d]);
   p--;
 }

}

int main() {

int a;
scanf("%d", &a);
fun(a);

} </syntaxhighlight> |type="()" /} +Print the binary of the input number -Compile error -Runtime error -Compiler dependent output ||The code is printing the binary equivalent of the input number. Suppose a is stored starting from address 1000. Since, we assume a little endian machine, the LSB of a will be stored at address 1000 and MSB will be stored at address 1003. So, we make a char pointer to 1003 and take out the MSB. Then using shift operator we get the most significant 4 bits from it and then the lest significant 4 bits. We repeat the same for the other three bytes of a. </quiz>