Pointer qn 1 - Revision history

Arjun Suresh: /* {{Template:Author|Arjun Suresh|{{arjunweb}} }} */

2015-03-18T13:22:59Z

‎{{Template:Author|Arjun Suresh|{{arjunweb}} }}

Arjun Suresh: /* {{Template:Author|Arjun Suresh|{{arjunweb}} }} */

2015-03-18T13:03:59Z

‎{{Template:Author|Arjun Suresh|{{arjunweb}} }}

Arjun Suresh: /* {{Template:Author|Arjun Suresh|{{arjunweb}} }} */

2015-03-18T12:14:01Z

‎{{Template:Author|Arjun Suresh|{{arjunweb}} }}

Arjun Suresh: /* {{Template:Author|Arjun Suresh|{{arjunweb}} }} */

2015-03-18T12:10:10Z

‎{{Template:Author|Arjun Suresh|{{arjunweb}} }}

Arjun Suresh: /* {{Template:Author|Arjun Suresh|{{arjunweb}} }} */

2015-03-18T12:01:14Z

‎{{Template:Author|Arjun Suresh|{{arjunweb}} }}

Arjun Suresh at 14:06, 14 April 2014

2014-04-14T14:06:41Z

Arjun Suresh at 16:43, 22 February 2014

2014-02-22T16:43:09Z

Arjun Suresh at 13:44, 22 February 2014

2014-02-22T13:44:23Z

Arjun Suresh at 13:44, 22 February 2014

2014-02-22T13:44:07Z

Arjun Suresh at 13:37, 22 February 2014

2014-02-22T13:37:11Z

@@ Line 33: / Line 33: @@
 Now, %d, *p.
-Here, *p is a character which is just a byte. So, *p value here is the byte at address 1000 which is 02 (in hex) and this is passed to printf. In C language, when a char is passed to a function actually it's int value as given by the char code (usually ASCII) is passed. So, here 02 is extended to 0002 and passed to printf. Now, in printf we used %d, and hence the output will be 2.
+Here, *p is a character which is just a byte. So, *p value here is the byte at address 1000 which is 02 (in hex) and this is passed to printf. In C language, when a char is passed to a function actually it's int value as given by the char code (usually ASCII) is passed. So, here 02 is sign-extended to 0002 and passed to printf. Now, in printf we used %d, and hence the output will be 2.
   When p is incremented, it will become to 1001, as p is a char pointer (for pointer addition, sizeof datatype pointed to is added and sizeof char is 1). So, now the value of *p is 00 and as explained above printf will print 0 for this.

@@ Line 40: / Line 40: @@
 So, output will be 0.
+On a big-endian machine output will be 0 0. So, correct answer should be architecture dependent.
 {{Template:FBD}}

@@ Line 29: / Line 29: @@
 So, the first two elements of the array will be stored in memory as follows:
-: 0 2 0 0 0 0 0 0 (hexa decimal representation and assuming int takes 4 bytes and assuming least significant byte is stored at highest address(little-endian))
+: 0 2 0 0 0 0 0 0 (hexa decimal representation and assuming int takes 4 bytes and assuming least significant byte is stored at lowest address(little-endian))
 : 0 3 0 0 0 0 0 0
 Now, %d, *p.
-Here, *p is a character which is just a byte. So, *p value here is the byte at address 1000 which is 00 (in hex) and this is passed to printf. In C language, when a char is passed to a function actually it's int value as given by the char code (usually ASCII) is passed. So, here 00 is extended to 0000 and passed to printf. Now, in printf we used %d, and hence the output will be 2.
+Here, *p is a character which is just a byte. So, *p value here is the byte at address 1000 which is 02 (in hex) and this is passed to printf. In C language, when a char is passed to a function actually it's int value as given by the char code (usually ASCII) is passed. So, here 02 is extended to 0002 and passed to printf. Now, in printf we used %d, and hence the output will be 2.
-  will print the first 4 bytes starting at 1000, that is 2. When p is incremented, it will go to 1001, as p is a char pointer.
+  When p is incremented, it will become to 1001, as p is a char pointer (for pointer addition, sizeof datatype pointed to is added and sizeof char is 1). So, now the value of *p is 00 and as explained above printf will print 0 for this.
 : 0 0 0 0 0 0 0 0

@@ Line 29: / Line 29: @@
 So, the first two elements of the array will be stored in memory as follows:
-: 0 0 0 0 0 0 0 2 (hexa decimal representation and assuming int takes 4 bytes and assuming least significant byte is stored at highest address(little-endian))
+: 0 2 0 0 0 0 0 0 (hexa decimal representation and assuming int takes 4 bytes and assuming least significant byte is stored at highest address(little-endian))
-: 0 0 0 0 0 0 0 3
+: 0 3 0 0 0 0 0 0
-Now, %d, *p will print the first 4 bytes starting at 1000, that is 2. When p is incremented, it will go to 1001, as p is a char pointer.
+Now, %d, *p.
 : 0 0 0 0 0 0 0 0

@@ Line 26: / Line 26: @@
 ==={{Template:Author|Arjun Suresh|{{arjunweb}} }}===
-Assume start location of arr s 1000.
+Assume start location of arr is 1000.
 So, the first two elements of the array will be stored in memory as follows:
-: 0 0 0 0 0 0 0 2 (hexa decimal representation and assuming int takes 4 bytes)
+: 0 0 0 0 0 0 0 2 (hexa decimal representation and assuming int takes 4 bytes and assuming least significant byte is stored at highest address(little-endian))
 : 0 0 0 0 0 0 0 3

@@ Line 1: / Line 1: @@
 ==What's the output?==
-<syntaxhighlight lang="c">
+<syntaxhighlight lang="c" name="pointer_1">
 int main()
+{

← Older revision		Revision as of 13:44, 22 February 2014
Line 37:		Line 37:
	So, output will be 0.		So, output will be 0.

−	~~===Solution===~~
−	~~Assume start location of arr s 1000.~~
−	~~So, the first two elements of the array will be stored in memory as follows:~~

−	~~1000: 0 0 0 0 0 0 0 2 (hexa decimal representation and assuming int takes 4 bytes)~~
−	~~1004: 0 0 0 0 0 0 0 3~~
−
−	~~Now, %d, *p will print the first 4 bytes starting at 1000, that is 2. When p is incremented, it will go to 1001, as p is a char pointer.~~
−	~~1001: 0 0 0 0 0 0 0 0~~
−
−	~~So, output will be 0.~~

	{{Template:FBD}}		{{Template:FBD}}